IOAA 2016 Problems

IOAA 2016 PROBLEMS

Table of Constants Page 1 of 1 Fundamental Constants

Speed of light light in vacuum Planck Constant Boltzmann Constant Stefan-Boltzmann Constant Charge of electron Universal Gravitational Constant Universal Gas Constant Avogadro Constant Wien's displacement law Mass of electron electron Mass of proton Mass of neutron Atomic Mass Unit (a.m.u.)

 ℎ

=

B     A   e p n 

=

=

= = = = = = = = = =

2.99 .998 × 10  m s − 6.62 .626 × 10 − J s 1.38 .381 × 10 − J K− 5.67 .670 × 10 − W m− K − 1.60 .602 × 10 − C 6.67 .674 × 10 − N m kg − 8.315 J mol − K− 6.02 .022 × 10  mol − 2.89 .898 × 10 − m K 9.10 .109 × 10 − kg 1.67 .673 × 10 − kg 1.67 .675 × 10 − kg 1.66 .661 × 10 − kg

Astronomical Data

1 parsec (pc) 1 astronomical unit (AU) Solar Mass Solar Radius Solar Luminosity Apparent magnitude magnitude of the Sun at mid-day

=

⊕ ⊙ ⊙ ⊙ ⊙

Solar Constant (at Earth) Apparent angular diameter of Sun Earth Mass Earth Radius 1 tropical year

= = = = = =

⊙ ⊕ ⊕

= = = = =

3.08 .086 × 10  m 1.49 .496 × 10  m 1.98 .989 × 10  kg 6.95 .955 × 10  m 3.82 .826 × 10  W 26.72 mag 1366 W m − 30′ 5.97 .972 × 10  kg 6.37 .371 × 10  m 365.242 solar days 3.15 .156 × 10  s

Theoretical Examination Page 1 of 1

Ins tructions tructions to the Contestants Contestants Total To tal time duration of this examination examination is five five hours. hour s. (2) This Th is examination examination consists of (1)

Question 1 to 5 - 10 marks each - 10 x 5 = 50 marks Question Question 6 to 10 - 20 marks each each - 20 x 5 = 100 marks Question Question 11 to 13 - 50 marks each - 50 x 3 = 150 marks Total = 300 marks (3)

Inside the envelope you will will find a. Question paper in English b. Question paper in your native language language (if applicable) applicable) c. Table of Constants in English

d. e. f. (4) Use

T able of Constants in your native language language (if applicable) applicable) A set of Summary Answersheets Cover Sheet only black or blue pen for writing. writing. For figures, figures, you may use pencils. pencils.

(5) T here is no negative marking. (6) Some marks will be deducted if the final final answer is given given without detail d etailed ed solution. (7) Some marks will will be deducted if if the final answers answer s have inapprop inappropriate riate number of significant significant digits, digits, no units or wrong units. (8) The necessary values of fundamental fundamental and astronomical constants should be taken from the Table of Constants provided to you. Instru ctions regarding using the answersheets answers heets (9) Instructions a. For each question question a separate separate Summary Answersheet has been been provided. Final Final answer(s) for each question / each part of the question must be written in the corresponding correspond ing box in in the Summary Answersheet. Write your contestant code and page number on each Summary Answersheet. b. You should also also show detailed solution in the blank space provided on each Summary Answersheet Answer sheet.. If necessary, you can ask for extra blank sheets from the invigil invigilators. ators. c. Use separate blank sheets for each question. question. d. Write your contestant code, question number and page number on FRONT side of each blank sheet. Write your contestant code and page number on the BACK side of each blank sheet. e. The page numbers should be continuous, i.e., if you use 20 sheets for the entire examination (including (including the Summary Answersheets), the page page numbers should s hould run from 1 to 40. f. Write only inside the boxed area. g. For the work that you do not want to to be evaluated, evaluated, cross that part out. (10) At the end of the examination a. On the Cover Sheet, clearly clearly write page numbers for each question. b. Check that you have written your contestant code on all pages. c. Put the Cover Sheet, all all Summary Answersheets, blank blank sheets and rough sheets inside inside the envelope. envelope. You may keep question paper and Table of Constants with you.

Theoretical Examination Page 1 of 6 (T1) True or False Determine if each of the following following statements statement s is True Tr ue or False. False. In the Summary Answersheet, Answersh eet, tick the correct answer (T RUE / FALSE) FALSE) for each statement. No justificati justifications ons are necessary necessary for this question. (T1.1) In a photograph of the clear sky on a Full Moon night night with a sufficiently long exposure, the colour of the sky would appear blue as in daytime.

2

(T1.2) An astronomer at Bhubaneswar marks the position position of the Sun on the sky at 05: 05: 00 UT every day of the year. If the Earth' s axis axis were perpendicular to its its orbital plane, these positi pos itions ons would trace an arc of a great great circle. circle.

2

(T1.3) If the orbital period p eriod of a certain minor body bo dy around the Sun in the ecliptic ecliptic plane is less than the orbital period of o f Uranus, then its its orbit must necessarily be b e fully fully inside the t he orbit of Uranus.

2

(T1.4) The centre of mass of the solar system is inside the Sun at all times.

2

(T1.5) A photon is is moving in free space. As the Universe expands, e xpands, its momentum decreases. decr eases.

2

(T2) Gases on o n Titan Titan Gas particl particles es in a planetary planetary atmosphere have a wide wide distribution distribution of speeds. If the r.m.s. (root mean square) thermal speed of particles of a particular gas exceeds exceeds 1/6 of the escape speed, then most of that gas will escape from the planet. What is the minimum atomic weight (relative atomic mass),  , of an ideal monatomic gas so that it remains in the atmosphere of Titan?

10

Given, mass of Titan  = 1.23 × 10 kg, radius of Titan  = 2575 2575 km km, surface temperature of Titan  = 93.7 93.7 K. (T3) Early Universe Cosmological models indicate that radiation energy density,  , in the Universe is proportional to (1 + ) 4 , and the matter energy density,  , is proportional to (1 + )3 , where  is the redshift. The dimensionless dimensionless density parameter, parameter , Ω, is given as Ω = / , where  is the critical energy en ergy density of the Universe. In the present Universe, the t he density density parameters paramet ers corresponding corres ponding to radiation and a nd matter, are − Ω = 10 and Ω  = 0.3, respectively. (T3.1) Calculate the redshift,  , at which which radiation and matter energy densities were equal.

3

(T3.2) Assuming that the radiation from the early Universe Universe has a blackbody blackbody spectrum with a temperature of 2.732 K, estimate estimate the temperature,  , of the radiation at redshift  .

4

(T3.3) Estimate the typical photon energy, ν (in eV), of the radiation as emitted at redshift  .

3

(T4) Shadows An observer in the northern hemisphere noticed that tha t the length length of the shortest shadow of a 1.000 m vertical stick on a day was 1.732 m. On the same day, the length length of the longest longest shadow of the same vertical stick s tick was measured to be 5.671 m .

10

Find the latitude,  , of the observer and declinati declination on of the Sun, ⊙, on that day. Assume the Sun to be a point source and ignore ignore atmospheric atmospheric refraction. refraction. (T5) GMRT beam transit Giant Metrewave Radio Telescope (GMRT), one of the world's largest radio telescopes at metre wavelengths, is located in western India (latitude: 19 ∘ 6′ N, longitude: 74∘ 3′ E). GMRT consists of 30 dish antennas, each with a diameter of 45.0 m . A singl singlee dish of GMRT was held fixed fixed with its axis axis ∘ pointing pointing at a zenith zenith angle of 39 42′ along the northern meridian such that a radio point source would pass along a diameter diameter of the beam, when it is transiting transiting the meridian. meridian. What is the duration  for which this source would be b e within within the FWHM (full width at half maximum) maximum) of the beam of a single single GMRT dish dish observing at 200 MHz ? Hint: The FWHM size of the beam of a radio dish operating at a given frequency corresponds to the angular resolution of the dish. Assume uniform illumination.

10

Theoretical Examination Page 2 of 6 (T6) Cepheid Pulsations The star  -Doradus -Doradu s is a Cepheid variabl var iablee star with a pulsation period of 9.84 days. We make a simplifying simplifying assumption that th at the star is brightest brightest when it is most contracted (radius being be ing ) and it is faintest when it is most expanded (radius being  ). For simpli simplicity, city, assume that the star maintains maintains its spherical shape and behaves as a perfect black body at every instant during d uring the entire cycle. The T he bolometric bolometric magnitude magnitude of the star varies from 3.46 to 4.08. From Doppler measurements, we know that during pulsation the stellar stellar surface s urface expands or contracts contract s at an average radial speed spee d of 12.8 km s − . Over the period of pulsation, pulsation, the peak of thermal radiati radiation on (intrinsic) (intrinsic) of the star varies from f rom 531.0 nm to 649.1 nm . (T6.1)

Find the ratio of radii of the star in its most contracted and most expanded states ( / ).

7

(T6.2)

Find the radii of the star (in metres) in its most contracted and most expanded states ( and  ).

3

(T6.3)

Calculate Calculate the flux of the star,  , when it is in its most expanded state.

5

(T6.4)

Find the distance to the star,  , in parsecs.

5

(T7) Telescope optics In a particular ideal refracting telescope of focal ratio /5 , the focal length length of the objective lens is 100 cm and that of the eyepiece is 1 cm . (T7.1)

What is the angular angular magnification, magnification,  , of the telescope? What is the length of the telescope,  , i.e. the distance between its objective and eyepiece?

4

An introduc introduction tion of a concave lens (Barlow lens) between the objective lens and the prime focus is a common way to increase increase the magni m agnification fication without a large large increase in the length length of the telescope. A Barlow lens of focal length 1 cm is now introduced between the objective and the eyepiece to double the magnification. (T7.2)

At what distance, B , from the prime focus must the Barlow lens be kept in order to obtain this desired double magnification?

6

(T7.3)

What is the increase, Δ , in in the th e length length of the telescope?

4

A telescope is now constructed with the same objective lens and a CCD detector placed at the prime focus (without any Barlow lens or eyepiece). The Th e size of each pixel pixel of the CCD detector is 10 µm. (T7.4)

What will will be the distance in pixels pixels between the centroids centr oids of o f the images images of the two stars , p , on

6

the CCD, if they are 20′′ apart on the sky? (T8) U-Band photometry A star has an apparent magnitude  = 15.0 in the U -band. -band. The U -band -band filter is ideal, i.e., it has perfect is completely completely opaque ( 0% transmission) transmission) outside outside the band. The (100%) transmission within the band and is filter is centered at 360 nm, and has a width of 80 nm. It is assumed that the star also has a flat energy spectrum with respect to frequency. T he conversion conversion between magnitude, magnitude,  , in in any band and flux density, − − − is given given by  , of a star in Jansky ( 1 J y = 1 × 10 W Hz m ) is

 = 3631 × 10−. Jy (T8.1)

Approximate Approximately ly how many U -band -band photons,  , from this th is star will will be incident incident normally on a  1 m area at the top of the Earth's atmosphere every second?

8

This star is being observed in the U -band -band using a ground based telescope, whose primary mirror has a diameter of 2.0 m. Atmospheric extinction extinction in U -band -band during the observation is 50% . You may assume that the seeing is diffraction limited. Average surface brightness of night sky in U -band -band was measured to  be 22.0 mag/arcsec . (T8.2)

What is the ratio,  , of number of photons received per second from the star to that received from the sky, when measured over a circular circular aperture of diameter 2′′ ?

(T8.3)

In practice, only 20% of U-band photons pho tons falling falling on the primary primary mirror are detected. How Ho w many photons,  , from the star are detected per second?

8 4

Theoretical Examination Page 3 of 6 (T9) Mars Orbiter Mission India's Mars Orbiter Mission (MOM) (M OM) was launched launched using the Polar Satellite Satellite Launch Lau nch Vehicle Vehicle (PSLV) (P SLV) on 5 November 2013 . The dry mass of MOM (body + instruments) was 500 kg and it carried fuel of mass initially y placed in an elli elliptical ptical orbit around arou nd the Earth with perigee at a height height of 264.1 km 852 kg . It was initiall and apogee at a height of 23903.6 km , above the surface of the Earth. After raising the orbit six times, MOM was transferred to a trans-Mars injection injection orbit (Hohmann orbit). The Th e first such suc h orbit-raising orbit-raising was performed perfo rmed by firing firing the t he engines engines for a very short time near the perigee. perigee. The Th e engines engines were fired to change the orbit without without changing the t he plane plane of the orbit and without without changin g  − its perigee. This gave a net impulse of 1.73 .73 × 10 kg m s to the satellite. Ignore the change in mass due to burning of fuel. (T9.1)

What is the height of the new apogee, ℎ above the surface of the Earth, after this engine burn?

(T9.2)

Find the eccentricity ( ) of the new new orbit after the burn and the new orbital period (  ) of MOM in hours.

14 6

(T10) Gravitational Lensing Telescope Einstein's General Theory T heory of Relativity Relativity predicts bending of light light around aroun d massive bodies. For simplicit simplicit y, we assume that the bending of lig light ht happens at a singl singlee point for each light light ray, as shown in the figure figure.. The angle of bending, b , is given given by

b =

  

where  is the Schwarzschild radius associated with that gravitational gravitational body. We call call , the distance of the incoming light light ray from the parallel -axis passing through the centre of the body, as the “impact parameter”.

A massive body thus thu s behaves somewhat somew hat like like a focusing focus ing lens. The light light rays coming from fr om infinite infinite distance beyond a massive body, and having the same impact impact parameter , converge at a point point along the axis, axis, at a distance  from the centre of the massive body. An observer at that point will will benefit from fro m huge amplification amplification due to this this gravitational gravitational focusing. T he massive body in this case is being used as a Gravitational Lensing T elescope for amplification amplification of distant distant signals. signals. (T10.1) Consider the possibility possibility of our Sun as a gravitationa gravitationall lensing lensing telescope. Calculate the shortest short est distance,  , from the centre of the Sun (in A.U. ) at which the light light rays can get get focused. focu sed.

6

(T10.2) Consider a small circular detector of radius  , kept at a distance  centered on the -axis and perpendicul perpendicular ar to it. it. Note that only the light light rays which which pass within a certain certain annulus annulus (ring) (ring) of width ℎ (where ℎ ≪ ⊙ ) around the Sun would would encounter the detector. The ampli amplifica fication tion factor at the detector is defined as the ratio of the intensity intensity of the light light incident incident on the detector in the presence of the Sun and the intensity intensity in the absence of the Sun.

8

Express the amplification amplification factor,  , at the detector in terms of ⊙ and  . (T10.3) Consider a spherical mass distribution, such as dark matter in a galaxy galaxy cluster, through th rough which light light rays can pass while while undergoi under going ng gravitationa gravitationall bending. Assume for simpli simplicity city that for the gravitational bending with impact parameter, , only only the mass () enclosed inside the radius  is relevant. What should be the mass distribution, () , such that the gravitational gravitational lens behaves like like an ideal optica optica l convex lens?

6

Theoretical Examination Page 4 of 6 (T11) Gravitational Waves The Th e first signal signal of gravitational gravitational waves was observed by two advanced LIGO detectors detec tors at Hanford and Livingston, USA in September 2015 . One of these measurements (strain vs time in seconds) is shown in the accompanying figure. figure. In this problem, we will will interpret this signal signal in terms of a small test mass  orbiting around arou nd a large mass  (i.e.,  ≪  ), by considering several models for the nature of the central mass.

The Th e test mass loses energy due to the emission emission of gravitationa gravitationall waves. As a result the th e orbit keeps on shrinking, shrinking, until the test mass reaches the surf ace of the object, or in the case of a black hole, the innermost innermos t stable circular orbit – ISCO – which is given given by ISCO = 3sch , where  is the Schwarzschild radius of the black hole. hole . This is the “epoch of merger". At this point, the amplitude of the gravitational wave is maximum, and so is its frequency, which is always twice the orbital frequency. In this problem, we will will only focus on the gravitationa gravitationall waves before the merger, when Kepler’s laws are assumed to be valid. valid. After the merger, the form of gravitational gravitational waves will drastically drastically change. (T11.1) Consider the observed gravitational waves shown in in the figure figure above. Estimate the time time period, calculate ate the frequency,  , of gravitational gravitational waves just before the epoch of  , and hence calcul merger.

3

(T11.2) For any main main sequence (MS) star, the radius radius of the star,  , and its mass,  , are related by a power law given as,

10

 where where 

(  )

∝ =

0.8

for ⊙ < 

=

1.0

for 0.08⊙ ≤  ≤ ⊙

If the central object were a main sequence star, star , write an expression for f or the maximum maximum frequency frequen cy of gravitationa gravitationall waves, wav es,  , in terms of mass of the star in units of solar masses ( /⊙ ) and  . (T11.3) Using the above result, determine the appropriate value of  that will will give give the maximu maximu m possible possible frequency of gravitati gravitational onal waves, ,x for any main sequence star. Evaluate this frequency.

9

(T11.4) White dwarf (WD) ( WD) stars have a maximum maximum mass of 1.44 ⊙ (known as the Chandrasekhar Chandrasekhar limit limit ) and obey the mass-radius relation  ∝  −/ . The radius of a solar mass white dwarf is equal

8

Theoretical Examination Page 5 of 6 to 6000 km . Find the highest highest frequency fr equency of emitted emitted gravitational waves, D,x , if the test mass is is orbiting a white dwarf. dwa rf. (T11.5) Neutron stars (NS) are a peculia peculiarr type of compact objects which which have masses between 1 and 3⊙ and radii in the range 10  15 km km . Find the range of frequencies of emitted gravitational waves, , and ,x, if if the test mass is is orbiti or biting ng a neutron star at a distanc distancee close to the neutron star radius.

8

(T11.6) If the test mass is orbiting orbiting a black hole ho le (BH), write the expression expression for the frequency of emitted emitte d gravitational waves, B , in terms of mass of the black hole, B , and the solar mass ⊙ .

7

(T11.7) Based only on the time time period (or frequency) of gravitati gravitational onal waves before the epoch of merger, merger, determine determine whether the central object can be a main main sequence star (MS), a white white dwarf (WD), a neutron star (NS), or a black hole (BH). Tick the correct option in the Summary Answersheet. Estimate Estimate the mass of this object, bj , in units of ⊙ .

5

(T12) AstroSat India astronomy satellite, AstroSat, launched in September 2015, has five different instruments.

In this question, we will discuss three of these instruments (SXT, LAXPC, CZTI), which point in the same direction direction and observe in X-ray wavelengths. T he details of these instrum instruments ents are given given in the table below. below. Instrument

SXT LAXPC

Band [keV] 0.3 – 0.3 – 80 80 3 – 3 – 80 80

Collecting Area [m2] 0.067 1.5

Effective Photon Detection Efficiency 60% 40%

CZTI

10 – 10 – 150 150

0.09

50%

Saturation Saturation level [counts] 15000 (total) 50000 (in any one counter) or 200000 (total) (total) ---

No. of Pixels 512 x 512 ---

4 x 4096

You should note that LAXPC energy range is divided into 8 different energy band counters of equal bandwidth bandwidth with no overlap. (T12.1) Some X-ray sources like Cas A have a prominent emission line at 0.01825 nm corresponding corresponding 13  to a radioactive transition of Ti. Suppose there exists exists a source which emits emits only one bright emission line corresponding to this transition. What should be the minimum relative velocity obse rved peak of this line line to get registered in a different differen t () of the source, which will make the observed energy band counter of LAXPC as compared to a source at rest? These instruments were used to observe an X-ray source (assumed to be a point source), whose energy spectrum followed the power law,

 ( ) =   −⁄

[in units of counts/keV/m /s ]

where  is the energy in keV ,  is a constant and () is photon flux density at that energy. energy. Photon flux  density, by definition, definition, is given given for per unit collecting collecting area (m ) per unit bandwidth (keV) and per unit

Theoretical Examination Page 6 of 6 time (seconds). From prior observations, we know that the source has a flux density of 10 counts/keV/m /s at 1 keV , when measured by a detector with 100% photon detection efficiency. The “counts” here mean the number of photons reported by the detector. As the source flux follows the power law law given above, abo ve, we know that for a given given energy range from fro m  (lower energy) to  (higher energy) the total photon flux () will be given by

 = 3 (

⁄

⁄

   )

[in units of counts/m /s ]

(T12.2) Estimate the incident flux density from the source at 1 keV, 5 keV, keV, 40 keV keV and 100 100 keV. Also estimate what w hat will will be b e the total count per unit bandwidth recorded reco rded by each of the instrume instruments nts at these energies energies for an exposure exposure time of 200 seconds.

8

(T12.3) For this source, calculate calculate the maximum maximum exposure time (  ), without suffering from saturation, for the CCD of SXT.

4

(T12.4) If the source became 3500 times brighter, brighter, calculate calculate the th e expected counts per second in LAXPC counter 1, counter 8 as well as total counts across the entire energy range. If we observe for longer longer period, will the counter saturate satu rate due to any individual individual counter or due to the total count? Tick the appropriate box in the Summary Answersheet.

8

(T12.5) Assume that the counts reported by CZT I due to random fluctuati fluctuations ons in electronics electronics are about all energy en ergy levels. levels. Any source is considered as 0.00014 counts per pixel per keV per second at all “detected” when the SNR (signal (signal to noise ratio) is at least least 3. What is minimum minimum exposure time, t ime,  , needed for the source above to be detected in CZTI? Note that the “noise” noise” in a detector is equal to the square root of the counts due to random fluctuations.

10

(T12.6) Let us consider consider the situation where the source shows show s variabili variability ty in number flux, flux, so that the factor  increases by 20% . AstroSat observed this source for 1 second before the change and 1 second after this this change change in brightness. brightness. Calcul Calculate ate the counts measured by SXT , LAXPC LAXPC and CZTI in both the observations. Which instrument is best suited to detect this change? Tick the appropriate box in the Summary Answersheet.

7

Theoretical Examination Examination

Page 1 of 27 (T1) True or False Determine Determine if each each of the following following statements statements is True or False. In the Summary Summary Answersheet, Answersheet, tick the correct answer (TRUE / FALSE) for each each statement statement.. No justifications justifications are necessary necessary for this question. (T1.1) In a photograph of the clear sky on a Full Moon night with a sufficiently long exposure, the colour of the sky would appear blue as in daytime. Solution: T The colour of the clear sky during night is the same as during daytime, since the spectrum of sunlight reflected by the Moon is almost the same as the spectrum of sunlight. Only the intensity is lower.

2

2.0

(T1.2) An astronomer at Bhubaneswar marks the position of the Sun on the sky at 05:00 UT every every day of the year. If the Earth’s axis were perpendicular perpendicular to its orbital plane, these positions would trace an arc of a great circle. Solution: T If the Earth’s axis were perpendicular to its orbital plane, the celestial equator will coincide with ecliptic and the Sun will remain along the celestial equator every day. However, as the Earth’s orbit is elliptical, the true sun would still lead or lag mean sun by a few minutes on different days of year.

2

2.0

(T1.3) If the orbital period of a certain minor body around the Sun in the ecliptic plane is less than the orbital period of Uranus, then its orbit must necessarily be fully inside the orbit of Uranus. Solution: F The semi-major axis of the orbit of the body will be less than that of Uranus. However the minor body’s orbit may have a high eccentricity, in which case it may go outside that of Uranus.

2

2.0

(T1.4) The centre of mass of the solar system is inside the Sun at all times. Solution: F The centre of mass of Sun-Jupiter pair is just outside the Sun. Thus, if all gas giants are on same side of the Sun, the centre of mass of Solar system is definitely outside the Sun.

2

2.0

2

(T1.5) A photon is moving in free space. As the Universe expands, its momentum decreases. Solution: T For photons the wavelength increases when the Universe expands.

2.0


Page 2 of 27 (T2) Gases on Titan Gas particle particless in a planet planetary ary atmosphe atmosphere re have have a wide wide distri distribut bution ion of speeds. speeds. If the r.m.s. r.m.s. (root mean square) thermal speed of particles of a particular gas exceeds 1 /6 of the escape speed, then most of that gas will escape from the planet. planet. What is the minimum minimum atomic weight weight (relative atomic atomic mass), A mass), A min , of an ideal monatomic gas so that it remains in the atmosphere of Titan? Given, mass of Titan M Titan M T = 1.23 Titan T T = 93 93..7 K.

10

Titan R T = 2575 km, surface surface temperature of × 1023 kg, radius of Titan R

Solution: As the gas is monatomic, 3 1 2 kB T T mg vrms 2 2   M g 2 3kB T T v N A rms

≈ ≈

∴

v rms

≈



3kB N A T T M g

4.0

50% deduction if 3/2 pre-factor is not used and 1/2 or 1 are used instead. 50% deduction Full credit if students writes the relation for vrms directly. To remain in atmosphere, atmosphere, vesc 1 vrms < = 6 6



3kB N A T T < M g ∴



2GM T RT



GM T 18 18R RT

54 54k kB N A T T RT GM T 54 1.381 10−23 6.022 1023 93 93..7 2.575 > 6.6741 10−11 1.23 1023 > 13 13..2 g

4.0

M g >

×

×

× ×

× ×

× ×

×

× 106 g

Thus, all gases with atomic weight more than Amin = 13 13..2 will be retained retained in in the atmospher atmospheree of Titan. Half mark for understanding that atomic mass has no units.

1.5

0.5

Alternative solution 3 kB T T 2 

2 ≈ 12 mgvrms

∴



v rms

∴

≈



3kB T T mg

4.0

54 54k kB T T RT GM T mg > 2 > 2..19 10−26 kg

m g >

×

mg 2.19 = ∴ A min = atomic mass unit 1.66

4.0 1.0 −26

× 10 × 10

kg −27 kg


Page 3 of 27

Amin = 13 13..2

1.0

Answers between 13.0 and 13.4 are acceptable with full credit.

(T3) Early Universe Cosmological models indicate that radiation energy density, ρ density, ρ r , in the Universe is proportional to 4 (1+ (1 + z ) , and the matter energy density, ρ density, ρ m, is proportional to (1+ z )3 , where z where z is is the redshift. The dimensionless density parameter, Ω, is given as Ω = ρ/ρ c , where ρc is the critical energy density of the Universe. Universe. In the present present Universe, Universe, the density parameters parameters corresponding corresponding to radiation radiation and −4 matter, are Ωr0 = 10 and Ωm0 = 0.3, respectively. 3

(T3.1) Calculate Calculate the redshift, redshift, z z e , at which radiation and matter energy densities were equal. Solution: ρm0 /ρc Ωm0 0 .3 = = −4 = 3000 ρr0 /ρc Ωr0 10 At z At z e , both matter density and radiation density were equal. ρr = ρ = ρ m ∴

ρ r0 (1 + ze )4 = ρ m0 (1 + ze )3 ρm0 1 + ze = = 3000 ρr0 ∴

z e

2.0

 3000

1.0

Only ze = 2999 and ze = 3000 are 3000 are acceptable answers. (T3.2) Assuming that the radiation from the early Universe has a blackbody spectrum with a temperature of 2. 2.732 K, estimate estimate the temperature, T temperature, T e , of the radiation at redshift z redshift z e .

4

Solution: As the Universe behaves like an ideal black body, the radiation density will be proportional to the fourth power of the temperature (Stefan’s law).

  T e T 0

4

ρre ρr0 ρr0 (1 + ze )4 = ρr0 =

2.0

4 T e = (1 + ze )4 2.732 T e = 1 + ze = 3000 2.732 T e = 3000 2.732

 

1.0

×

T e = 8200K 8100

1.0

8200 gives gives 1.0; 8200 1.0; 8200 < < T ≤ 9000 9000 gives gives 0.5; else 0. ≤ T ≤ 8200 e

e

(T3.3) Estimate the typical photon energy, E ν ν (in eV), of the radiation as emitted at redshift ze .

3


Page 4 of 27

Solution: Wien’s law: λmax = = E ν ν = = E ν ν =

0.002898mK T e 0.002898 m = 354 354 nm 8200 hc λmax 6.62 10−34 3 108 5.62 10−19 J = eV 354 10−9 1.602 10−19 3.5 eV

×

×

× ×

1.0

× ×

1.0 1.0

Alternative solution: E ν ν = k B T e = 1.38

−23

× 10 ×

1.13 10−19 8200 8200 J = eV 1.602 10−19

× ×

E ν ν = 0.71eV

2.0 1.0

Use Use of eith either er Wien’ Wien’ss law law or E = kB T T gets full full credit credit.. E ν ν = 3kB T /2 or similar similar gets no credit credit.. Answers Answers with E ν T also get ν = 3kB T or E ν ν = 2.7kB T full credit.

(T4) Shadows An observer in the northern hemisphere noticed that the length of the shortest shadow of a 1 .000m vertical stick on a day was 1. 1 .732 m. On the same day day, the length of the longest longest shadow shadow of the same vertical stick was measured to be 5. 5 .671m.

10

Find the latitude, latitude, φ φ,, of the observer and declination of the Sun, δ  , on that day. Assume the Sun to be a point source and ignore atmospheric refraction. Solution: As the longest shadow of the Sun on the given day is of finite length, the Sun is circumpolar for this observer on this day.

φ

90 90

− δ



− δ



A θ O

θ2

θ1

S

In the figure above, the left panel shows the shadow OS shadow OS formed formed by stick OA stick OA (of (of length 1. 1.000 m), m), and the right panel shows the Sun’s location in two cases.

2.0


Page 5 of 27

For an altitude θ altitude θ of the Sun, OA 1.000m tan θ = = OS OS ∴

cot θ = OS = OS (in (in metres) metres)

1.0

Let θ Let θ 1 and θ and θ 2 be altitude in two extreme cases. θ1 = 180◦ ◦

cot(90 ∴

◦

◦

− φ − (90 − δ ) = 90 − φ + δ 

− φ + δ ) = 1.732 tan(φ tan(φ − δ ) = 1.732 φ − δ = tan 1 (1. (1.732) = 60  

−

◦



θ2 = φ = φ cot(φ cot(φ

∴

1.0



− 90 tan(φ tan(φ − 90

◦

= 1.047 047 rad

◦

◦

− (90 − δ ) = φ − 90

1.5

+ δ 

1.0

+ 90◦ = 100◦ = 1.745 745 rad

1.5



+ δ  ) = 5.671 1 ◦ + δ  ) = 5.671

φ + δ  = tan−1 Solving,

  1 5.671

φ = 80◦ = 1.396rad

1.0

δ  = 20◦ = 0.349rad

1.0 ◦

±0.5 is allowed. One can also solve the question by manipulating tan(φ tan( φ − δ ) and tan(φ tan(φ + δ + δ ), to get tan(φ tan(φ) Given high accuracy of shadow length, only





and tan(δ tan(δ  ).

(T5) GMRT beam transit Giant Metrewave Radio Telescope (GMRT), one of the world’s largest radio telescopes at metre wavel wavelengths engths,, is located in western India (latitude: (latitude: 19◦ 6 N, longitu longitude: de: 74◦ 3 E). GMRT consists of 30 dish antennas, each with a diameter of 45 .0 m. A single single dish of GMRT GMRT was held fixed fixed with ◦  its axis pointing at a zenith angle of 39 42 along the northern meridian such that a radio point source would pass along a diameter of the beam, when it is transiting the meridian.

10

What is the duration T duration T transit transit for which this source would be within the FWHM (full width at half maximum) maximum) of the beam of a single GMRT GMRT dish observing observing at 200 MHz? Hint: The FWHM size of the beam of a radio dish operating at a given frequency corresponds to the angular resolution of the dish. Assume uniform illumination.

Solution: As the dish is pointed towards northern meridian at zenith angle of 39 .7◦ , altitude of the centre of the beam is a = 90 90..00◦

◦

◦

90..00 − 39 39..70 − z = 90

= 50 50..30◦

1.0


Page 6 of 27

Thus, declination of the source should be, δ = = 90 90..00

◦

◦

90..00 − 50 50..30 − a + φ = 90

+ 19. 19.10◦ = 58 58..80◦

2.0

Declination = ZA + Latitude also gets full credit. FWHM beam size (for uniform illumination) will be given by 1.22 22λ λ θ = D 1.22 22cc 1.22 2.998 108 = = Dν 45 45..0 2 108 = 0.0406 0406 rad rad

×

× ×

1.5

×

θ = 2.33◦ θ 3.99min T transit transit = cos δ 2.33 3.99min = cos58. cos58.8 T transit 17..9min transit = 17

1.5

×

3.0

×

1.0

•

Use of 4 min per degree is also acceptable.

•

Missing cos δ gets δ gets a penalty of 2.0.

(T6) Cepheid Pulsation The star β -Doradu -Doraduss is a Cephei Cepheid d variable ariable star with a pulsat pulsation ion period of 9.8 9.844 days. days. We mak makee a simplifying assumption that the star is brightest when it is most contracted (radius being R1 ) and it is faintest when it is most expanded (radius being R 2 ). For simplicity, assume that the star maintains its spherical shape and behaves as a perfect black body at every instant during the entire cycle. The bolometric magnitude of the star varies from 3.46 to 4.08. From Doppler measurements, we know that during pulsation the stellar surface expands or contracts at an average radial speed of 12. 12.8 k m s−1 . Over the period of pulsation, the peak of thermal radiation (intrinsic) of the star varies from 531. 531.0 nm to to 649. 649.1nm. (T6.1) Find the ratio of radii of the star in its most contracted and most expanded states (R1 /R2 ).

7

Solution: We first find flux ratio and then use Stefan’s law to compare the fluxes. m1

− m2 = −2.5log ∴

  F 1 F 2

1.0

F 1 4(m1 −m2 ) 4(3.46−4.08) = 10−0.4(m = 10−0.4(3. F 2 = 1.77

1.0

Li = 4πRi2 σT i4

1.0

∴ F 1 =

4πR12 σT 14 4πD 2

F 2 =

4πR 22 σT 24 4πD 2

1.0


Page 7 of 27

F 1 R2 = 12 F 2 R2 R1 = R2

4

× T T 14 2

   × F 1 F 2

T 2 T 1

2

From Wien’s displacement law, ∴

F 1 F 2

=

1.77

λ1 λ2

1.0

2

   ×   √ ×

R1 = R2

T 2 λ1 = . T 1 λ2

531.0 531. 649..1 649

1.0 2

R1 = 0.890 R2

Acceptable range:

1.0 010.. ±0.010

(T6.2) Find the radii of the star (in metres) in its most contracted and most expanded states (R1 and R and R 2 ).

3

Solution: R2

= v × P /2 − R1 = v 9.84 R2 − R1 = 12 12..8 × 103 × 86400 × m 2 (1 − 0.890)R 890)R2 = 5.441 × 109 m R2 = 4.95 × 1010 m R1 = 4.41 × 1010 m Acceptable range: ±0.02 × 1010 m for both. ∴

2.0

0.5 0.5

5

(T6.3) Calculate the flux of the star, F 2 , when it is in its most expanded state. Solution: To get the absolute value of flux (F ( F 2 ) we must compare it with observed flux of the Sun. F 2 m2 m = 2.5log F 

−

∴

−

 

4(m2 −m ) F 2 = F = F  10−0.4(m L 4(4.08+26. 08+26.72) = 10−0.4(4. 2 4πa ⊕

3.0

× 3.826 × 1026 × 4.7863 × 10 = 4π (1. (1.496 × 1011 )2 F 2 = 6.51 × 10 10 W m 2 Acceptable range: ±0.04 × 10 10 W m 2 .

−13

−

−

−

−

W m−2 2.0


Page 8 of 27 5

(T6.4) Find the distance to the star, D star , in parsecs. Solution: Dstar =

L2 = 4πF 2



2.898

× 10



Wien’s law: T 2 =

Dstar = 4.95 ∴

R22 σT 24 = R 2 T 22 F 2 −3

σ F 2



2.0

mK

1.0

λ2

× 10

10

 ×

2.898 649..1 649

× 10 × 10

−3 −9

  2

5.670 10−8 6.51 10−10

× ×

298 pc × 1018 m = 298 298 ± 2pc (depends on truncation).

Dstar = 9.208

Acceptable range:

2.0

(T7) Telescope optics In a particular ideal refracting telescope of focal ratio f /5, the focal length of the objective lens is 100 cm and that of the eyepi eyepiece ece is 1 cm. (T7.1) What is the angular magnification, m0 , of the teles telesco cope? pe? What What is the length length of the the telescope, L telescope, L 0 , i.e. the distance between its objective and eyepiece? Solution: The magnification will be given by, f o m0 = f e 100 = = 100 1 The magnification is m0 = 100 Length of the telescope will be L0 = f = f o + f e = 100 + 1 = 101 cm

4

1.0 1.0

1.0 1.0

The telescope length will be L0 = 101 101 cm Exact answer required for credit. An introduction of a concave lens (Barlow lens) between the objective lens and the prime focus is a common way to increase the magnification without a large increase in the length of the telescope. A Barlo Barlow w lens lens of focal length length 1 cm is now now introdu introduced ced between between the objectiv objectivee and the eyepi eyepiece ece to double the magnification magnification.. (T7.2) At what distance, dB , from the prime focus must the Barlow lens be kept in order to obtain this desired desired double magnification? magnification? Solution: We use the following sign convention. Lens is the origin. Direction along the direction

6


Page 9 of 27

1 1 1 = (f is f is positive for convex f v u v lens and negative negative for concave concave lens). Magnification Magnification is m = . Solutions Solutions using using other u sign conventions are acceptable. Let v Let v be image distance from the Barlow lens. 1 1 1 = f B v u Distance of Barlow lens, dB , before the prime focus is same as the object distance, u, in this case. of light is taken as positive. The lens formula is

−

−

1 1 = f B v

1.0 1.0

− d1

Also, m Also, m B = 2 = ∴

∴

B

v v = u dB

1 2 = dB v 1 1 = 1 v

0.5 1.0

2 − v − −1 = −1

v v = 1 cm v 1 cm dB = = = 0.5 cm 2 2

2.5

The positive sign for d for dB indicates indicates that the Barlow Barlow lens was introduced introduced 0.5 cm befo before re the prime focus. (T7.3) What is the increase, ∆L ∆ L, in the length of the telescope?

4

Solution: The increase in the length will be, ∆L = v = v

− dB = 1.0 − 0.5 = 0.5 cm

2.0 2.0

Thus, Thus, the length length will be increased increased by by ∆ L = 0.5 cm

A telescope is now constructed with the same objective lens and a CCD detector placed at the prime focus (without any Barlow lens or eyepiece). The size of each pixel of the CCD detector is 10 µm. (T7.4) What will be the distance in pixels between the centroids of the images of the two stars, np , on the CCD, if they are 20  apart on the sky? Solution: Plate scale at prime focus is given by, 1 1rad s = = = 0.206 265 arcsec/ arcsec/µm f o 1m

6

2.0


Page 10 of 27

Since each pixel is 10 µm in size, sp = 10

arcsec/ m = 2.06 arcsec/pixel arcsec/pixel × 0.206 arcsec/

2.0

µ

Two stars will be separated by,

20 np = pixels 10 pixels 2.06 Accepta Acceptable ble range: range: 9.5 to 10.5 pixels. pixels.

2.0



(T8) U-band photometry A star has an apparent apparent magnitude magnitude m U = 15 15..0 in the U the U -band. -band. The U The U -band -band filter is ideal, i.e., it has perfect (100%) transmission within the band and is completely opaque (0% transmission) outside the band. band. The filter filter is center centered ed at 360 nm, and has a width width of 80 nm. It is assumed assumed that that the star also has a flat energy spectrum with respect to frequency frequency.. The conversion conversion between between magnitude, magnitude, m, m , −26 −1 −2 in any band and flux density, f , f , of a star star in Jansk Jansky y (1 Jy = 1 10 W Hz m ) is given by f = = 3631

× 10

−0.4m

×

Jy

(T8.1) Approximately how many U many U -band -band photons, N photons, N 0 , from this star will be incident normally 2 on a 1 m area at the top of the Earth’s atmosphere every second?

8

Solution: The U The U -band -band is defined as (360 40) nm. Thus, Thus, the maximum, maximum, minimum and average average frequencies of the band will be, c ν max = 9.369 1014 Hz max = λmax ν min 1014 Hz min = 7.495

±

×

× 14 ν avg avg = 8.432 × 10 Hz ∆ν = = ν max max − ν min min = 1.874 × 1014 Hz 0.4 15 f st1 st1 = 3631 × 10 −

×

= 3.631 631 mJy mJy = 3. 3.631

Now, N Now, N 0

2.0 −29

× 10

W Hz−1 m−2

× hν avg avg = ∆ν × f st1 st1 × A × ∆t

2.0 2.0

where, A where, A = = 1 m2 & ∆t = 1 s

∴ N 0 =



1.874 1014 3.631 10−29 6.626 10−34 8.432 1014 12180

× ×

× ×

× ×

Exact calculation including integration is accepted with full credit (exact answer: answer: 12190). Accepted range: 12180 200 200.. Using flat spectrum for ∆ for ∆λ λ instead of ∆ν ∆ ν is is considered a major conceptual error, and will incur penalty of 2.0 marks.

±

This star is being observed in the U -band U -band using a ground based telescope, whose primary mirror has a diameter of 2. 2 .0 m. Atmos Atmosphe pheric ric extinct extinction ion in U -band U -band during the observation observation is 50%. You may ma y assume assume that the seeing seeing is diffractio diffraction n limite limited. d. Av Avera erage ge surfac surfacee brigh brightne tness ss of night night sky in U -band U -band was measured to be 22.0 mag/arcsec2 .

2.0


Page 11 of 27 8

(T8.2) What is the ratio, R, of number of photons received per second from the star to that received from the sky, when measured over a circular aperture of diameter 2  ? Solution: Let us call sky flux per square arcsec as Φ and total sky flux for the given aperture as φ as φ sky . Let total star flux be φ st . φsky = AΦ = AΦ = = π π ∴

arcsec))2 × Φ = πΦ = πΦ × (1 arcsec

   

Φ φsky Φ  = 22 22..0 + 2. 2 .5log10 π  Φ = 22 22..0 2.5log10 (π (π )

m sky = 22 22..0 + 2. 2 .5log10

−

3.0 1.0

msky = 20 20..76 mag

1.0

As extinction is 50% φst2 0.5φst1 R = = = 0.5 φsky φsky

1.0 (20.76 × 10(20.

−15)/ 15)/2.5

 100 ±5.

2.0

Accepted range: A student may also calculate number of photons incident per second per metre square in this case and then compare it with the answer in the first case to get the correct ratio. (T8.3) In practice, practice, only 20% of U -band U -band photons photons falling on the primary mirror are detected. detected. How many photons, N photons, N t , from the star are detected per second?

4

Solution:

× 1 m2 = N 0 × 0.5 × 0.2 × At N t = 12180 × 0.5 × 0.2 × π N t  3813 Accepted range (3813 ± 50). 50). N t

2.0 2

  2.0 2

= 1233π 1233π

1.0 1.0

(T9) Mars Orbiter Mission India’ India’ss Mars Mars Orbiter Orbiter Missio Mission n (MOM) (MOM) was was launc launched hed using using the Polar Polar Satell Satellite ite Launc Launch h Vehicle ehicle (PSL (PSLV) V) on 5 No Nov vem ember ber 2013. 2013. The The dry dry mass mass of MO MOM M (body (body + inst instru rume men nts) ts) was 500 500 kg and and it carried carried fuel fuel of mass 852 kg. It was initiall initially y placed placed in an elliptic elliptical al orbit around around the Earth with perigee perigee at a heigh heightt of 264 km and apogee at a heigh heightt of 23 904 km, above above the surface surface of the Earth. Earth. After raising the orbit six times, MOM was transferred to a trans-Mars injection orbit (Hohmann orbit). The first such orbit-raising was performed by firing the engines for a very short time near the perigee. The engines were were fired to change change the orbit without changing changing the plane of the orbit and without changing its perigee. This gave a net impulse of 1 .73 105 k g m s−1 to the satellite. Ignore the change in mass due to burning of fuel.

×

(T9.1) What is the height of the new apogee, ha , above the surface of the Earth, after this

14


Page 12 of 27 engine engine burn? Solution: Let apogee and perigee distances be r a and rp respectively. rp = R = R ⊕ + hpi = (6371 + 264) 264) km = 6635 km

1.0

ra = R = R ⊕ + hia = (6371 (6371 + 23904 23904)) km = 30 275 km

1.0

Conservation of energy and angular momentum gives GmM ⊕ E = = rp + ra

−

Total energy at perigee 1 GmM ⊕ mvp 2 = E = E = 2 rp

−

1 2 GM ⊕ vp = 2 rp ∴

v p = =

1

  × 2

−

− GmM r + r

⊕

p

rp rp + ra

a



GM ⊕ ra ra + rp rp

2

2.0

6.674 10−11 6.635 106

× ×

= 9.929kms−1

× 5.972 × 1024 × 3.0275 × 107 (3.0275 + 6. 6.635 × 106 ) × (3.

As the engine burn is just 41. 41 .6 s, we assume that the entire entire impulse is applied instaninstan5 −1 taneou taneously sly at p erigee erigee.. The impulse impulse is J = 1.73 10 k g m s . No Note te that that the total total mass of MOM must include the fuel, so we have to use m = 500 + 852 = 1352 kg. Change in velocity due to impulse at perigee is

1.0

×

J 1.73 105 ∆v = = = 128 128..0 m s−1 m 1352 The new velocity will be given given by (we use  symbol to denote quantities after the first orbit-raising maneuvre)

×

vp = v = v p + ∆v ∆ v = 10 10..06kms−1 The perigee remains unchanged. So we get r p = r = r p . Since the satellite is moving faster, the new apogee will be higher. 

vp =

1.0

1.0 1.0



ra 2GM ⊕   rp (ra + rp )

rp 2GM ⊕ = ∴ 1 + ra (vp )2 rp =

1.0

2

×

2.0

× 6.674 × 10 11 × 5.972 × 1024 = 1.188 (10. (10.06 × 103 )2 × 6.635 × 106 −

6635 = 35 380 380 km 0.188 ha = 35380 6371 ra =

−

= 29009km

2.0

1.0


Page 13 of 27

Acceptable range:

150 km ±150 6

(T9.2) Find eccentricity (e ( e) of the new orbit after the burn and new orbital period ( P ) P ) of MOM in hours. Solution: As seen above, rp 1 e = 0 . 188 = ra 1+e 1 0.188 ∴ e = e = = 0.683 1 + 0. 0 .188 Acceptable range: 0.002 The new orbital semi-major axis and orbital period will be,

−

−

1.0 1.0

±

ra + rp 2 35380 + 6635 = = 20933km 2

a =

P = 2π = 2π

 

1.0 1.0

a 3 GM ⊕ (2. (2.0933 6.674 10−11

× Acceptable range: ±0.1 h

1.0

× 107)3 × 5.972 × 1024 =

30 136 136 s = 8.37 h

(T10) Gravitational Lensing Telescope Einste Einstein’ in’ss General General Theory of Relati Relativit vity y predic predicts ts bending bending of light light around around massiv massivee bodies. bodies. For simplicity, we assume that the bending of light happens at a single point for each light ray, as shown in the figure. The angle of bending, θ b , is given by 2Rsch θb = r where Rsch is the Schwarz Schwarzsch schild ild radius associated with that gravitational gravitational body bo dy.. We call r , the distance of the incoming light ray from the parallel x parallel x-axis -axis passing through the centre of the body, as the “impact parameter”. θb r x-axis

A massive body thus behaves somewhat somewhat like a focusing lens. The light rays coming coming from infinite infinite distance beyond a massive body, and having the same impact parameter r , converge at a point along the axis, at a distance f r from the centre centre of the massiv massivee body. body. An observer observer at that point point

1.0


Page 14 of 27 will benefit from huge amplification amplification due to this gravitational gravitational focusing. The massive massive body bo dy in this case is being used as a Gravitational Lensing Telescope for amplification of distant signals. (T10.1) Consid Consider er the possibili possibility ty of our Sun as a gravit gravitatio ational nal lensing lensing telesc telescope. ope. Calcul Calculate ate the shortest distance, f distance, f min min , from the centre of the Sun (in A.U.) at which the light rays can get focused.

6

Solution: The rays rays trave travelli lling ng closer to the gravitat gravitation ional al body will bend more. Thus, Thus, we get shortest convergence point where the rays just grazing the solar surface will meet each other. θb R θb f min min

2Rsch R R f min min 2 2 c2 R R ∴ f min = min = 2Rsch 4GM  (6. (6.955 108 2.998 108 )2 = m 4 6.674 10−11 1.989 1030 8.188 1013 13 = 8.188 10 m = AU 1.496 1011 f min 547.3 AU min = 547. θb =

2.0



2.0

× × × × × × × × × ×

2.0

(T10.2) Consider a small circular detector of radius a, kept at a distance f min min centred on the x-axis and perpendicular to it. Note that only the light rays which pass within a certain annulus (ring) of width h width h (where (where h h R ) around the Sun would encounter the detector. The amplification amplification factor at the detector is defined as the ratio of the intensity intensity of the light light incident on the detector in the presence of the Sun and the intensity in the absence of the Sun. Express Express the amplification amplification factor, A factor, A m , at the detector in terms of R R  and a and a..

8



Solution: The following figure needs to be drawn. h

1.0

θ2 Detector a

R f min min f 2


Page 15 of 27

The light bending from the surface of the Sun ( r = R  ) will intersect the detector at its centre, as it is kept at f min min . The boundary of the detector will be intersected by a light ray with r = R + h + h.. This ray will intersect the x-axis x -axis at a distance f distance f 2 . (R + h)2 f 2 = 2Rsch

2.0

Same argument as in Part 1 For small angles, a = (f 2 =



− f min min )θ2 R2 (R + h)2 − 2R 2R 



sch

sch

 2h



2Rsch 2R h + h2 = (R + h) R + h 2.0

Let original intensity of the incoming radiation be I be I 0 . The flux at the detector in the presence of Sun is Φ  = I = I 0 2π R h The flux at the detector in the absence of Sun is Φ 0 = I = I 0 π a2 The amplification is therefore Am =

Φ I 0 2πR  h R = = 2 Φ0 I 0 πa a

1.0 1.0

1.0

(T10.3) Consider a spherical mass distribution, such as a dark matter cluster, through which light light rays can pass while undergoing undergoing gravitational gravitational bending. Assume Assume for simplicit simplicity y that for the gravitational bending with impact parameter, r , only the mass M ( M (r) enclosed inside the radius r radius r is relevant.

6

What should be the mass distribution, M ( M (r ), such that the gravitational lens behaves like an ideal optical convex lens ? Solution:

r2

r1

θ2 θ1

f 1 = f 2

All rays should focus at the same spot. This should be evident from figure drawn on answersheet or otherwise. Let there be two rays with impact parameters r1 and r and r2 . The corresponding corresponding distances distances of focus will be ri2 ri2 c2 f i = = 2rschi 4GM ( GM (ri ) Same argument as in Part 1

2.0

2.0


Page 16 of 27

The rquirement f rquirement f 1 = f = f 2 implies r12 M ( M (r1 ) = M ( M (r2 ) r22 The required mass distribution is: M ( M (r )

1.0

∝ r2

1.0

(T11) Gravitational Waves The first signal of gravitational waves was observed by two advanced LIGO detectors at Hanford and Livingston, USA in September 2015. One of these measurements (strain vs time in seconds) is shown in the accompanying accompanying figure. In this problem, we will interpret interpret this signal in terms of a small test mass m orbiting around a large mass M M (i.e., m M ), M ), by considering several models for the nature of the central mass.



The test mass loses energy due to the emission of gravitational waves. As a result the orbit keeps on shrinking, until the test mass reaches the surface of the object, or in the case of a black hole, the innermost stable circular orbit – ISCO – which is given by RISCO = 3Rsch, where Rsch is the Schwa Schwarzs rzsch child ild radius radius of the black black hole. This This is the “epoch “epoch of merger merger”. ”. At this point, point, the amplitude of the gravitational wave is maximum, and so is its frequency, which is always twice the orbital frequency frequency. In this problem, we will only focus on the gravitational gravitational waves waves before the merger, merger, when Kepler’s laws are assumed assumed to be valid. After the merger, the form of gravitation gravitational al waves will drastically change. (T11.1) Consider Consider the observed observed gravitational gravitational waves waves shown in the figure above. above. Estimate Estimate the time period, T 0 , and hence calculate the frequency, f 0 , of gravitational waves just before the epoch of merger.

3


Page 17 of 27

Solution: From the graph, just before the peak of emission, the time period of gravitational waves is approximately (0. (0.007 0.004 004)) s. T 0

≈

±

0.007s

2.0

Acceptable Acceptable range: range: 0.003 to 0.011 s That is, the frequency of the gravitational waves is f 0 142..86Hz . 142 Acceptable Acceptable range: range: 333.33 to 90.91 Hz Answer given in terms of angular frequency with correct value and units gains full credit.

≈

(T11.2) For any main sequence (MS) star, the radius of the star, RMS, and its mass, M MS MS , are related by a power law given as,

1.0

10

α ∝ (M MS MS )

RMS

where α = 0.8 for M  < M MS MS = 1.0 for 0.08 08M M 

≤ M MS MS ≤ M



If the central object were a main sequence star, write an expression for the maximum frequency of gravitational waves, f waves, f MS MS , in terms of mass of the star in units of solar masses (M MS and α.. MS /M  ) and α Solution: Since m Since m

M then, by Kepler’s third law  M then,

1 f orbital orbital = 2π



GM r3

4.0

Hence the frequency of the gravitational waves is 1 f grav grav = 2f orbital orbital = π



GM r3

1.0

The frequency will be maximum when r = R = R MS .

1.0

For main sequence stars, α M MS MS M  M MS MS ∴ R MS = R  M 

RMS = R

            

1 ∴ f MS MS = π

GM MS MS 3 R

α

M  M MS MS

1.0 3α/2 α/2

1 = π

GM  3 R

M  M MS MS

(3α (3α−1)/ 1)/2

1 f MS MS = π

GM  3 R

M MS MS M 

(1−3α)/2

(T11.3) Using the above result, determine the appropriate value of α of α that that will give the maximum possible frequency of gravitational waves, f MS, MS,max for any main sequence star. Evaluate

3.0

9


Page 18 of 27 this frequency. Solution: For possible values of α α given in the question, the exponent 1−23α is negative. negative. Thus, Thus, if M MS Thus, for highest possible frequency frequency MS > M  , the frequency will be smaller. Thus, coming from a main sequence star, you should take lowest possible mass i.e. α = 1.0 4.0 1 f MS, MS,max = π =

1 π

( 1−32×1 ) M MS MS M 

    GM  3 R GM  3 R

M × M



2.0

MS MS

The frequency of gravitational waves will be given by, f MS, MS,max =

1 π



6.674

× 1.989 × 1030 × 1 0.08 (6. (6.955 × 108 )3

× 10

−11

f MS, MS,max = 2.5mHz

3.0

Answer given in terms of angular frequency with correct value and units gains full credit. (T11.4) White dwarf (WD) stars have a maximum mass of 1. 1 .44 44M M  (known as the Chandrasekhar −1/3 limit) and obey the mass-radius relation R M . The radius radius of a solar solar mass white white dwarf dwarf is equal equal to 6000 km. Find Find the highest highest frequen frequency cy of emitte emitted d gravit gravitati ational onal wave waves, s, f WD, WD,max , if the test mass is orbiting a white dwarf.

8

∝

Solution: The maximum frequency would be when r = R = R WD . We use the notation RW D for the radius radius of a sola solarr ma mass ss white white dwarf. dwarf. Then Then for white dwarfs M W D 3 3 RW = R D W D M  f WD WD =

1 π

=

1 π

 

1.0

1.0

GM WD WD 3 RW D GM  M W D 3 RW D M 

2.0

For maximum frequency, we have to take highest white dwarf mass. f WD, WD,max = 2.600

1030 × × × 103)3 × 1.44 6 95..96 × 103 × 1.44 × 95

−6

× 10 = 2.600 × 10

2.0



1.989 (6000

−

f WD, WD,max = 0.359Hz

2.0

Answer given in terms of angular frequency with correct value and units gains full credit. (T11.5) Neutron stars (NS) are a peculiar type of compact objects which have masses between

8


Page 19 of 27 1 and 3M 3M  and radii radii in the range 10 – 15 km. Find Find the range of frequen frequencie ciess of emitted emitted gravitational waves, f waves, f NS, NS,min and f NS, NS,max , if the test mass is orbiting a neutron star at a distance close to the neutron star radius. Solution: 1 GM NS NS f grav grav = 3 π RNS



The lowest possible frequency is when M NS NS is lowest and R NS is the highest. For M or M NS = M  and R and R = 15 km km,, we get NS = M f NS, kHzz NS,min = 1.996 kH Similarly, the largest possible frequency is when M NS NS is the largest and RNS is the smallest. For M or M NS and R = 10 km km,, we get NS = 3M  and R f NS, 352 kH kHzz NS,max = 6.352

2.0 2.0 2.0 2.0

Answer given in terms of angular frequency with correct value and units gains full credit. 7

(T11.6) If the test mass is orbiting a black hole (BH), write the expression for the frequency of emitted gravitational waves, f waves, f BH BH , in terms of mass of the black hole, M BH BH , and the solar mass M  . Solution: For black holes, we have to consider R ISCO . Hence the equation will be, f BH BH =

1 π

×



GM  3 27 27R Rsch −

f BH kHzz BH = 4.396 kH

× M M



2.0

2.0

BH BH

× M M



3.0

BH BH

Answer given in terms of angular frequency with correct value and units gains full credit. 5

(T11.7) Based only on the time period (or frequency) of gravitational waves before the epoch of merger, determine whether the central object can be a main sequence star (MS), a white dwarf (WD), a neutron star (NS), or a black hole (BH). Tick the correct option in the Summary Answersheet. Estimate the mass of this object, M obj M  . obj , in units of M Solution: We found found the freque frequency ncy of the LIGO-det LIGO-detect ected ed wave wave to be 166 166..67Hz just before merger. merger. As per our analysis analysis above, above, only only black black holes holes can lead to emission emission in this frequency range. Black Hole By using corresponding corresponding expression, expression, 4396 M obj M  31 31M M  obj = 142..86 142 Any answer between 13 to 50 will get full credit.

≈

2.0 3.0


Page 20 of 27 (T12) Exoplanets Two major methods of detection of exoplanets (planets around stars other than the Sun) are the radial velocity velocity (or so-called “wobble”) “wobble”) method and the transit transit method. In this problem, we find out how a combination of the results of these two methods can reveal a lot of information about an orbiting exoplanet and its host star. Throughout this problem, we consider the case of a planet of mass M p and radius R radius R p moving in a circular orbit of radius a radius a around a star of mass M mass M s (M s M p ) and radius R radius R s . The normal to the orbital plane of the planet is inclined at angle i with respect to the line of sight (i ( i = 90◦ would mean “edge on” orbit). We assume that there is no other planet orbiting the star and R s a.





“Wobble” Method: When a planet and a star orbit each other around their barycentre, the star is seen to move slightly, or “wobble”, since the centre of mass of the star is not coincident with the barycentre of the star-pla star-planet net system. system. As a result result,, the light light receiv received ed from the star undergoe undergoess a small small Dopple Dopplerr shift related to the velocity of this wobble. The line of sight velocity, v , of the star can be determined from the Doppler shift of a known spectral line, and its periodic variation with time, t, is shown in the schematic diagram below. In the diagram, the two measurable quantities in this method, namely, the orbital period P and maximum line of sight velocity v 0 are shown. l

v

l

P

v0 t

(T12.1) Derive Derive expressions expressions for the orbital radius (a ( a) and orbital speed (v ( vp ) of the planet in terms of M M s and P and P ..

3

Solution: Kepler’s law: a =



GM s 2 P 4π 2



1/3

1.0

Gravitational force provides centripetal acceleration: vp = vp =

   GM s a

2πGM s P

0.5 1/3

1.5


Page 21 of 27 4

(T12.2) Obtain a lower limit on the mass of the planet, M planet, M p, min in terms of M s , v 0 and v and v p . Solution: Momentum conservation: M p vp = M = M s vs

1.5

Observed quantity is v 0 = v = v s sin i. Thus, M p, min = M = M p sin i =

M s vs sin i M s v0 = vp vp

This is a lower limit on M p .

2.5

Transit Method: As a planet orbits its host star, for orientations of the orbital plane that are close to “edge-on” (i 90◦ ), it will pass periodically, or “transit”, in front of the stellar disc as seen by the observer. This would cause a tiny decrease in the observed stellar flux which can be measured. The schematic diagram below (NOT drawn to scale) shows the situation from the observer’s perspective and the resulting resulting transit light curve curve (normalised (normalised flux, f flux, f ,, vs time, t) t ) for a uniformly bright stellar disc.

≈

Rs

bRs 1

2

3

4

Rp f 1 ∆ tF tT t If the inclination angle i is exactly 90 ◦ , the planet would be seen to cross the stellar disc along a diameter. diameter. For other values of i, the transit occurs along a chord, whose centre lies at a distance bRs from the centre centre of the stellar stellar disc, as shown. shown. The no-trans no-transit it flux is normalise normalised d to 1 and the maximum dip during the transit is given by ∆.


Page 22 of 27 The four significant points in the transit are the first, second, third and fourth contacts, marked by the positions 1 to 4, respectively, in the figure above. The time interval during the second and third contacts is denoted by tF , when when the disc of the planet planet overl overlaps aps the stellar stellar disc fully. fully. The time interval between the first and fourth contacts is denoted by t T . These points are also marked in the schematic diagram below showing a “side-on” view of the orbit (NOT drawn to scale).

i

4 3

Observer

a 1

2

The measurable quantities in the transit method are P , P , t T , t F and ∆. (T12.3) Find the constraint on i in terms of Rs and a for the transit to be visible at all to the distant observer.

2

Solution: bRs = a = a cos i Therefore, for visibility, 0

1.0

≤ b ≤ 1 ⇒ i ≥ cos

−1

(Rs /a) /a)

1.0

(T12.4) Express ∆ in terms of R R s and Rp .

1

Solution: Blackbody brightness is proportional to area. Since the observer is far away from the star-planet system, size of silhouette of planet on stellar disc is independent of a of a..

⇒

∆=

Rp Rs

 

2

1.0

(T12.5) Express t Express t T and t and t F in terms of R R s , R p , a, a , P and b and b.. Solution: Circular Circular orbit uniform orbital speed t aφ φ = = P 2πa 2π where φ where φ is the angle subtended by the planet at the centre of the star during transit (over time t time t). ).

8

⇒

⇒

1.0


Page 23 of 27

4

Rs + Rp

φ14

Rs − Rp

a

3

l14

φ23

l23

bRs 1

2

3

l14 /2

4

1

2

l23 /2

(l23 /2)2 = (Rs l23 = 2Rs

− Rp)2 − (bRs)2

 − (1

sin(φ sin(φ23 /2) =

Rp /Rs )2

− b2

2.0

l23 /2 a

2.0

     −  −        − −       − 

sin ⇒ φ23 = 2 sin

l23 Rs = 2 sin sin−1 2a a

−1

P P Rs tF = φ23 = sin−1 2π π a Similarly,

P −1 Rs tT = sin π a

1

Rp Rs

1

Rp Rs

2

b2

1.0

2

b2

1.0

2

R p 1+ Rs

b2

1.0

(T12.6) In the approxim approximatio ation n of an orbit orbit much much larger larger than the stella stellarr radius, radius, show show that the parameter b parameter b is given by

  

b = 1 + ∆

−2

  −

1+

√

∆ 1

Solution: Since R Since R s a, use sin−1 x



   

tT

≈

P Rs π a

tF

≈

P Rs π a

   

tF tT tF tT

−

1/2

    2

≈ x.

R p 1+ Rs

1

2

Rp Rs

5

2.0

 −   −  2

b2

2

b2


Page 24 of 27

√

Dividing, Dividing, and putting R putting R p /Rs = tF = tT

⇒ b =

√ (1 − ∆)2 − b2 √ (1 + ∆)2 − b2

   

(1

−

√

∆)2

∆,

1/2

   −   − tF tT tF tT

1

1.0

2

(1 + 2

√

1/2

∆)2

  

  

= 1+∆

√

−2

  −

1+

∆ 1

tF tT tF tT

2

1/2

    2

2.0

Expres Expression sionss lackin lacking g the use of appro approxima ximation tion Rs /a 1, but otherwis otherwise e correct will get a penalty of 2.0. Use of appro approxima ximation tion with proper justifi justificati cation on at a later later stage stage than than at the first step will get full credit.



(T12.7) Use the result of part (T12.6) to obtain an expression for the ratio a/Rs in terms of measurable measurable transit parameters, parameters, using a suitable suitable approximati approximation. on.

3

Solution:

   

P Rs tT = π a

R p 1+ Rs

 −  2

b2

1.0

Either substitution of b or elimination of b gets 1.0. Substituting b Substituting b and R p /Rs ,

  

P Rs tT = π a

   √ − −  √      − 

⇒ tT = P π Ras ⇒

(1 +

∆)2

1

1/2

√

  −

1+

∆+2 ∆ 1

tF tT tF tT

     2

2

4 ∆

1

tF tT

2

a 2P ∆ P ∆1/4 = 2 2 )1/2 Rs π (tT tF

−

2.0

Expres Expression sionss lackin lacking g the use of appro approxima ximation tion Rs /a 1, but otherwis otherwise e correct will get a penalty of 1.0. If penalty has already been imposed in part (T12.6), no further penalty for lack of approximation.



(T12.8) Combine the results of the wobble method and the transit method to determine the M s stellar mean density ρ density ρ s in terms of t of t T , t F , ∆ and P and P .. 4πR s3 /3

≡

6


Page 25 of 27

Solution: From part (T12.7) 2P ∆ P ∆1/4 a = R = R s 2 2 )1/2 π (tT tF

−

From part (T12.1) a =



GM s 2 P 4π 2

1/3



Combining,



3

2P ∆ P ∆1/4

GM s 2 P = Rs 2 4π 2 π (tT

− tF2 )1/2



3.0

Ident Identifyi ifying ng the two two equatio equations ns for combinin combining g gets gets credit credit.. No credit credit for writing only one equation or irrelevant equations. 2

3

3/4

M s 3 4π 8P ∆ = ⇒ ρs ≡ 4πR 3 /3 4π P 2 G π 3 (t2 − t2 )3/2 s

⇒

T

2.0

F

24 P (∆) P (∆)3/4 ρs = 2 2 2 )3/2 π G (tT tF

1.0

−

Rocky or gaseous: Let us consider an edge-on (i ( i = 90◦ ) star-planet system (circular orbit for the planet), as seen from from the the Earth Earth.. It is know known n that that the host star star is of ma mass ss 1. 1 .00 00M M  . Transits ransits are observed observed with a period (P (P )) of 50.0 days and total transit duration (t ( tT ) of 1.0 1.00 0 hour. The transi transitt depth (∆) is 0.0064. The same system is also observed in the wobble method to have a maximum line of sight velocity of 0. 0.400ms−1 . (T12.9) Find the orbital radius a radius a of the planet in units of AU and in metres.

2

Solution: From Kepler’s third law (with same mass of host star): a = a⊕ a =

2/3

    P P ⊕

50 50..0 365..242 365

= 0.266

1.0 2/3

× 1 AU =

× 1.496 × 1011 m =

0.266AU 3.97

0.5

× 1010 m

0.5

(T12.10) Find the ratio t ratio t F /tT of the system.

2

Solution: Edge-on b = 0

⇒

tF = tT



√ (1 − ∆)2 − b2 √ (1 + ∆)2 − b2

1.0



1/2

√ − √ ∆ =

1 = 1+

∆

0.851 0.85199

1.0


Page 26 of 27 8

(T12.11) Obtain the mass M mass M p and radius R radius R p of the planet in terms of the mass (M ( M ⊕ ) and radius (R⊕ ) of the Earth Earth respect respectiv ively ely.. Is the com composi positio tion n of the planet planet likely likely to be rocky rocky or gaseous? Tick the box for ROCKY or GASEOUS in the Summary Answersheet. Solution: From parts (T12.1) and (T12.9)





10 11 × 1.989 × 1030 × vp = = 57 57..798kms 1 10 3.97 × 10 Assumption of small planet (M (M p  M s ) is valid because ∆ is very small; less dense GM  = a

−

6.674

−

1.0

planet would make the assumption stronger!

M s v0 1.989 1030 0.400 M p = = M ⊕ = 2.30 2.30 M ⊕ vp 5.7798 104 5.972 1024

×

From part (T12.4),

×

×

×

2.0

×

√

Rp = R = R s ∆ From part (T12.7), a 2P ∆ P ∆1/4 2P ∆ P ∆1/4 = = 2 2 )1/2 Rs π (tT tF πt T (1 (tF /tT )2 )1/2

−

∴

R s =

−

aπt T (1

− (tF/tT)2)1/2

2P ∆ P ∆1/4

Combining, Rp =

aπt T (1

− (tF/tT)2)1/2∆1/2

2P ∆ P ∆1/4 aπt T (1 (tF /tT )2 )1/2 ∆1/4 = 2P 1 10 3.97 10 π 24 (1 0.85192 )1/2 = 2 50 50..0 6.371 106 = 1.21 1.21 R⊕

−

×

× × × − × ×

×

2.0 (0.0064)1/4 × (0. R

⊕

1.0

Mean density ρp =

M p 2.30 = ρ⊕ = 1.3ρ⊕ 4πR p3 /3 (1. (1.21)3

Since mean mean density density is higher than that of Earth, the planet is Rocky .

1.0 1.0

Transit light curves with starspots and limb darkening: (T12.12) Consider a planetary transit with i = 90◦ around a star which has a starspot on its equator, comparable to the size of the planet, Rp . The The rotati rotation on period period of the the star is 2P . P . Draw schematic schematic diagrams of the transit light light curve for five successiv successivee transits of the planet planet (in the templates templates provided provided in the Summary Answershee Answersheet). t). The no-transit no-transit flux for each each transit may be normalised normalised to unity unity independently independently.. Assume Assume that the planet does not encounter the starspot on the first transit but does in the second.

4


Page 27 of 27

Solution: f

Transit no. 1

f

Transit no. 2

t

f

Transit no. 3

t

f

Transit no. 4

t

f

t

Transit no. 5

t •

No change change in the first: first: 0.5

•

Spike Spike in second second (width (width and phase of spike spike is arbitr arbitrary) ary):: 1.0

•

Heigh Heightt of spike (almost) (almost) equal equal to maximum maximum dip: 0.5

•

No change change in third: third: 0.5

•

Spike again again in fourth: fourth: 0.5

•

Same phase of spike spike in second second and fourth: fourth: 0.5

•

No change in 5th: 0.5

(T12.13) Throughout Throughout the problem problem we have have considered considered a uniformly uniformly bright stellar disc. Howeve However, r, real stellar discs have have limb darkening. darkening. Draw a schematic schematic transit light light curve when limb darkening is present in the host star. Solution: f

t

Non-flat Non-flat bottom with central central minimum minimum gets 2.0. Curvature Curvature of ingress and egress are tolerated.

2

Data Analysis Examination Page 1 of 4 (D1)

Binary Pulsar Through systematic searches during the past decades, astronomers have found a large number of millisecond pulsars (spin period ). Majority of these pulsars are found in binaries, with nearly circular orbits.

< 10 ms ms

() ( )  (0 ≤  ≤ 2 )  where  =  

For a pulsar in a binary orbit, the measured pulsar spin period and the measured line-of-sight acceleration both vary systematical system atically ly due to orbital motion. mot ion. For circular circular orbits, this variation can c an be described mathematicall math ematically y in terms of orbital orbital phase ph ase as,

()

() =  + t   () = t  

where  is is the th e orbital period of the binary, of the orbit.

 4 where t =  

 is the



intrinsic intrinsic spin period period of the pulsar pulsar and is is the radius radius





The Th e following following table gives gives one such set of measurements of o f and at different heliocentric heliocentric epochs, expressed in truncated Modified Julian Julian Days (tMJD), i.e. number of days since MJD . No.

T (tMJD)

P (μs)

a (m s -2)

1 2 3 4 5 6 7

5740.654 5740.703 5746.100 5746.675 5981.811 5983.932 6005.893

7587.8889 7587.8334 7588.4100 7588.5810 7587.8836 7587.8552 7589.1029

- 0.92 ± 0.08 - 0.24 ± 0.08 - 1.68 ± 0.04 + 1.67 ± 0.06 + 0.72 ± 0.06 - 0.44 ± 0.08 + 0.52 ± 0.08

8 9

6040.857 6335.904

7589.1350 7589.1358

+ 0.00 ± 0.04 + 0.00 ± 0.02

()

= 2,44 2,440,0,00 0000

,

()

By plotting as a function of , we can obtain a parametric curve. As evident from the relations above, this curve in the period-acc period-acceleration eleration plane is an ellipse. ellipse.



In this this problem, we estimate estimate the th e intrinsic intrinsic spin period,  , the orbital period, per iod, by an analysis analysis of this data set, assuming assuming a circular circular orbit.

 , and the orbital radius, ,

(D1.1) Plot the data, including including error bars, in the period-acceleration period- acceleration plane (mark your graph as “D1.1”). “D1.1 ”).

7

(D1.2) Draw an ellipse ellipse that appears to be a best fit to the data (on the same graph “D1. 1”).

2

 , t and t, including including error margins. Write expressions for  and  in terms of  , t , t . Calculate approximate value of B and  based on your estimations estimations made in (D1.3), includin g

(D1.3) From the plot, estimate

7

(D1.4)

4

(D1.5)

6

error margins.



(D1.6) Calculate orbital phase, , corresponding corresp onding to the epochs of the following following five observations in the above table table:: data rows 1, 4, 6, 8, 9. (D1.7) Refine the estimate of the orbital period, way:

B , using the results in part (D1.6) in the following 

(D1.7a) First determine the initial initial epoch,  , which corresponds to the nearest epoch of zero orbital phase before the first observation. (D1.7b) The expected time, given by,

4

calc , of the estimated orbital phase angle of each observation is

calc =  +  + 3 ∘   ,

2 7

Data Analysis Examination Page 2 of 4



where n is the number of full cycle of orbital phases phas es that may have elapsed between betwe en  and (or calc ). Estimate Estimate n and calc for each of the five observations in part (D1.6). Note down difference difference − between observed and calc . Enter these calculations calculations in the table given in the Summary Answersheet.



(D1.7c) Plot









OC against  (mark your graph as “D1.7”).

(D1.7d) Determine the refined values values of the initial initial epoch,



4

 ,, and the orbital period,  ,.

(D2) Distance to the Moon Geocentric Geocentric ephemerides ephemerides of the Moon for September 2015 are give given n in the form of a table. table. Each reading was taken at 00:00 UT. Date Sep 01 Sep 02 Sep 03 Sep 04 Sep 05 Sep 06 Sep 07 Sep 08 Sep 09 Sep 10 Sep 11 Sep 12 Sep 13 Sep 14 Sep 15 Sep 16 Sep 17 Sep 18 Sep 19 Sep 20 Sep 21 Sep 22 Sep 23 Sep 24 Sep 25 Sep 26 Sep 27 Sep 28 Sep 29 Sep 30

h 0 1 2 3 4 5 6 7 7 8 9 10 11 11 12 13 14 14 15 16 17 18 19 20 21 22 23 0 1 2

R.A. m 36 33 30 27 23 19 14 7 59 49 37 23 9 54 39 25 11 58 47 38 31 26 22 19 16 14 12 10 9 7

(α) s 46.02 51.34 45.03 28.48 52.28 37.25 19.23 35.58 11.04 0.93 11.42 57.77 41.86 49.80 50.01 11.64 23.13 50.47 54.94 50.31 40.04 15.63 17.51 19.45 55.43 46.33 43.63 48.32 5.89 39.02

Dec. (δ) ' '' 3 6 16.8 7 32 26.1 11 25 31.1 14 32 4.3 16 43 18.2 17 55 4.4 18 7 26.6 17 23 55.6 15 50 33.0 13 34 55.6 10 45 27.7 7 30 47.7 3 59 28.8 0 19 50.2 -3 20 3.7 -6 52 18.8 -10 9 4.4 -13 2 24.7 -15 24 14.6 -17 6 22.8 -18 0 52.3 -18 0 41.7 -17 0 50.6 -14 59 38.0 -11 59 59.6 -8 10 18.3 -3 44 28.7 0 58 58.2 5 38 54.3 9 54 16.1

∘

Angular Angular Size (θ) '' 1991.2 1974.0 1950.7 1923.9 1896.3 1869.8 1845.5 1824.3 1806.5 1792.0 1780.6 1772.2 1766.5 1763.7 1763.8 1767.0 1773.8 1784.6 1799.6 1819.1 1843.0 1870.6 1900.9 1931.9 1961.1 1985.5 2002.0 2008.3 2003.6 1988.4

Phase (ϕ) 0.927 0.852 0.759 0.655 0.546 0.438 0.336 0.243 0.163 0.097 0.047 0.015 0.001 0.005 0.026 0.065 0.120 0.189 0.270 0.363 0.463 0.567 0.672 0.772 0.861 0.933 0.981 1.000 0.988 0.947

Elongation Of Moon 148.6∘ W 134.7 ∘ W 121.1∘ W 107.9∘ W 95.2∘ W 82.8∘ W 70.7∘ W 59.0∘ W 47.5∘ W 36.2∘ W 25.1∘ W 14.1∘ W 3.3∘ W 7.8 ∘ E 18.6∘ E 29.5∘ E 40.4∘ E 51.4∘ E 62.5∘ E 73.9∘ E 85.6∘ E 97.6∘ E 110.0 ∘ E 122.8∘ E 136.2∘ E 150.0 ∘ E 164.0∘ E 178.3∘ E 167.4 ∘ W 153.2∘ W

T he composite graphic1 below shows multiple multiple snapshots of the Moon taken at different times times during the total lunar lunar eclipse, eclipse, which occurred occur red in this this month. For each shot, the centre of frame was coinciding coinciding with the central north-south line of o f umbra. For this problem, assume that the observer is at the centre of the Earth and angular size refers to angular diameter diameter of the object / shadow.

1

Credit: NASA’s Scientific Scientific Visualization Visualization Studio

7


(D2.1) In September 2015, apogee of the lunar orbit is closest to New Moon / First First Quarter / Full Moon / T hird Quarter. Tick the correct answer in the Summary Answersheet. No justificat justification ion for your answer is necessary.

3

(D2.2) In September 2015, 20 15, the ascending node of lunar lunar orbit with respect to the eclipti eclipticc is closest closest to New Moon / First First Quarter / Full Moon / T hird Quarter. T ick the correct answer in the Summary Answersheet. Answersheet . No justification justification for your answer is necessary.

4

, of the lunar orbit from the given given data. Estimate the th e angular angular size of the umbra, uba , in terms of the angular size of the Moon, Moon .

(D2.3) Estimate the eccentricity,

4

(D2.4)

8

Show your working wor king on the image image given on the backside of the Summary Answersheet. Answersheet .



=

(D2.5) The angle subtended by the Sun at Earth on the day of the lunar eclipse is known to be un . In the figure below, b elow,   and   are rays coming from diametrically opposite ends en ds of the solar disk. The figure is not to scale.

1915.0′′





Calculate the angular size of the penumbra,

9

pnuba , inin terms of Moon . Assume the observer

to be at the centre of the Earth.



(D2.6) Let Eat be angula angularr size size of the Earth as seen from the centre of the Moon. Calcul Calculate ate the angular angular size of the Moon, Moon , as would be seen from the centre of the Earth on the eclipse day in terms of Eat .

5

(D2.7) Estimate the radius of the Moon,

3

(D2.8)

4



(D2.9)



Moon , in km from the results above. Estimate Estimate the shortest distance, pig , and the farthest distance, apog , to the Moon. Use appropriate data from September 10 to estimate the distance, un , to the Sun from the Earth.

(D3) Type IA Supernovae Supernovae of type Ia are considered very important for the measurements of large extragalactic distances. The T he brightening brightening and subsequent dimming of these explosions explosions follow a characteristic ligh lightt curve, which helps in in identifying identifying these as supernovae superno vae of type Ia. Light Light curves of all all type typ e Ia supernovae superno vae can be fit to the same model lig light ht curve, when they are scaled appropriatel appropr iately. y. In order to achieve this, we first first have to express the t he lig light ht curves in the reference frame fram e of the host galaxy galaxy by taking taking care of the cosmological cosmological stretching stretch ing/dilation /dilation of all all observed time intervals, ob , by a factor of . The Th e time interval interval in the rest frame of the host galaxy galaxy is denoted by gal .

(1 + )

Δ

Δ

10

Data Analysis Examination Page 4 of 4 The Th e rest frame light light curve of a supernova supern ova changes c hanges by two magnitudes magnitudes compared to the peak in in a time interval intervalss by a factor of s (i.e.   after the peak. If we further scale the time interval gal) such that the scaled value of supernov ae, the light light curves turn tur n out to have the same  isis the same for all supernovae, shape. It also turns out that is is related linearly linearly to the absolute absolute magnitude, magnitude, pa , at the peak luminosity

Δ

Δ

Δ =Δ





for the supernova. That is, we can write

 =  +  pa , where  and  are constants. Knowing Know ing the scaling scaling factor, f actor, one can determine absolute magni m agnitudes tudes

of

supernovae at unknown distances distances from the above linea linearr equation. equation.



The Th e table below contains data for three supernovae, including their distance moduli mod uli,, (for the first two), their recession speed, , and their apparent magnitudes, different times. times. The time ob ob , at different pa shows number of days from the date at which the respective supernova reached peak brightness.







Δ ≡  

The Th e observed magnitudes have already already been corrected for interstellar interstellar as well well as atmospheric atmosph eric extinction. extinction. Name μ (mag) cz (km s -1) Δtobs (days) -15.00 -10.00 -5.00 0.00 5.00 10.00 15.00 20.00 25.00 30.00

SN2006TD 34.27 4515 mobs (mag) 19.41 17.48 16.12 15.74 16.06 16.72 17.53 18.08 18.43 18.64

SN2006IS 35.64 9426 mobs (mag) 18.35 17.26 16.42 16.17 16.41 16.82 17.37 17.91 18.39 18.73

SN2005LZ 12060 mobs (mag) 20.18 18.79 17.85 17.58 17.72 18.24 18.98 19.62 20.16 20.48

Δ

(D3.1) Compute values for all three supernovae, supernov ae, and fill fill them in in the given given blank boxes in the gal values data tables on the BACK side of the Summary Answersheet. On a graph paper, plot the points and draw the three light light curves in the rest frame (mark your yo ur graph as “D “ D3.1”).



(D3.2) Take Ta ke the scaling scaling factor, factor ,  , for the supernova factors,  and 3 , for the other two supernovae calculating  for them.



Δ



SN2006IS to be 1.00. Calculate the scaling SN2006TD and SN2005LZ , respective respectively, ly, by

Δ

Δ

(D3.3) Compute the scaled time differences,  , for all three supernovae. Write the values for  in the same data tables tables on the Summary Answersheet. Answersheet. On another graph graph paper, plot the points and draw the 3 light light curves to verify that th they ey now have an identical identical profile (mark your graph as “D3.3”). (D3.4) Calculate Calculate the absolute magnitudes magnitudes at peak brightness,

SN2006IS . Use these values to calculate  and . (D3.5) Calculate Calculate the absolute magnitude magnitude at peak brightness, brightness,

SN2005LZ . 

5

14

pa,, for SN2006TD and pa, , for

6

pa,3 , and distance modulus,  3, for

4

 . Further, estimate

6

(D3.6) Use the distance modulus 3 to estimate estimate the value value of Hubble's Hu bble's constant, consta nt, the characteristic age of the universe, H .



15

Data Analysis Examination Page 1 of 20 (D1) Binary Pulsar Through systematic searches during the past decades, astronomers have found a large number of millisecond pulsars (spin period < 10 ms). ms). Majority Majority of these these pulsars pulsars are found found in binarie binaries, s, with nearly circular orbits. For a pulsar in a binary orbit, the measured pulsar spin period ( P ) P ) and the measured line-of-sight acceleration (a (a) both b oth vary systematicall systematically y due to orbital motion. motion. For circular orbits, this variation ariation can be described mathematically in terms of orbital phase φ (0 ( 0 ≤ φ ≤ 2π ) as, 2πP 0 r P ( P (φ) = P 0 + P t cos φ where P t = cP B 4π 2 r a(φ) = −at sin φ where at = P B2 where P B is the orbital period of the binary, P 0 is the intrinsic spin period of the pulsar and r is the radius of the orbit. The following table gives one such set of measurements of P of P and a and a at at different heliocentric epochs, T , T , expr expres esse sed d in trun trunca cate ted d Modifi Modified ed Juli Julian an Days Days (tMJ (tMJD), D), i.e. number umber of days days sinc sincee MJD MJD = 2,440,000.

No. 1 2 3 4 5 6 7 8 9

T (t ( tMJD) 57 57440.6 0.654 57 57440.7 0.703 57 57446.1 6.100 5746.675 5981.811 59 59883.9 3.932 6005.893 6040.857 6335.904

P (µs) 758 5877.888 8889 758 5877.833 8334 758 5888.410 4100 7588.5810 7587.8836 758 5877.855 8552 7589.1029 7589.1350 7589.1358

a (m s−2 ) −0.92 ± 0.08 −0.24 ± 0.08 −1.68 ± 0.04 +1.67 ± 0.06 +0.72 ± 0.06 −0.44 ± 0.08 +0.52 ± 0.08 +0.00 ± 0.04 +0.00 ± 0.02

By plotting a(φ) as a function of P ( P (φ), we can obtain obtain a paramet parametric ric curve. curve. As evident evident from the relations above, this curve in the period-acceleration plane is an ellipse. In this problem, we estimate the intrinsic spin period, P 0 , the orbital period, P B , and the orbital radius, r, by an analysis of this data set, assuming a circular orbit. (D1.1) Plot the data, including error bars, in the period-acceleration plane (mark your graph as “D1.1”). Solution: Graph Number : D1.1

7


•

Plot uses more than 50% of graph paper: 0.5

•

Axes labels (P ( P and a): 0.5

•

Dimension Dimensionss of axes: axes: 0.5

•

Ticks and values values on axes (or scale written explicitly): explicitly): 0.5

•

Points correctly plotted: Points plotted 9 8 Marks given 4. 0 3. 5

7 3. 0

6 2. 0

5 1. 0

< 5 0

Correctness Correctness of points: deduction deduction of 0.5 for each each wrong point. •

Errorbars Errorbars on points (at least least 5): 1.0

(D1.2) Draw an ellipse that appears to be a best fit to the data (on the same graph “D1.1”). Solution: See above • •

Ellipti Elliptical cal curve curve with with visual visual best b est fit: 1.0 Curve symmetric about a = 0 line: line: 0.5

2


•

Curve symmetric about some value of P (P ≈ 7588 7588..48 48): ): 0.5 7

(D1.3) From the plot, estimate P estimate P 0 , P t and a and a t , including error margins. Solution: Values are determined from lengths of axes of ellipse and mid-point of P -axis. P -axis. Error margins may be determined determined by estimating estimating extreme ellipses covering covering the points with errorbars. errorbars. Any reasonable reasonable method of estimating estimating error margins will be accepted. accepted. 2P t = (1. (1.34 ± 0.04) µs

[(13.4 ± 0.4) cm on graph] graph]

P t = (0. (0.67 ± 0.02) µs

2.0

P 0 = (7588. (7588.48 ± 0.02) µs

2.0

2at = (3. (3.42 ± 0.12)ms−2

[(17. [(17.1 ± 0.6) cm on graph] graph]

3.0

at = (1. (1.71 ± 0.06)ms−2 •

•

•

Marking table: Parameter H al alf credit Mini Minim mum P t (µs) 0. 0.59 δP t (µs) 0 .0 1 P 0 (µs) 7 5 8 8 . 38 δP 0 (µs) 0 .0 1 −2 at (m s ) 1 .6 1 −2 δa t (m s ) 0 .0 4

Full credit Mini Minim mum Maxi Maxim mum 0. 6 3 0 .7 1 0. 0 2 0 .0 4 7 5 8 8 . 43 75 88 . 5 3 0. 0 2 0 .0 4 1. 6 5 1 .7 7 0. 0 5 0 .0 7

Half credit Maxi Maxim mum 0. 7 5 0. 0 5 75 88 . 5 8 0. 0 5 1. 8 1 0. 0 8

Wrong values due to wrong/poor plot/fit in (D1.1) and (D1.2) WILL BE penalised. Error Error estimat estimation ion is based based on graph graph drawin drawing. g. Quoted Quoted values values correcorrespond to the envelope of possible ellipses drawn to include all points with errorbars. errorbars. Any reasonable reasonable method to estimate error to b e given credit.

(D1.4) Write expressions for P B and r and r in terms of P P 0 , P t , a t .

4

Solution: We can easily recover the orbital period (P ( P B ) and the radius of the orbit (r ( r) in a circular orbit: 4π 2 at = 2 r P B P B2 at 4π 2 2πP 0 r P t = × P B c 2πP 0 P B2 at P 0 P B at = = × 2 P B c 4π 2πc

r = ∴ r =

1.0


∴

P B =

P t 2πc P 0 at

   

at r = 4π 2 r =

∴

1.0

P t 2πc P 0 at

P t P 0

2

2

1.0

c2 at

1.0

Alternativ Alternative e algebraic algebraic routes accepted. accepted. Each of P B and r carry 2.0 marks. (D1.5) Calculate approximate value of P B and r based on your estimations made in (D1.3), including including error margins. margins.

6

Solution: P t 2πc P 0 at 0.67 2π × 2.998 × 108 = s × 7588..48 7588 1.71 = 96 96 260 260 s = 1. 1.12570d

P B =

∆P B = P = P B

1.0

              2

∆P t P t

+

0.02 0.67

= 1.12570 ×

2

∆P 0 P 0

+

2

2

∆at at

2

0.02 7588..48 7588

+

+

0.06 1.71

1.0

2

d

= 1.12570 × 0.0461  0.052d ∴

P B = (1. (1.13 ± 0.05)d

r = = r = ∴ r =

2

1.0

c2 at

                 P t P 0

0.67 7588..48 7588

2

2.998 × 108 × 1.71

2

m

4.09739 × 108 m = 2.73890 × 10−3 AU

∆r = r = r

2∆P t 2∆P P t

2

+

−3

= 2.73890 × 10

×

2∆P 0 2∆P P 0

2

+

2 × 0.02 0.67

∆at at 2

+

1.0

2

1.0 2 × 0.02 7588..48 7588

2

2

   +

0.06 1.71

AU

= 2.73890 × 10−3 × 0.06925AU  0.19 × 10−3 AU (2.74 ± 0.19) × 10−3 AU r = (2.

1.0

Errors in P B and r can be also estimated as maximum possible (worst case) error. In such case, errors would be about 1.5 times the standard error calculated above (δP B = 0.07 07,, δr = δr = 0.25). (D1.6) Calculate orbital phase, φ phase, φ,, corresponding to the epochs of the following five observations

4

Data Analysis Examination Page 5 of 20 in the above table: data rows 1, 4, 6, 8, 9. Solution: Using these newly determined orbital parameters, we can calculate the angular orbital phase for each data point, i.e., for each pair of acceleration and period measured ( P , P , a). φ = tan−1



a P t − at P − P 0



Care has to be taken to choose the value of the phase from among φ, φ , π ± φ, 2π − φ, depending on the sign of cos φ and sin φ. Sr. T P a φ −2 no. (t ( tMJD) (µs) (m s ) 1 5740.6 0.654 75 75887.8 7.8889 −0.92 148.62◦ 4 5746.675 7588.5810 +1.67 278.77◦ 6 5983.9 3.932 75 75887.8 7.8552 −0.44 164.57◦ 8 6040.857 7589.1350 +0.00 0.00◦ 9 6335.904 7589.1358 +0.00 0.00◦ • • •

•

Credit Credit for each each correct correct value value:: 1.0 for first three, three, 0.5 for last two. two. Credit for π ± φ or 2π − φ is 0.5 per value. All values wrong due to wrong expression for φ gets a maximum of 1.0 mark. Values in radians accepted.

(D1.7) (D1.7) Refine Refine the estimate estimate of the orbital orbital period, period, P B , using the results in part (D1.6) in the following way: (D1.7a) First determine the initial epoch, T 0 , which corresponds to the nearest epoch of zero phase before the first observation. Solution: T 1 − T 0 φ1 φ1 = = T 1 − P B ⇒ T 0 = T P B 2π 2π 148.62◦ 148. T 0 = 5740. 5740.654 − 125700 tMJD tMJD × 1.1257 360◦ T 0 = 5740. 5740.189 tMJD tMJD

2

1.0

1.0

Tolerance: ±0.002 tMJD. Using P 0 instead of P B gets zero. (D1.7b) The expected time, T calc calc , of the estimated phase of each observation is given by





φ P B , 360◦ where n is the number of full cycle of orbital phases elapsed between T 0 and T calc Estimate n and and T T calc observations in part (D1.6). Note calc . Estimate n calc for each of the five observ down difference difference T O–C Enter these calculatio calculations ns O–C between observed T and T calc calc . Enter in the table given in the Summary Answersheet. T calc = T 0 + n + calc = T

Solution:

7




φ T calc = T 0 + n + calc = T 360◦



P B

where n where n = Intege Integerr part of [(T [(T − T 0 )/P B ]. Sr. T φ n Tcalc T O–C O–C No. (tMJD) (MJD) (days) ◦ 1 5740.654 148.62 0 5740.654 0.000 ◦ 4 5746.675 278.77 5 5746 5746.6 .689 89 −0.014 6 5983.932 164.57◦ 216 5983.855 0.077 ◦ 8 6040.857 0.00 267 6040.751 0.106 ◦ 9 6335.904 0.00 529 6335.684 0.220 Deduction for each wrong/missing value of n, T calc O–C : 0.5 calc and T O–C No double penalty in one row. (D1.7c) Plot T Plot T O–C against n (mark your graph as “D1.7”). O–C against n

4

Solution: Graph Number : D1.7

•

Plot uses more than 50% of graph paper: paper: 0.5

•

Axes labels (T (T O–C dimensions: 0.5 O–C and n) including dimensions:

•

Ticks and values values on axes (or scale written explicitly): explicitly): 0.5

•

Points Points correctly plotted: plotted: 0.5 for each each point

•

Goodness of linear fit credited in next part

(D1.7d) Determine the refined values of the initial epoch, T 0,r , and the orbital period, P B,r B,r .

7


Solution: A linear fit to the plot of T O–C O–C vs n gives the offset of period per cycle (slope) and the shift in the zero-phase point (intercept). This concept, which may be b e evident evident in the subsequent subsequent calculation, gains the credit, explicit statement is not necessary. From a linear fit, Slope = 0. 0.00 0000 43 d/n

Interc terceept = −0.010d

•

Credit Credit for good visual visual linear linear fit: 1.0

•

Correct values values of slope and intercept: intercept: 1.0 each

•

Tolerance: ±0.00002 00002 in in slope and ±0.002 002 in intercept.

T 0,r = 5740. 5740.189 − 0.010 010 = 574 57400.179 tMJD tMJD

2.0

3.0

1.0

P B,r (1.12570 + 0. 0.000 00043) 43) d B,r = (1. = 1.12613d P B = 1.1261d Incorrect sign of correction applied carries penalty of 0.5 for each quantity.

1.0

Data Analysis Examination Page 8 of 20 (D2) Distance to the Moon Geocen Geocentric tric ephemer ephemeride idess of the Moon for Septem September ber 2015 are given given in the form of a table. table. Each Each reading was taken at 00:00 UT. Date Sep Sep Sep Sep Sep Sep Sep Sep Sep Sep Sep Sep Sep Sep Sep Sep Sep Sep Sep Sep Sep Sep Sep Sep Sep Sep Sep Sep Sep Sep

01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

h 0 1 2 3 4 5 6 7 7 8 9 10 11 11 12 13 14 14 15 16 17 18 19 20 21 22 23 0 1 2

R.A. m 36 33 30 27 23 19 14 7 59 49 37 23 9 54 39 25 11 58 47 38 31 26 22 19 16 14 12 10 9 7

(α) s 46.02 51.34 45.03 28.48 52.28 37.25 19.23 35.58 11.04 0.93 11.42 57.77 41.86 49.80 50.01 11.64 23.13 50.47 54.94 50.31 40.04 15.63 17.51 19.45 55.43 46.33 43.63 48.32 5.89 39.02

Dec. (δ )

◦



3 7 11 14 16 17 18 17 15 13 10 7 3 0 -3 -6 -1 -10 -13 -15 -17 -18 -18 -17 -14 -11 -8 -3 0 5 9

6 32 32 25 32 32 43 55 55 7 23 50 34 45 30 30 59 59 19 19 20 20 52 9 2 24 6 0 0 0 59 59 10 44 58 58 38 38 54 54

Angular Size (θ) 



16.8 26.1 26 31.1 4.3 18.2 4.4 26.6 26 55.6 33.0 55.6 27.7 47.7 47 28.8 28 50.2 50 3.7 18.8 4.4 24.7 14.6 22.8 52.3 41.7 50.6 38.0 59.6 18.3 28.7 58.2 58 54.3 54 16.1 16

1991.2 1974.0 1950.7 1923.9 1896.3 1869.8 1845.5 1824.3 1806.5 1792.0 1780.6 1772.2 1766.5 1763.7 1763.8 1767.0 1773.8 1784.6 1799.6 1819.1 1843.0 1870.6 1900.9 1931.9 1961.1 1985.5 2002.0 2008.3 2003.6 1988.4

Phase (φ) 0.927 0.852 0.759 0.655 0.546 0.438 0.336 0.243 0.163 0.097 0.047 0.015 0.001 0.005 0.026 0.065 0.120 0.189 0.270 0.363 0.463 0.567 0.672 0.772 0.861 0.933 0.981 1.000 0.988 0.947

Elon Elonga gati tion on of Moon 148.6◦ W 134.7◦ W 121.1◦ W 107.9◦ W 95.2◦ W 82.8◦ W 70.7◦ W 59.0◦ W 47.5◦ W 36.2◦ W 25.1◦ W 14.1◦ W 3.3◦ W 7.8◦ E 18.6◦ E 29.5◦ E 40.4◦ E 51.4◦ E 62.5◦ E 73.9◦ E 85.6◦ E 97.6◦ E 110.0◦ E 122.8◦ E 136.2◦ E 150.0◦ E 164.0◦ E 178.3◦ E 167.4◦ W 153.2◦ W

Data Analysis Examination Page 9 of 20 The composite graphic1 below shows multiple snapshots of the Moon taken at different times during the total lunar eclipse, which which occurre o ccurred d in this month. month. For each shot, the centre centre of frame was coinciding with the central north-south line of umbra. For this problem, assume that the observer is at the centre of the Earth and angular size refers to angular diameter of the relevant object / shadow.

(D2.1) In September 2015, apogee of the lunar orbit is closest to New Moon / First Quarter / Full Moon / Third Quarter. Tick Tick the correct correct answer in the Summary Answershee Answersheet. t. No justification justification for your answer answer is necessary. Solution: From the table we see that the angular size of Moon is smallest close to the New Moon day day. Thus, Thus, the answe answerr is New Moon . Justification is NOT necessary for full credit.

3

3.0

(D2.2) In September 2015, the ascending node of lunar orbit with respect to the ecliptic is closest to New Moon / First Quarter / Full Moon / Third Quarter. Tick Tick the correct correct answer in the Summary Answershee Answersheet. t. No justification justification for your answer answer is necessary. Solution: As there is an eclipse happening in this month, the lunar nodes are close to Full Moon day and New Moon day. day. Next Next we notice notice that lowest lowest declinat declination ion of Moon is ◦ just 18 . This This means that after after the New Moon day, day, the orbit orbit of Moon is above the eclipt ecliptic. ic. In other words words,, the ascendin ascending g node is near the New Moon . Justification is NOT necessary for full credit.

4

4.0

(D2.3) Estiamte the eccentricity, e, e , of the lunar orbit from the given data. Solution: The largest angular size of the Moon in the ephemerides is 2008. 2008 .3 and the smallest angular size is 1763. 1763.7 . The distance distance is inve inverse rsely ly proportional proportional to the angular angular size. size. Hence ratio of distance at perigee to the distance at apogee is: rperigee a0 1−e  Ratio = =  × rapogee 1+e a0   1

Credit: NASA’s Scientific Visualization Visualization Studio

4

2.0


∴

1 − e 1763 1763..7 = = 0.87821 1+e 2008.3 2008. 1 − 0.87821 e = = 0.064846 ∴ e = 1 + 0. 0 .87821 e  0.065

2.0

Rounding off is done to account for the fact that our data are not continuous, hence exact angular sizes at perigee and apogee are not known. A non-rounded answer will also receive full credit. (D2.4) Estimate the angular size of the umbra, θumbra , in terms of the angular size of the Moon, θMoon . Show Show your your working working on the image given given on the backsi backside de of the Summary Summary Answersheet. Solution: The following construction construction is shown. shown.

Only two chords are necessary to determine the centre. The credit is divided in two parts for the drawing: •

•

Realisation that centre of umbra circle needs to be determined to find θumbra : 1.5 Accurate determination of centre of umbra circle by geometric construction: struction: 2.5 Determination Determination of centre centre of hand-draw hand-drawn n circle: maximum maximum 1.5

8

5.0


•

Measuri Measuring ng diameters diameters of umbra umbra and Moon: 0.5 each

By estimating approximate centre of the shadow in the image, we find out, θumbra dumbra = θMoon dMoon 9 .1 = = 2.76 3 .3 76θθMoon ∴ θumbra  2.76

2.0

1.0

Acceptable range: ±0.10 10.. (D2.5) The angle subtended by the Sun at Earth on the day of the lunar eclipse is known to be θSun = 1915. 1915.0 . In the figure below, S 1 R1 and S and S 2 R2 are rays coming from diametrically opposite ends of the solar disk. The figure is not to scale. S 1

R2

θSun

R1

S 2 Earth

Moon

Calculate the angular size of the penumbra, θpenumbra , in terms of θMoon . Assu Assume me the the observer to be at the centre of the Earth. Solution: The following diagram needs to be drawn. A

P

θ1 θ2

B J

Q

Moon’s path Angular size of umbra is θ is θ umbra = 2BOQ BO Q Angular size of penumbra is θ is θ penumbra = 2AOQ We have QA = QA = QP QP + P A = OE + OE + P E tan E tan θ1

E R⊕ O

S 1 θSun

S 2

9


R⊕ + dMoonθ1 θSun ≈ R⊕ + dMoon 2 ≈

since P since P A  P E since θ since θ 1 ≈ θ2 ≈ θSun /2

2.0

and QB = QB = QP − P B = OE − P E tan E tan θ2 R⊕ − dMoonθ2 θSun ≈ R⊕ − dMoon 2 ≈

∴ θ penumbra =

2.0

2AOQ = AOQ = 2 tan tan−1 R⊕ + dMoon

=2

dMoon

  QA OQ

≈

2

QA since Q since QA A  OQ OQ

θSun 2 = 2R⊕ + θ Sun dMoon

1.0

and −1

θumbra = 2BOQ BO Q = 2 tan tan

=2

R⊕ − dMoon dMoon

  QB OQ

≈

2

QB OQ

θSun 2 = 2R⊕ − θ Sun dMoon

1.0

Subtracting, θpenumbra − θumbra = 2θSun ⇒ θpenumbra = θ = θumbra + 2θ 2 θSun

1.0

We have, θumbra = 2.76 76θθMoon

and θ and θ Sun = 1915. 1915.0

From the given data, θ data, θ Moon= 2008. 2008.3 . Therefore, 1915..0 1915 θpenumbra = 2.76 76θθMoon + 2 θMoon 2008..3 2008 θpenumbra = 4.67θ 4.67θMoon Acceptable Acceptable range: 4.57θ 4.57θMoon to 4.77θ 4.77θMoon .

Alternative solution:

In the figure below, rays H EA and I F C C are coming from one edge of solar disk and rays H F D and GEB are coming from the opposite opposite edge. The observ observer er (O ( O) is assumed to be at the centre of the Earth. The Moon travels along the path ABCD during the course of eclipse.

1.0

1.0


G

A θSun

B J

E

θSun

O

θSun

C

θSun

F

H θSun

D

I

Moon’s path A θSun

E

B J

O

θSun

C θSun

F

D Moon’s path From figure, =  EJ F = θ Sun AEB =  GEH = H F I =  DF C = θumbra =  BOC BO C = = 2.76 76θθMoon θpenumbra =  AOD

3.0 1.0 1.0

AOD =  AOB + AOB + BOC BO C + + C OD AOD = AOB =  AEB AOB =

1.0

OD =  C F D C OD =

1.0

θpenumbra  AEB + AEB + θumbra + C F D = 2θSun + 2. 2 .76 76θθMoon = 2 × 1915 1915..0 + 2. 2.76 × 2008 2008..3 θpenumbra = 9372. 9372.9 = 4.67 67θθMoon

2.0

(D2.6) Let θEarth be angular angular size of the Earth as seen seen from the centre centre of the Moon. Calcul Calculate ate the angular size of the Moon, θMoon, as would be seen from the centre of the Earth on the eclipse day in terms of θ of θ Earth .

5

Solution: From the Moon, 2R⊕ θEarth = dMoon

1.0


From part (D2.5), θumbra + θpenumbra = 2θEarth ∴ θ Earth =

2.0

θumbra + θpenumbra 2.76 + 4. 4.67 = θMoon = 3.72 72θθMoon 2 2

0.269θEarth θMoon= = 0.269θ

2.0

Alternative solution: Let us say that the Moon is at position of B . Thus, Thus, angular angular size size of Earth Earth as seen seen from this position will be, (see figure in the previous part) θEarth =  EB F =  BF D

2.0

=  BF C + + C F D  θumbra + θSun

1.0

The angular size of the Full Moon on 28 September as seen in the table is 2008 .3 . θEarth = 2.76 × 2008 2008..3 + 1915. 1915.0 = 7453. 7453.0 θMoon = 0.269 269θθEarth =

θEarth 3.72

2.0

(D2.7) Estimate the radius of the Moon, R Moon , in km from the results above. Solution: Thus, the radius of Moon will be, R⊕ RMoon = 3.72 6371 RMoon = 3.72 RMoon  171 17133 km

3

1.0 2.0

Acceptable range: ±20km 20km.. (D2.8) Estimate Estimate the shortest shortest distance, distance, r r perigee , and the farthest distance, r apogee , to the Moon.

4

Solution: The shortest and longest distances will be, 2 × 1713 × 206265 rperigee = 2008..3 2008 rperigee = 3.52 × 105 km rapogee =

2.0

2 × 1713 × 206265 1763..7 1763

rapogee = 4.01 × 105 km (D2.9) Use appropriate data from September 10 to estimate the distance, d distance, d Sun, to the Sun from the Earth.

2.0

10


Solution: Moon S Sun

φM φS

M E Earth

φE

Moon’s orbit On September 10, phase of Moon is 0. 0.097 and elongation of Moon is 36. 36.2◦ . Angular size of the Moon on this day is 1792. 1792 .0 . Therefo Therefore, re, distance distance to Moon (from Earth) on September 10 is 2 × 1713 × 206265 dMoon,10 = 1792..0 1792 = 3.94 × 105 km Let

2.0

= φ M EM S = ES M = φ S SE M = φ E

36 36..2◦ 1 + cos φM phas phasee = 2 −1 (2 × phase − 1) ∴ φ M = cos ∴ φ E =

1.0 2.0

= cos−1 (2 × 0.097 − 1) = cos−1 (−0.806) = 143. 143.71◦

2.0

φS = 180◦ − φE − φM = 180◦ − 36 36..2◦ − 143 143..71◦ = 0.09◦ Now using sine rule, dSun sin φM = dMoon,10 sin φS 3.94 × 108 × sin143. sin143.71◦ ∴ d Sun = sin0. sin0.09◦ dSun = 1.48 × 1011 m

1.0

2.0

1.0

Data Analysis Examination Page 16 of 20 (D3) Type IA Supernovae Supernovae of type Ia are considered very important for the measurements of large extragalactic distances. distances. The brightening brightening and subsequen subsequentt dimming dimming of these explosions explosions follow a characteris characteristic tic light curve, which helps in identifying these as supernovae of type Ia. Light curves of all type Ia supernovae can be fit to the same model light curve, when they are scaled scaled appropria appropriatel tely y. In order order to achie achieve ve this, we first have have to expres expresss the light light curve curvess in the reference frame of the host galaxy by taking care of the cosmological stretching/dilation of all observed time intervals, ∆t ∆ tobs , by a factor of (1 + z + z). ). The time inter interv val in the rest rest frame frame of the host galaxy is denoted by ∆t ∆ tgal . The rest frame light curve of a supernova changes by two magnitudes compared to the peak in a time interval ∆t ∆t0 after after the peak. If we further further scale scale the time interv intervals by a factor factor of s (i.e. ∆ts = s∆tgal ) such that the scaled value of ∆t ∆ t0 is the same for all supernovae, the light curves turn turn out to hav have the the same same shape. shape. It also turns turns out that s is related linearly to the absolute magnitude, M magnitude, M peak peak , at the peak luminosity for the supernova. That is, we can write s = a = a + bM peak peak , where a where a and b are constants. Knowing the scaling factor, one can determine absolute magnitudes of supernovae at unknown distances from the above linear equation. The table below contains data for three supernovae, including their distance moduli, µ (for the first two), their recession recession speed, cz, cz , and their apparent magnitudes, m magnitudes, m obs , at different times. The time ∆t ∆tobs ≡ t − tpeak shows number of days from the date at which the respective supernova reached reached peak brightness brightness.. The observed observed magnitudes magnitudes have already been corrected corrected for interstel interstellar lar as well as atmospheric extinction. Name SN2006TD SN2006IS SN2005LZ µ (mag) 34.27 35.64 cz (kms−1 ) 4515 9426 12060 ∆tobs (days) mobs (mag) mobs (mag) mobs (mag) 15..00 19.41 18.35 20.18 −15 10..00 17.48 17.26 18.79 −10 16.12 16.42 17.85 −5.00 0.00 15.74 16.17 17.58 5.00 16.06 16.41 17.72 10 10..00 16.72 16.82 18.24 15 15..00 17.53 17.37 18.98 20 20..00 18.08 17.91 19.62 25 25..00 18.43 18.39 20.16 30 30..00 18.64 18.73 20.48 (D3.1) Compute ∆t ∆tgal values for all three supernovae, and fill them in the given blank boxes in the data tables on the BACK side of the Summary Answershe Answersheet. et. On a graph paper, plot the points and draw the three light curves in the rest frame (mark your graph as “D3.1”).

15

Solution: Redshifts for the three supernovae are z are z 1 = 0.0151, 0151, z z 2 = 0.0314 and z and z 3 = 0.0402.

1.5

Filling in the three tables (∆t (∆ tgal , third column)

3.5


SN2006TD

SN2006IS

SN2005LZ

∆tobs

mobs

∆tgal

∆ts

∆tobs

mobs

∆tgal

∆ts

∆tobs

mobs

∆tgal

∆ts

(d)

(mag)

(d)

(d)

(d)

(mag)

(d)

(d)

(d)

(mag)

(d)

(d)

−15.00

19.41

−14.78

−20.00

−15.00

18.35

−14.54

−14.54

−15.00

20.18

−14.42

−17.03

−10.00

17.48

−9.85

−13.34

−10.00

17.26

−9.70

−9.70

−10.00

18.79

−9.61

−11.35

−5.00

16.12

−4.93

−6.67

−5.00

16.42

−4.85

−4.85

−5.00

17.85

−4.81

−5.68

0.00

15.74

0.00

0.00

0.00

16.17

0.00

0.00

0.00

17.58

0.00

0.00

5.00

16.06

4.93

6.67

5.00

16.41

4.85

4.85

5.00

17.72

4.81

5.68

10.00

16.72

9.85

13.34

10.00

16.82

9.70

9.70

10.00

18.24

9.61

11.35

15.00

17.53

14.78

20.00

15.00

17.37

14.54

14.54

15.00

18.98

14.42

17.03

20.00

18.08

19.70

26.67

20.00

17.91

19.39

19.39

20.00

19.62

19.23

22.70

25.00

18.43

24.63

33.34

25.00

18.39

24.24

24.24

25.00

20.16

24.03

28.38

30.00

18.64

29.56

40.01

30.00

18.73

29.09

29.09

30.00

20.48

28.84

34.06

Full marks of 3.5 for all correct values. Penalty for incorrect values (3 ×7 independent values): Incorrect

1-3

4-6

7-9

10-12

13-15

16-18

19-21

Deduction

0.5

1.0

1.5

2.0

2.5

3.0

3.5

The light curves in galaxy frame would appear as follows Graph Number: D3.1

•


10.0


• • • •

•

Both axes labels (∆ (∆tgal and mobs ) present present:: 0.5 Both Both dimens dimensions ions of axes axes (days (days and mag) presen present: t: 0.5 Ticks and values values on axes (or scale written explicitly): explicitly): 0.5 Points correctly plotted: All points correctly correctly plotted: 5.0 Penalty for incorrect or missing points: Incorrect

1

2-4

5-7

8-10

11-13

14-16

17-19

20-22

23-25

26-30

Deduction

0

0 .5 0.

1 .0 1.

1.5

2.0

2.5

3.0

3.5

4.0

5.0

Smooth Smooth curve curve through through points: points: 1.0 per curve curve

(D3.2) Take the scaling factor, s factor, s 2 , for the supernova SN2006IS to be 1. 1.00. Calculate the scaling factors, s factors, s 1 and s and s 3 , for the other two supernovae SN2006TD and SN 2005LZ, respectively, by calculating ∆t ∆t0 for them. Solution: From the graph D3.1, SN2006IS took 22. 22.0 d to fade fade by 2 magnitu magnitudes des.. That is, ∆t ∆t0 (SN2006IS) = 22. 22.0 d. Similarly, ∆t ∆t0 (SN2006TD) (SN2006TD) = 16. 16.4 d. And ∆t ∆t0 (SN2005LZ) = 18. 18.8 d. Acceptable range: ±1.0 days Thus, stretching factors for these two supernovae are s1 =

22 22..2 = 1.354 16 16..4

s3 =

22 22..2 = 1.181 18 18..8

5

3. 0

2.0

(D3.3) Compute the scaled time differences, ∆t ∆ ts , for all three supernovae. Write the values for ∆ts in the same data tables on the Summary Summary Answers Answershee heet. t. On anothe anotherr graph graph paper, paper, plot the points and draw 3 light curves to verify that they now have an identical profile (mark your graph as “D3.3”).

14

Solution: Filling the scaled values in the fourth column of the table (∆ ts in table above) Full marks of 3.5 for all correct values. Penalty for incorrect values (3 ×7 independent values): Incorrect

1-3

4-6

7-9

10-12

13-15

16-18

19-21

Deduction

0.5

1.0

1.5

2.0

2.5

3.0

3.5

The scaled light curves would appear as follows, Graph Number: D3.3

3.5


10.5 •


•

Both axes labels (∆ (∆ts and mobs ) present: present: 0.5

• • •

Both Both dimens dimensions ions of axes axes (days (days and mag) presen present: t: 0.5 Ticks and values values on axes (or scale written explicitly): explicitly): 0.5 Points correctly plotted: All points correctly correctly plotted: 5.0 Penalty for incorrect or missing points: Incorrect

1

2-4

5-7

8-10

11-13

14-16

17-19

20-22

23-25

26-30

Deduction

0

0 .5 0.

1 .0 1.

1.5

2.0

2.5

3.0

3.5

4.0

5.0

•

Smooth Smooth curve curve through through points: points: 1.0 per curve curve

•

The curves should show identical profiles.

0. 5

(D3.4) Calcul Calculate ate the absolu absolute te mag magnit nitude udess at peak brigh brightne tness, ss, M peak,1 SN2006TD TD and and peak,1 , for SN2006 M peak,2 b . peak,2 , for SN2006IS. Use these values to calculate a and b.

6

Solution: To get a get a and a nd b b,, M peak,1 = m peak,1 − µ1 = 15 15..74 − 34 34..27mag peak,1 = m = −18 18..53mag M peak,2 = m peak,2 − µ2 = 16 16..17 − 35 35..64mag peak,2 = m = −19 19..47mag s 1 − s2 1.354 − 1 0.354 b = = mag −1 = mag−1 ∴ b = M peak,1 18..53 − (−19 19..47) 0.94 −18 peak,1 − M peak,2 peak,2

2.0


b = 0.376 37622 mag−1

2.0

a = s = s 2 − bM peak,2 19..47) = 1 + 7. 7.325 peak,2 = 1 − 0.3762 × (−19 a = 8.325

2.0

No penalty for missing mag−1 in b. (D3.5) Calculate the absolute magnitude at peak brightness, M peak,3 modulus, µ 3 , peak,3 , and distance modulus, µ for SN2005LZ. SN2005LZ.

4

Solution: s3 = a = a + bM peak,3 peak,3 s3 − a 1.181 − 8.325 −7.144 = mag = mag ∴ M peak,3 peak,3 = b 0.3762 0.3762 M peak,3 18..99mag peak,3 = −18

2.0

Distance modulus to SN2005LZ is µ3 = m = m peak,3 − M peak,3 17..58 − (−18 18..99) ma magg peak,3 = 17 µ3 = 36 36..57mag

2.0

(D3.6) Use the distance modulus µ3 to estimate the value of Hubble’s constant, H 0 . Further, urther, estimate the characteristic age of the universe, T H .

6

Solution: Distance to SN2005LZ is µ3 µ3 d = 10( 5 +1) pc = 10( 5 +1−6) Mpc 3

36.57 = 10( 5 −5) Mpc = 102.314 Mpc

206Mpc cz3 12060 H 0 = = km s−1 Mpc−1 d3 206 

58..5 k m s−1 Mpc−1 H 0 = 58

4.0

1 3.086 × 1022 = yr H 0 58 58..5 × 103 × 3.156 × 107 T H = 16 16..7Gyr

2.0

T H =

Extra factor of 2/3 allowed in the value of T H .

Observatio Observ ational nal Examinati Ex amination on (OM) Page 1 of 1

Instructions

(1)

Total duration of the examination examination is

(2)

Your answersheet consists of an A3 size sheet on which a rectangular sky map (Mercator Projection) is printed. printed. In Mercator projection projection the circles circles of constant R.A. and circle circless of constant declina declination tion appear as straight lines. lines. Please make sure that you write your contestant code cod e on the map sheet in in the box provided for the purpose.

(3)

Some inaccuracy inaccuracy in marking of a position will will be tolerated, but the contestant will will only get partial credit.

(4)

After the examination, examination, fold your map, put it back inside inside the envelope and an d hand it it over to the volunteers.

(5)

Markings with w ith incorrec incorrect/missing t/missing codes will will not get get any marks.

30 minutes.

Observational Examinati Ex amination on (OM) Page 1 of 1

(OM1) Mark any 5 (five) (five) of the following following stars on the map by putting a circle circle (O) around the appropriate app ropriate star s tar and writing its code next next to it. If I f you mark more than 5 stars, only on ly the first first 5 in in serial ser ial order will will be considered considered . Code Co de S1 S2 S3 S4

Name Caph Asellus Australis Austr alis Acrux Alphard

Bayer Name β Cas δ Cnc α Cru α Hya

Code Co de S5 S6 S7 S8

Name Sheliak Albireo Albireo Rasalhague Kaus Australis Australis

Bayer Name β Lyr β Cyg α Oph ϵ Sgr

(OM2) Mark location of any 3 (three) of the following following galaxi galaxies es on the map by putting a ‘+’ sign sign at appropriate appropriate place place in the map and writing writing its code next next to it. If you mark more than 3 galaxi galaxies, es, only the first 3 in serial order will be considered. Code G1 G2 G3

Name T riangulum riangulum Galaxy Galaxy Whirlpool Galaxy Galaxy Southern Souther n Pinwheel Galaxy Galaxy

M number M 33 M 51 M 83

Code G4 G5

20

Name Virgo Virgo A Sombrero Sombrer o Galaxy Galaxy

15

M number M 87 M 104

(OM3) Draw ecliptic ecliptic on the map and label it as ‘E’.

5

(OM4) Show position of Autumnal Equinox (descending (descend ing node nod e of the ecliptic) ecliptic) on the map by a ‘+’ sign sign and label it as ‘A’.

5

(OM5) Draw local meridian for Bhubaneswar on Winter Solstice day (22 nd December) December ) at local local midnight midnight and labe labell it as ‘M’.

5

Observational Examinat Ex amination ion (OP) Page 1 of 1

Instructions A.

Logistical (1) T otal duration of the examination examination is

30 minutes.

(2) After you take your designated designated seat inside inside planetarium dome, you will will be given 5 minutes to read all the questions. P lanetarium will be turned on only after this time. You will will be given given 5 minutes minutes for familiarisatio familiarisation n with the sky after it it has been turned on. (3) At your seat you will will find a writing writing board, a torch and a pen. Please use this pen to mark final answe rs on the maps provided (Map 1 and Map 2). Answers marked by other writing writing instruments will will not be considered. (4) At any time time during the examinati examination, on, you may stand up at your place place or turn around. However, you are not allowed allowed to move out of your place. (5) You must maintain strict silence throughout the examination. (6) After your examination examination is over, keep writing board, torch torc h and pen at the same place. place. Fold Fo ld your map, put it back inside inside the envelope envelope and hand it over to the volunteers. B.

Academic (1) Your Summary Answersheet consists consists of an A4 A4 size size sheet on which which sky maps are printed printed on both sides (Map 1 and Map 2). Please make sure that you write your contestant code on both sides of the sheet in the boxes provided for the purpose. (2) To mark the answers put a question next to it.

‘

+

’

sign at appropriate place on the map and write code as given in the

(3) Some inaccuracy inaccuracy in marking of a position will will be b e tolerated, tolerated, with an appropriate penalty. (4) Start of a new part of each question will be announced. announced . Simultaneously, Simultaneously, a corresponding corresp onding code will wi ll be displayed displayed on the dome. Please pay attention. attention.

Observational Examinat Ex amination ion (OP) Page 1 of 1 (OP1)

Eight Eight well w ell known historical h istorical supernovae super novae will will appear ap pear in the projected sky one at a time time (not ( not necessarily in chronologic chronological al order). You have to identify identify the appropriate map (Map 1 / Map 2) where a particular particular supernova superno va belongs and mark it in in the corresponding map with ‘+’ sign sign and write codes ‘S1’ to ‘ S8’ besides besides it. Each supernova supern ova code will will be projected proj ected on dome for 10 seconds, followed followed by appearance appearance of supernova for 60 seconds and then 20 seconds for you to mark the answers.

40

(OP1.1) For S1, S1, S2, S3, S3, S4 and S5, the projected sky corresponds to the sky as seen from Rio de Janeiro on the midnight midnight of 21st May. (OP1.2) For S6,S7 and S8, the projected sky corresponds to the sky as seen from Beijing Beijing on the midnigh midnigh t th of 20 November. There T here will will be b e a gap of two minute after aft er S5 for change over and adaptation to new sky. (OP2) We are now projecting sky of another planet. The T he sky will will be slowly slowly rotated for fo r 5 minutes . Identify Ident ify the visible visible celestial pole p ole of this planet and mark ma rk it it with a ‘+’ sign sign and label it as ‘ P’ on the appropriate appropriate map (Map 1 / Map 2).

10

Observational Examinatio E xamination n (OT ( OT)) Page 1 of 1

Instructions

(1) Total duration of the examination examination is 25 minutes. minutes. There Th ere will will be a bell every five minutes. m inutes. (2) T he examiner examiner at your station will will give give you a writing board, bo ard, a torch and pen. Please use only the given given pen to mark final answers on the map provided in the Summary Answersheet. Answers marked using any other writing writing instrument will will not no t be considered. considered. (3) After your examination examination is over, over, return the writing writing board, b oard, torch, pen and the envelope including your answersheet to the examiner.

Observational Examination (OT) Page 1 of 1 When you arrive at your observing station, sta tion, DO NOT disturb the telescope before attempting the th e first first question (OT1). (OT1) T he telescope telescope is already already set to a deep sky object. Identify the object and tick the correct box in the Summary Answersheet. Note: You can use any technique to identify the object. However, if you disturb the telescope, you will NOT be helped to bring it back to the original original position.

10

(OT2) (OT2.1) Point the telescope to M45. Show the object to the examiner. examiner. Note: 1. After 5 minutes, 1 mark will will be deducted for a delay delay of every minute (or part thereof) in pointing pointing the telescope. 2. You have a single single chance to be evaluated. If your pointing pointing is incorrect incorrect the examin examiner er will will change the t he pointing pointing to M45 for the next next part of the question.

5

(OT2.2) Your Summary Answersheet shows telescopic field of M45. In the image, seven ( 7) brightest stars of the cluster are replaced by ‘ + ’ sign. sign. Compare Compar e the image image with the field field you see in the telescope and number the ‘ +’ marks from 1 to 7 in the order of decreasing decreasing brightness brightness (brighte (brighte st is 1 and faintest is 7) of the corresponding corresponding stars.

15

(OT3) The examiner will give you a moon filter, an eyepiece with a cross-wire and a stopwatch. Point the telescope towards to wards the Moon. Moon . Attach Attach the filter filter to the th e telescope. telescope. On the surface of the Moon, you yo u will will see se e several “seas” (mar ia) (mar ia) which are nearly circular circular in shape. Estimate the diameter of Mare Serenitatis, Serenitatis, Sr , labelled as “ 1” in the figure below, below , as a fraction of the lunar diameter, oon , by measuring measuring the telescope telescope drift times, oon and Sr , for the Moon and the mare, respectively. respectively.

20

Group Examination Page 1 of 2 (G1) A spacecraft of mass m and velocity ⃗ approaches a massive planet of mass M and and orbital velocity   , as measured by an inertial inertial observer ob server.. We consider a special case, where the incoming trajectory tr ajectory of the spacecraft spacecraf t is designed in a way such that velocity velocity vector of the planet does not change direction direction due to the gravitational gravitational boost given given to the spacecraft. In this case, the gravitational gravitational boost to the velocity the t he spacecraft spacecraf t can be estimated using conservation laws by measuring asymptotic asymptot ic velocity velocity of the spacecraft before and after the interaction interaction and angle of approach of the spacecraft.

(G1.1) What will will be the final final velocity ( ⃗f ) of the spacecraft, if ⃗ and  are exactly exactly anti-parallel (see Figure 1).

3

(G1.2) Simplify Simplify the expression expression for the case where  <<  .

1

(G1.3) If angle between ⃗ and  Figure 2), use results above to write  is  and  <<  (see Figure expression for the magnitude of final velocity ( f ).

3

(G1.4) Table on the last page gives data of Voyager- 2 spacecraft for a few months in in the year 1979 as it passed close to Jupiter. Assume that the observer is located at the centre of the Sun. The distance from the observer is given in AU and  is heliocentric heliocentric ecli e cliptic ptic longitude longitude in degrees. degrees. Assume all objects obje cts to be in in the ecliptic ecliptic plane. Assume that the orbit of the Earth to be circular. Plot appropriate column against the date of observation to find the date at which the spacecraft was closest closest to the Jupiter, and label the th e graph graph as G1.4. G1. 4.

8

(G1.5) Find the Earth-Jupiter distance, ( − J ) on the day of the encounter.

4

(G1.6) On the day of the encounter, around what standard time ( td ) had the Jupiter transited the meridian in the sky of Bhubaneswar Bhubanes war ( 20.27∘ N; 85.84∘ E; UT + 05: 30)?

6

(G1.7) Speed of the spacecraft (in km s 1) as measured by the same observer on some dates before the encounter and some dates after the encounter are given below. Here day n is the date of encounter. Use these data to find the orbital speed of Jupiter (  ) on the date of encounter and angle  .

12

date

n-45

n-35

n-25

n-15

n-5

n

vtot

10.1408

10.0187

9.9078

9.8389

10.2516

25.5150

date

n+5 21.8636

n+15 21.7022

n+25 21.5580

n+35 21.3812

n+45 21.2365

vtot

(G1.8) Find eccentricity, J , of Jupiter's orbit.

8

(G1.9) Find heliocentric ecliptic longitude, p , of Jupiter's perihelion perihelion point.

5

Group Examination Page 2 of 2 Month

June June June June June June June June June June June June June June June June June June June June June June June June June June June June June June July July July July July July July July July July July July July July July July

Date

λ

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

(∘ ) 135.8870 13 5.8870 135.9339 13 5.9339 135.9806 13 5.9806 136.0272 13 6.0272 136.0736 13 6.0736 136.1200 13 6.1200 136.1662 13 6.1662 136.2122 13 6.2122 136.2582 13 6.2582 136.3040 13 6.3040 136.3496 13 6.3496 136.3951 13 6.3951 136.4405 13 6.4405 136.4857 13 6.4857 136.5307 13 6.5307 136.5756 13 6.5756 136.6202 13 6.6202 136.6647 13 6.6647 136.7090 13 6.7090 136.7532 13 6.7532 136.7970 13 6.7970 136.8407 13 6.8407 136.8841 13 6.8841 136.9273 13 6.9273 136.9702 13 6.9702 137.0127 13 7.0127 137.0550 13 7.0550 137.0969 13 7.0969 137.1384 13 7.1384 137.1795 13 7.1795 137.2200 13 7.2200 137.2600 13 7.2600 137.2993 13 7.2993 137.3378 13 7.3378 137.3754 13 7.3754 137.4118 13 7.4118 137.4467 13 7.4467 137.4798 13 7.4798 137.5116 13 7.5116 137.5628 13 7.5628 137.6898 13 7.6898 137.8266 13 7.8266 137.9599 13 7.9599 138.0903 13 8.0903 138.2186 13 8.2186 138.3453 13 8.3453

Distance (AU) 5.1589731906 5.1 589731906 5.1629499712 5.1 629499712 5.1669246607 5.1 669246607 5.1708975373 5.1 708975373 5.1748689006 5.1 748689006 5.1788390741 5.1 788390741 5.1828084082 5.1 828084082 5.1867772826 5.1 867772826 5.1907461105 5.1 907461105 5.1947153428 5.1 947153428 5.1986854723 5.1 986854723 5.2026570402 5.2 026570402 5.2066306418 5.2 066306418 5.2106069354 5.2 106069354 5.2145866506 5.2 145866506 5.2185705999 5.2 185705999 5.2225596924 5.2 225596924 5.2265549493 5.2 265549493 5.2305575243 5.2 305575243 5.2345687280 5.2 345687280 5.2385900582 5.2 385900582 5.2426232385 5.2 426232385 5.2466702671 5.2 466702671 5.2507334797 5.2 507334797 5.2548156324 5.2 548156324 5.2589200110 5.2 589200110 5.2630505798 5.2 630505798 5.2672121872 5.2 672121872 5.2714108557 5.2 714108557 5.2756542053 5.2 756542053 5.2799520895 5.2 799520895 5.2843175880 5.2 843175880 5.2887686308 5.2 887686308 5.2933308160 5.2 933308160 5.2980426654 5.2 980426654 5.3029664212 5.3 029664212 5.3082133835 5.3 082133835 5.3140161793 5.3 140161793 5.3210070441 5.3 210070441 5.3312091210 5.3 312091210 5.3405592121 5.3 405592121 5.3466522674 5.3 466522674 5.3516661563 5.3 516661563 5.3561848203 5.3 561848203 5.3604205657 5.3 604205657 5.3644742164 5.3 644742164

Month

July July July July July July July July July July July July July July July August August August August August August August August August August August August August August August August August August August August August August August August August August August August August August August August August August August August August August August August August August August August August August August August August August August August August August August August August August August August August August

Date

λ

17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31

(∘ ) 138 .4707 138 .5949 138 .7183 138 .8409 138 .9628 139 .0841 139 .2048 139 .3250 139 .4448 139 .5641 139 .6831 139 .8016 139 .9198 140 .0377 140 .1553 140 .2725 140 .3895 140 .5062 140 .6225 140 .7387 140 .8546 140 .9702 141 .0856 141 .2007 141 .3157 141 .4303 141 .5448 141 .6591 141 .7731 141 .8869 142 .0006 142 .1140 142 .2272 142 .3402 142 .4530 142 .5657 142 .6781 142 .7904 142 .9024 143 .0143 143 .1260 143 .2375 143 .3488 143 .4599 143 .5709 143 .6817

Distance (AU) 5.3684017790 5.3 684017790 5.3722377051 5.3 722377051 5.3760047603 5.3 760047603 5.3797188059 5.3 797188059 5.3833913528 5.3 833913528 5.3870310297 5.3 870310297 5.3906444770 5.3 906444770 5.3942369174 5.3 942369174 5.3978125344 5.3 978125344 5.4013747321 5.4 013747321 5.4049263181 5.4 049263181 5.4084696349 5.4 084696349 5.4120066575 5.4 120066575 5.4155390662 5.4 155390662 5.4190683021 5.4 190683021 5.4225956100 5.4 225956100 5.4261220723 5.4 261220723 5.4296486357 5.4 296486357 5.4331761326 5.4 331761326 5.4367052982 5.4 367052982 5.4402367851 5.4 402367851 5.4437711745 5.4 437711745 5.4473089863 5.4 473089863 5.4508506867 5.4 508506867 5.4543966955 5.4 543966955 5.4579473912 5.4 579473912 5.4615031166 5.4 615031166 5.4650641822 5.4 650641822 5.4686308707 5.4 686308707 5.4722034391 5.4 722034391 5.4757821220 5.4 757821220 5.4793671340 5.4 793671340 5.4829586711 5.4 829586711 5.4865569133 5.4 865569133 5.4901620256 5.4 901620256 5.4937741595 5.4 937741595 5.4973934544 5.4 973934544 5.5010200385 5.5 010200385 5.5046540300 5.5 046540300 5.5082955377 5.5 082955377 5.5119446617 5.5 119446617 5.5156014948 5.5 156014948 5.5192661222 5.5 192661222 5.5229386226 5.5 229386226 5.5266190687 5.5 266190687 5.5303075275 5.5 303075275

Group Examination Page 1 of 8 (G1) A spacecraft spacecraft of mass mass m and m and velocity  velocity v approaches a massive planet of mass M and M and orbital velocity u, as measured by an inertial observer. We consider a special case, where the incoming trajectory of the spacecraft is designed in a way such that velocity vector of the planet does not change direct direction ion due to the gravita gravitation tional al boost bo ost given given to the spacecr spacecraft. aft. In this case, case, the amount amount of gravitational boost to the velocity the spacecraft can be roughly estimated using conservation laws by measuring asymptotic velocity of the spacecraft before and after the interaction and angle of approach of the spacecraft.

Planet

Planet

u

u

v spacecraft θ

v spacecraft

Figure 1

Figure 2

Figure 3

(G1.1) What will be the final velocity ( v f ) of the spacecraft, if  v and  and u are exactly anti-parallel (see Figure 1).

3

Solution:

Let v f and u f be the final velocity of the spacecraft and the planet respectively. As the planet For anti-parallel anti-parallel case, using conservation conservation of linear momentum, momentum, M u + m v = M = M u f + mv f ∴

Mu

mv = M M u + mv − mv = m u = u − (v + v ) M f

f

f

f

Now, using conservation of energy, 2 M u2 + mv 2 = M uf + mvf 2 2 m 2 m m 2 u2 + v = u (vf + v ) + v M M M f m m m  v 2 =   (vf + v )2 2u m  (vf + v ) + m  v 2      u2 +  u2 +  f M M M M M     m 0= (vf + v )2 2u(vf + v ) + (v ( vf 2 v 2 ) M m   2u    (v    0= (vf + v )  (v  (v  )(vf f + v ) f + v ) +  f + v )(v M m 0= (vf + v ) 2u + vf v M m m = 2u + 1 v ∴ vf 1 + M M m 2u + 1 M v vf = m 1 + M

−



·

−

−

−





−

− −

− v)

−   −  

Alternat Alternativ ive e solution solution in COM frame frame

(G1.2) Simplify the expression for the case where m

M .  M .

1

Group Examination Page 2 of 8

Solution:

If m m

M ,  M , v ≈ 2u + v f

(G1.3) If angle between v and u is θ and m M M (see Figure 2), use results above to write expression for the magnitude of final velocity (v ( vf ).

−



3

Solution:

As velocity vector of the planet is not changing direction, there is no momentum transfer in direction perpendicular to u. We will resolve resolve v and v f into components parallel and perpendicular to  to u. vx =

−v cos θ

vf x = 2u + v cos θ

vy = v sin θ vf y = v sin θ

vf 2 = v f 2x + vf 2y = (2u (2u + v cos θ)2 + (v (v sin θ)2 = 4u2 + 4uv 4uv cos θ + v 2 cos2 θ + v 2 sin2 θ = 4u2 + 4uv 4uv cos θ + v 2 ∴

vf =



4u2 + v 2 + 4uv 4uv cos θ

(G1.4) Table on the last page gives data of Voyager-2 spacecraft for a few months in the year 1979 as it passed passed close to Jupiter. Assume Assume that the observer observer is located lo cated at the centre of the Sun. Sun. The distanc distancee from the observ observer er is given given in AU and λ is heliocentric ecliptic longit longitude ude in degrees. degrees. Assume Assume all objects to b e in the eclipti eclipticc plane. Assume Assume the orbit of the Earth to be circular. circular. Plot appropriate appropriate column against the date of observation observation to find the date at which the spacecraft was closest to the Jupiter, and label the graph as G1.4.

8

Solution:

From rom the the graph graph,, it can be infe inferr rred ed that that the the enco encoun unte terr with with Jupi Jupite terr occur occured ed on 10th July July (day (day 191) and its distance distance from the Sun on that day is 5.33121 5.33121 AU (G1.5) Find the Earth-Jupiter distance, (d (dE −J ) on the day of the encounter.

4


Solution:

The day number number of Vernal Vernal Equinox is 80. Thus, Thus, ecliptic longitude longitude of the Sun as seen from the Earth on the day of encounter will be, λ = (191

◦

◦

365..25 = 109. 109.4045 − 80) ∗ 360 /365

Thus, the ecliptic longitude of the Earth as seen from the Sun on the day of encounter will be, λ⊕ = 180◦ + 109. 109.4045◦ = 289. 289.4045◦ Applying Applying cosine cosine rule, d⊕−J = =

 

2 d2⊕ + dJ

12

cos∆λ − 2d d cos∆λ + 5. 5.3312 − 2 × 1 × 5.3312 × cos(289. cos(289.4045 − 137 137..5628 ) ⊕

J

◦

2

◦

= 6.230 23088 AU

i.e. i.e. the the Eart Earth h is 6.2308 AU from Jupit Jupiter er on that day day.

(G1.6) On the day of the encounter, around what standard time (t ( tstd ) had the Jupiter transited ◦ the meridian in the sky of Bhubaneswar (20 .27 N; 85. 85.84◦ E; UT + 05:30)?

Solution:

Thus, the angle of eastern elongation for Jupiter (SE J ) on that day would be, ξ = = sin−1



5.3312

◦

◦

sin(289.4045 − 137 137..5628 ) × sin(289. 6.2308

= 23 23..8146◦



It would rise 95 minutes after the Sun rise, i.e. around 7:35am. It would transit the meridi meridian an after around around 6 hours hours i.e. around around 13:35 local local time or 13:2 13:22 2 IST . For more precise answer, R.A. of Jupiter on the day of encounter is approximately, λJ geocentric = 109. 109.4045◦ + 23. 23.8146◦ = 133. 133.2191◦ geocentric tan αJ = tan λJ geocentric cos  geocentric = tan 133 133.2191◦ cos23◦ 26 ∴

αJ = 135. 135.68◦ = 9h 3m

Thus, Thus, it will will culmin culminate ate at that that sidere sidereal al time. On that day, day, sidere sidereal al time at noon is 07:24 (111 days from V.E. times times 4 minute minutes). s). Thus, Thus, it will will culmin culminate ate 1 hour hour 39 minute minutess after the local local noon i.e. at 13:39 local local time or at about 13:2 13:26 6 IST .

6


To V.E.

S

δλ

S

E

φ1

A

D u

J

J

B Figure 4

φ2

Figure 5

C

(G1.7) Speed Speed of the spacec spacecraft raft (in km s−1 ) as measured by the same observer on some dates before the encounter and some dates after the encounter are given below. Here day n is the date of encounter. Use these data to find the orbital speed of Jupiter ( u) on the date of encounter and angle θ angle θ.. date vtot date vtot

n 45 10.1 0.1408 n+5 21.8 21.863 6366

−

n 35 10.01 01887 n + 15 21.7 21.702 0222

n 25 9.9078 078 n + 25 21.5 21.558 5800

−

n 15 9.8 9.8389 n + 35 21.3 21.381 8122

−

−

n 5 10.2 0.2516 n + 45 21.2 21.236 3655

−

n 25.515 5150

Solution:

In Figure 5, path of Voyager-2 is shown as A-B-C. The Sun is shown as S and the Jupiter is shown as J. From the data we note that r is increasing continuously. The same should should be reflected reflected in the diagram. For practical purpose, J and B are the same points. The direction of velocity vector of Jupiter is given by J D. In the figure, = δλ λ1 ASB = δ

= δλ λ2 ASB = δ

δ λ ASC = δλ

= θ ABD = θ

ABC = θ 1

DBC = ABC

= φ 1 SAB = φ

= φ 2 SC B = φ

= θ − θ − ABD = θ 1

Now the lines originating from the Sun indicate radial direction on the respective dates. Let us take speed of the spacecraft sufficiently far from the day 190, to avoid any influence of Jupiter in initial and final velocity estimation. We can choose dates 35 days on either either side side of July July 10 i.e. i.e. June June 5 and August August 14. δλ 1 = 137. 137.5628◦ l(AB) AB ) = =

◦

◦

136..0736 = 1.4892 − 136 l(SA SA)) + l(SB ) − 2 × l(SA SA)) × l(SB ) × cos δλ 5.17487 + 5. 5.33121 − 2 × 5.17487 × 5.33121 × cos1. cos1.4892

 

2

2

2

1

◦

2

= 0.20755au φ1 = sin

−1



l(SB )sin δλ 1 l(AB) AB )



= sin

−1



5.33121 sin1. sin1.4892◦ 0.20755

×



12


= sin−1 (0. (0.66755) φ1 = 41 41..8783◦ or 138 138.1217◦ δλ 2 = 141. 141.2007◦ l(BC ) BC ) = =

◦

◦

137..5628 = 3.6379 − 137 l(SC ) SC ) + l(SB ) − 2 × l(SC ) SC ) × l(SB ) × cos δλ 5.45085 + 5. 5.33121 − 2 × 5.45085 × 5.33121 × cos3. cos3.6379

 

2

2

2

2

◦

2

= 0.36253au φ2 = sin



−1

l(SB )sin δλ 2 l(BC ) BC )

= sin−1 (0. (0.66755)



= sin

−1



5.33121 sin3. sin3.6379◦ 0.36253

×



φ2 = 68 68..9199◦ or 111 111.0801◦ from the figure, φ figure, φ 1 should be obtuse and φ 2 may be acute. In δλ = δλ = λ λ 2

◦

141.2007 − 136 136..0736 − λ = 141. 1

SABC

◦

= 5.1271◦ ∴

θ1 = 360◦ ◦

= 360

− δλ − φ − φ 138..1217 − 68 68..9199 − 5.1271 − 138 1

2

◦

◦

◦

θ1 = 147. 147.8313◦ In

SB C , we notice = SB C =

180◦ ◦

= 180

− φ − δλ 68..9199 − 3.6379 − 68 2

2

◦

◦

= 107. 107.4422◦ vy v sin θ tan DBC DB C = f = 2u vxf v cos θ + 2u sin θ tan(θ tan(θ1 θ) = cos θ + 2 uv 2u sin θ = cos θ ∴ v tan(θ tan(θ1 θ)

−

− − We use this expression to find |u|.

vf 2 = 4u2 + v 2 + 4uv 4uv cos θ

vf 2

4u 2 4u 4 u = 2 +1+ cos θ 2 v v v vf 2 2u 2 2u = +1+2 v v v

      =

−

   − −

−

−

sin θ tan(θ tan(θ1 θ)

−

cos θ

cos θ 2

+1+2



sin θ tan(θ tan(θ1 θ)

− − cos θ



cos θ

2     sin θ 2sin θ  cos θ 2sin θ  cos θ 2 = + cos θ + 1 + 2cos2 θ    θ1 θ )  θ1 θ) tan(θ tan(θ1 θ) tan(θ tan( tan(θ tan(   sin2 θ = + 1 cos2 θ 2 tan (θ1 θ) sin2 θ = + sin2 θ = sin2 θ cot2 (θ1 θ) + 1 2 tan (θ1 θ)

−

− −

−



−




sin2 θ sin2 (θ1 θ) vf sin θ sin θ = = ∴ v sin(θ sin(θ1 θ ) sin θ1 cos θ cos θ1 sin θ v = sin θ1 cot θ cos θ1 ∴ vf sin θ1 tan θ = v + cos θ1 vf =

−

−

−

−

sin 147 147.8313◦ = 10.0187 = ◦ + cos 147. 147 . 8313 21.3812

θ = 180◦

∴

◦

−1.4088

= 125. 125.3672◦

54..6328 − 54

vf 2 = 4u2 + v 2 + 4uv 4uv cos θ 21 21..38122 = 4u2 + 10. 10.01872 + 4u 4u 10 10..0187cos125. 0187cos125.3672◦ (457. (457.1557 100 100..3743) 0 = u 2 5.7990 7990u u 4 2 0 = u 5.7990 7990u u 89 89..1953

×

− −

− − √ 5.7990 + 5.7990

u =

∴

2

−

+4

2

89..1953 × 89

−1

= 12 12..7789kms

Jupite Jupiter’s r’s orbital orbital veloci velocity ty on the day of encoun encounter ter is 12. 12.779kms−1 and the angle between the initial velocity velocity of the spacecraft and Jupiter’s velocity velocity vectors is 125 ◦ 22 . (G1.8) Find eccentricity eccentricity, e J , of Jupiter’s orbit.

8

Solution:

The angle between between  r and  and u on the day of encounter will be, ψ =

SB C SBC

− (θ − θ) = 107. 107.4422 − 147 147..8313 1

◦

◦

+ 125. 125.3672◦

= 84 84..9781◦

Now we use angular momentum momentum conservation conservation to estimate estimate eccentricity eccentricity.. If u p and r p represent perihelion velocity and perihelion distance of Jupiter, r p u p = a = a J (1 =



   GM  aJ

− e)

GM  aJ (1

r p u p = r = ru u sin ψ ∴

1

2

−e

=

1+e 1 e

−

2

−e )

r 2 u2 sin2 ψ GM  aJ 2

5.331212 1.496 1011 12..7789 103 sin2 84 12 84..9781◦ = 6.6741 10−11 1.9891 1030 5.20260 = 0.99761

×

∴

e =

×

×

× ×

√ 1 − 0.99761 = 0.0.0489



×

×

×




The eccentri eccentricity city of Jupiter’s Jupiter’s orbit is 0.0489 . (G1.9) Find heliocentric ecliptic longitude, λ longitude, λ p , of Jupiter’s perihelion point. Solution:

To estimate longitude of perihelion, one should estimate true anomaly of Jupiter on that day. a(1 e2 ) r = 1 + e cos Θ a(1 e2 ) 5.20260 0.99761 04899 cos cos Θ = 1= ∴ 0.048 r 5.33121 = 0.02646

−

−

−

−

Θ = 122. 122.754◦ Thus, the longitude of perihelion of Jupiter is, λ p = λ = λ J

− Θ

= 137. 137.5628◦ λ p = 14 14..809◦

◦

122..754 − 122

×

−1

5

Group Examination Page 8 of 8 Mon Mo nth

Date ate

λ ( ) 135.8870 135.9339 135.9806 136.0272 136.0736 136.1200 136.1662 136.2122 136.2582 136.3040 136.3496 136.3951 136.4405 136.4857 136.5307 136.5756 136.6202 136.6647 136.7090 136.7532 136.7970 136.8407 136.8841 136.9273 13 136.9702 13 137.0127 13 137.0550 13 137.0969 13 137.1384 13 137.1795 137.2200 137.2600 137.2993 137.3378 137.3754 137.4118 137.4467 137.4798 137.5116 13 137.5628 13 137.6898 13 137.8266 13 137.9599 13 138.0903 13 138.2186 13 138.3453 o

June June June June June June June June June June June June June June June June June June June June June June June June June June June June June June July July July July July July July July July July July July July July July July

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Distance

(AU) 5.1589731906 5.1629499712 5.1669246607 5.1708975373 5.1748689006 5.1788390741 5.1828084082 5.1867772826 5.1907461105 5.1947153428 5.1986854723 5.2026570402 5.2066306418 5.2106069354 5.2145866506 5.2185705999 5.2225596924 5.2265549493 5.2305575243 5.2345687280 5.2385900582 5.2426232385 5.2466702671 5.2507334797 5. 5.2548156324 5. 5.2589200110 5. 5.2630505798 5. 5.2672121872 5. 5.2714108557 5. 5.2756542053 5.2799520895 5.2843175880 5.2887686308 5.2933308160 5.2980426654 5.3029664212 5.3082133835 5.3140161793 5.3210070441 5. 5.3312091210 5. 5.3405592121 5. 5.3466522674 5. 5.3516661563 5. 5.3561848203 5. 5.3604205657 5. 5.3644742164

Month

Da Date

λ ( ) 138.4707 138.5949 138.7183 138.8409 138.9628 139.0841 139.2048 139.3250 139.4448 139.5641 139.6831 139.8016 139.9198 140.0377 140.1553 140.2725 140.3895 140.5062 140.6225 140.7387 140.8546 140.9702 141.0856 141.2007 14 141.3157 14 141.4303 14 141.5448 14 141.6591 14 141.7731 14 141.8869 142.0006 142.1140 142.2272 142.3402 142.4530 142.5657 142.6781 142.7904 142.9024 14 143.0143 14 143.1260 14 143.2375 14 143.3488 14 143.4599 14 143.5709 14 143.6817 o

July July July July July July July July July July July July July July July August August August August August August August August August August August August August August August August August August August August August August August August August August August August August August August

17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31

Distance

(AU) 5.3684017790 5.3722377051 5.3760047603 5.3797188059 5.3833913528 5.3870310297 5.390644477 5.3942369174 5.3978125344 5.4013747321 5.4049263181 5.4084696349 5.4120066575 5.4155390662 5.4190683021 5.4225956100 5.4261220723 5.4296486357 5.4331761326 5.4367052982 5.4402367851 5.4437711745 5.4473089863 5.4508506867 5. 5.4543966955 5. 5.4579473912 5. 5.4615031166 5. 5.4650641822 5. 5.4686308707 5. 5.4722034391 5.4757821220 5.4793671340 5.4829586711 5.4865569133 5.4901620256 5.4937741595 5.4973934544 5.5010200385 5.5046540300 5. 5.5082955377 5. 5.5119446617 5. 5.5156014948 5. 5.5192661222 5. 5.5229386226 5. 5.5266190687 5. 5.5303075275

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