Introduction to Orbital Mechanics AE 1350
Newton’s Law of Universal Gravitation • Any two bodies attract one another with a force – Proportional to the product of their masses – Inversely proportional to the square of the distance between them
v − GMm ˆ Fg = r 2 r
M
v Fg
v Fg
.
m
where G = 6.674 x 10-11 Nm2/kg2
This assumes a spherical mass distribution of the gravitational bodies
The N-Body Problem • At any given time, a body in space is being acted upon by several gravitational forces, and other forces like drag, solar pressure, thrust, etc. • If we consider just the gravitational forces, assuming each body has a spherical mass distribution: Let’s study the motion of body mi
r1 is the position of the body 1 relative to our origin r1i is the position of the body i relative to body 1
The N-Body Problem • The vector sum of all such gravitational forces action on body mi is: n
F gi = ∑ j =1 j ≠i
− Gmi m j rji
2
n
F g i = −Gmi ∑ j =1 j ≠i
rˆji
mj rji
2
rˆji
N-Body Equation of Motion • Newton’s 2nd Law:
v &rv&i = Fi mi
i = 1…n
• If gravitational forces are the only forces: v n m v&& Fg i j ri = = −G ∑ 2 rˆji mi j =1 r ji
i = 1…n
i≠ j
• This is a 2nd order, nonlinear vector differential equation that does not have a general closed-form solution, normally solved via numerical integration
Relative Motion of 2 Bodies Body 1: Earth Body 2: Spacecraft Body 3: Moon Body 4: Sun Body 5: Mars Body 6: …
n m v&& j r1 = −G ∑ 2 rˆj1 j = 2 r j1 n m v&& j r2 = −G ∑ 2 rˆj2 j =1 r j2 j≠2
As we defined earlier:
v v v r12 = r2 − r1 &rv& = &rv& − &rv& 12 2 1
v where r12 is the position of body 2 relative to body 1 v and &r&12 is the acceleration of body 2 relative to body 1
− rˆ12
n m n m ⎡ m1 ⎤ m2 v&& j j r12 = −G ⎢ 2 rˆ12 + ∑ 2 rˆj2 − 2 rˆ21 − ∑ 2 rˆj1 ⎥ r21 j = 3 r j2 j = 3 r j1 ⎢⎣ r12 ⎥⎦ n ⎛ rˆj2 rˆj1 ⎞ − G (m1 + m2 ) v&& r12 = rˆ12 − G ∑ m j ⎜ 2 − 2 ⎟ 2 ⎜r ⎟ r12 r j =3 j2 j1 ⎝ ⎠
Relative Motion of 2 Bodies This be used to describe the acceleration (motion) of the spacecraft relative to the Earth, the 2nd term on the RHS accounts for the “perturbing effects” of other bodies (1)
&rv&12
(2)
n ⎛ rˆj2 rˆj1 ⎞ − G (m1 + m2 ) v&& ⎜ ⎟ ˆ − − r12 = r G m ∑ 12 j 2 2 2 ⎟ ⎜ r r r12 j =3 j1 ⎠ ⎝ j2
Acceleration of body 2 relative to body 1
(3)
(4)
(1) two body term which is a function of m1, m2, r12 (2) gravitational acceleration caused by masses 3…n, “Perturbing Term” (3) gravitational acceleration of body 2 due to masses 3…n (4) gravitational acceleration of body 1 due to masses 3…n
Acceleration Contributions for a Satellite in Earth Orbit • Acceleration contributions in G’s (9.81 m/s2) for a 200-NM Earth satellite: Earth Earth oblateness Sun Moon Jupiter Venus Saturn Mars
0.89 (Spherical) 10-3 6x10-4 3x10-6 3x10-8 2x10-8 2x10-9 7x10-10
In this table, relative acceleration is defined as: which is equivalent to the term used:
− Gm j rj
2
rˆj
− Gm j rˆj2 r j2
2
Relative Motion of 2 Bodies • When we simplify to only 2 bodies: – M = m1 (example: mass of the Earth) – m = m2 (example: spacecraft) – r = r12
G (M + m )rˆ v&& r =− 2 r
• This 2nd order vector differential equation has a closed form solution!
Relative Motion of 2 Bodies • For most 2-body problems, we have a spacecraft orbiting a much more massive planetary body: m << M
G (m + M ) ≈ GM = μ
⎛ km 3 ⎞ μ is in units of ⎜⎜ 2 ⎟⎟ ⎝ s ⎠
• This allows us to write the more compact:
v&& − μ r = 2 rˆ r
Total Energy is Conserved • Kinetic energy: KE = 1/2mv2 • Potential energy: ΔPE = − ∫ F ( x )dx xf
xi
ΔPE = − ∫
r
ref
Work done equals the negative of the increase in PE
v r − GMm Fg dr = − ∫ rˆdr 2 ref r r
− GMm ΔPE = r ref PE r − PE ref
− GMm GMm + = r rref
Total (Specific) Energy is Conserved • Define PE = 0 at ref
rref = ∞ therefore,
GMm − μm = PE = − r r
• Total Energy per unit mass: KE + PE v 2 μ ε= = − m 2 r 2
v μ ε= − 2 r
(PE will always be negative!)
(Specific) Angular Momentum • Angular momentum per unit mass:
v v v h = r ×v
• Since h is the cross-product of r and v, it must always be perpendicular to the plane containing r and v
v r
Two-Body Motion • Motion conserves angular momentum and energy – An object moving under the influence of a single, spherical, and constant-mass gravitational field
• Therefore: – Has constant specific angular momentum vector (h), both magnitude and direction: • Motion in a constant plane • Our 3D problem just became a 2D problem!
– Does not lose or gain energy, but simply exchanges PE for KE and vice versa (constant ε) • Types of solutions characterized by ε
Two-Body Orbit Equation • The exact analytic solution to the two-body equation of motion &rv& = − μ rˆ r2 is the conic section equation:
h2 r=
μ
⎛ ⎞ B ⎜⎜ 1 + cos θ ⎟⎟ μ ⎝ ⎠
• This is the 2-body trajectory equation, written in polar coordinates where – r is a function of θ and constants of motion (h, μ, B) – θ is defined as the angle between the B and r vectors – B is a vector constant of integration that has yet to be defined
General Equation of a Conic Section • This is also the general equation of a conic section, written in polar coordinates with the origin located at p one focus: r = 1 + e cos θ
2 p = h /μ • So let’s we define: e = B/μ
θ
p is termed the semi-latus rectum e is termed the eccentricity, sets type of conic section we have
Conic Sections • The prime focus, F, marks the location of the gravitational body • 2p is the width of each curve at the focus (latus rectum) • 2a is the length of the chord passing through the foci (major axis) • 2c is the distance between the two foci
(
p = a 1 − e2
)
e = c/a 17
Summary of Two-Body Motion (so far) • h vector is a constant, so motion takes place in a single plane, the orbital plane, fixed in inertial space • The family of curves called conic sections represents the possible paths for the orbiting object • The focus of the conic orbit is located at the center of the gravitational mass • We know ε = constant; PE < 0 and PE -> 0 as r -> ∞, so – KE (and orbiting object speed) decreases as the orbiting object gains altitude – Speed increases as the orbiting object loses altitude
Apses • The extreme points along a conic’s major axis are referred to as “apses” – periapsis, θ = 0o (closest point), rp – apoapsis, θ = 180o (farthest point), ra p rp = = a (1 − e ) 1+ e p ra = = a (1 + e ) 1− e
• For Earth, these would be the perigee and apogee
Angular Momentum and Energy • At rp and ra
(velocity and radius are at right angles here)
h = rp v p = ra va
• At periapsis, the energy can be calculated as: vp
2
μ ε= − 2 rp
where:
2rp μ h 2 − 2rp μ h2 ε= − = 2 2 2 2rp 2rp 2rp
h 2 − 2μ(a[1 − e]) ε= 2 2(a[1 − e])
(
h 2 = pμ = aμ 1 − e 2
)
Energy (continued) So:
aμ(1 + e )(1 − e ) − 2aμ(1 − e ) ε= 2 2a 2 (1 − e ) μ(1 + e ) − 2μ ε= 2a (1 − e ) μe − μ ε= 2a (1 − e ) μ(e − 1) ε= 2a (1 − e ) μ ε=− 2a
Circle Ellipse Parabola Hyperbola
a>0, ε<0 a>0, ε<0 a=∞, ε=0 a<0, ε>0
Constant, valid for all conic sections
Eccentricity • Now, since h determines p, and ε determines a, the two together determine e
(
p = a 1 − e2
)
h2 − μ = 1 − e2 μ 2ε 2h 2 ε 2 = e −1 2 μ 2h 2 ε 2 = e −1 2 μ
(
⎡ 2h ε ⎤ e = ⎢1 + 2 ⎥ μ ⎦ ⎣ 2
)
1/2
= 1 when ε = 0
Conic Section Parameters e
a
Circle Ellipse Parabola
0 0
>0 >0 ∞
ε <0 <0 =0
Hyperbola
>1
<0
>0
Dominated by PE PE = KE V∞ = 0 KE>PE V∞ > 0
Elliptical Orbits • Planets in our solar system, most spacecraft • An ellipse can be defined geometrically as the locus of points transcribed by a piece of string anchored at the two foci and traced out for all θ For θ = 0o or 180o, r + r’= 2a By definition rp + ra = 2a
Orbital Period • Orbital Period is defined as the time required for the spacecraft to travel once around its orbit (P) – Tangential velocity component, V cos γ = rθ& – Use the fact that, h = rv cos γ h = r 2θ& 2 dθ h=r dt 2 r dt = dθ h
γ V cos γ = rθ& θ
So: h dθ = 2 dt r
Orbital Period From geometry, we see that:
dθ
dA =
1 2 r dθ 2
2 dθ = 2 dA r So, 2 dt = dA h
This is a mathematical statement of Kepler’s 2nd law (note: h = constant) (Planet orbits sweep out equal area per unit time as they orbit) Over the course of one full orbit,
∫
p
0
dt =
2 2 dA = πab ∫ h h 2abπ P= h
Orbital Period • Using the fact that: h = pμ • So,
P=
2aπ ap pμ
2a 3/2 π P= μ
• This is a mathematical statement of Kepler’s 3rd law, • It also shows that the period of an elliptical orbit depends only on the length of the semi-major axis (a, the average of rp and ra)
Kepler’s Laws (1609-1619) 1. A satellite describes an elliptical path around its center of attraction, which is located at one of the two focus points of the ellipse 2. Per unit time, area swept by the radius vector joining the satellite and the center of attraction is constant 3. The periods of any two satellites revolving around the same center body are proportional to their (3/2) power of their semi-major axes
Orbits! • • • • •
Low Earth Orbits Geosynchronous Orbits Transfer Orbits Gravitational Slingshot Lagrangian Points
Classical Orbit Elements a Semi-major axis: a constant defining the size of the orbit in a plane e Eccentricity: a constant defining the type/shape of the orbit in a plane i Inclination: the angle between the North pole (planet’s angular momentum vector) and the angular momentum vector of the orbit Ω Longitude of ascending node: longitude spacecraft goes north across the equator ω Argument of periapsis: angle in the plane of the spacecraft orbit between the ascending node and periapsis, measured in the direction of spacecraft motion ν True anomaly: the angle in the plane of the spacecraft orbit between the periapsis and the current spacecraft position, measured in the direction of s/c motion
6 numbers to define position and velocity in 3D …but only one is changing
Circular Orbits • A circular orbit is a special case of an elliptical orbit where e = 0 and r = constant = p = a, 2r 3/2 π P= μ
since r = a
• We know energy is only a function of a: v2 μ − μ ε= − = 2 r 2a • For a circular orbit where r = a, this becomes: v 2 − μ 2μ μ = + = 2 2r 2r 2r 1/2 and Note that V decreases as r μ ⎛ ⎞ increases and V increases v=⎜ ⎟ as r decreases ⎝r⎠
Circular Orbits: Gravity Provides Centrifugal Force Centrifugal force = mv2/r r
m Force due to gravity = GMm/r2
GMm/r2 = mv2/r GM/r = v2 ⎛ μ⎞ v=⎜ ⎟ ⎝r⎠
1/2
Two ways to get same answer!
Example • Calculate the velocity of an artificial satellite orbiting the earth in a circular orbit at an altitude of 150 miles above the Earth's surface – – – –
r = (3,960 + 150 mile) x 5,280 ft/mile = 21,700,800 ft v2 = GM / r v2 = 1.408 x 1016 / 21,700,800 ft2/s2 v = 25,470 ft/s
Low Earth Orbit (LEO) • Defined as all orbits below geostationary • Lowest energy required to achieve • Almost all LEO missions between – The atmosphere ~400,000 ft (so the orbit will last a while…) – The Van-Allen Belts ~400 nm (radiation trapped in the magnetic field of the Earth, a dangerous place to be)
Low inclination LEO • Inclination of orbit = launch latitude is the minimum required-energy orbit • So the inclination of many US manned missions have been approximately the latitude of Kennedy Space Center (KSC), about 28 degrees, with launches almost due East • The lower the latitude, the lower the required energy due to the spin of the Earth – Cheaper to launch from Florida than North Carolina…
Polar LEO • Polar orbits have an inclination near 90 degrees • Have the advantage that they pass over the entire planet at regular intervals – Which has many uses…
• Takes more energy to get there than for low inclination orbits
Geostationary Orbit (GEO) • Circular • Equatorial (inclination is zero) • Radius is such that angular rate of orbit is the same as the angular rate of the Earth • Radius is about 23,000 nm, period 23.934 hours • Because it’s only one radius, and one inclination – it’s getting very crowded up there!
GOES-8 Typical Image
Orbital Transfers: Elliptical Orbits • Hohmann transfer requires: – minimum delta-V – maximum time
Escape Speed, Parabolic Orbit • Escape Speed is minimum velocity required such that the spacecraft will reach ∞ at 0 speed (will never return) • At ∞, ε = 0 => -μ/(2a) = 0, a = ∞ => parabolic orbit • So, to calculate the velocity required to escape a body’s gravitational force 2 vesc μ =− =0 2 r 2μ Independent vesc = of direction! r • Note that: vesc = 2vcirc and that as r increases, vesc decreases • Orbital speed is “halfway to anywhere” (actually, 70% of the way…)
Hyperbolic Orbits • A hyperbolic path is required when we define an orbit that (theoretically) reaches infinity with some residual velocity
Gravitational Slingshot, Hyperbolic Orbit • Use of gravity of an object to alter the path and/or speed of a spacecraft
(wikipedia)
Method of Patched Conics • Divide complex n-body problem trajectory into a series of 2-body problems • Spacecraft moves relative to body in depending on which sphere of influence it is in – Size of sphere depends on mass and orbit radius of body
• Example, five 2-body problems to Jupiter: – – – – –
Launch from Earth, attain Earth escape velocity Elliptical orbit around sun Hyperbolic trajectory around Mars (slingshot) Elliptical orbit around sun Hyperbolic entry into Jupiter atmosphere
Restricted Three-Body Problem • Planar motion of a spacecraft relative to two other masses (which orbit each other) e.g., motion of spacecraft with respect to the Earth/Sun or Earth/Moon
Sample orbit Seen in frame rotating with two bodies (animation)
• Complex, but 5 equilibrium points for the spacecraft exist…
Lagrangian Points L4
Earth
Moon L1
Usually Stable (depending on masses)
L2
L3
L5
N-Body Problem, Some Interesting Orbits
tuvalu.santafe.edu/~moore/gallery.html
