CELESTIAL MECHANICS

Judhistira Aria Utama, M.Si.Lab. Bumi & Antariksa

Jurusan Pendidikan FisikaFPMIPA - UPI

Source: Astronomy: Principles and Practice, A.E. Roy & D. Clark, 4th Edition

Elliptic Orbit

• Then the line AA is the major axis of the ellipse, C is the centre and, therefore, CA and CA’ are the semi-major axes. Likewise BB is the minor axis, with CB and CB’ the semi-minor axes. If a and b denote the lengths of the semi-major and semi-minor axes respectively, then

b2 = a2(1 − e2) and e = CS/CA

Find the perihelion & aphelion distance!

The velocity of a planet in its orbit

• Let VP & VA be the velocities at perihelion A and aphelion A’ respectively. We may write:

V = rω

where ω is the angular velocity.

• From Kepler’s second law:

r2ω = h

• Hence, at perihelion and aphelion only, we have

V =h/r

• For perihelion:

VP =h/a(1 − e)

• For aphelion:

VA =h/a(1 + e)

Measuring the mass of a planet

• Let a planet, P, with orbital semi-major axis a, sidereal period of revolution T and mass m possess a satellite, P1, that moves in an orbit about P with semi-major axis a1 and sidereal period of revolution T1. Let the masses of the Sun and satellite be M and m1 respectively.

• From Kepler’s third law for circular orbit:

where μ = G(m1 + m2)

• For the planet and the Sun we have:

μ = G(M + m)

• For the satellite and planet we have:

μ = G(m + m1)

• Dividing the later with previous one, we have:

Transfer between circular, coplanar orbits about the Sun (Hohmann Transfer)

• The semi-major axis, α, of the transfer orbit.

From previous figure, it can be seen that:

AB = 2α = a1 + a2

Hence,

α = (a1 + a2) / 2

• The eccentricity, e, of the transfer orbit.

SA = a1 = α(1 − e)

SB = a2 = α(1 + e)

Hence,

e = (a2 − a1) / (a2 + a1)

• Any time thrust (∆V) is added to a circular orbit  increases eccentricity.

• Adding ∆V (thrust) at perigee increases eccen-ticity.

• Adding ∆V at apogee decreases eccentricity (apogee kick).

• The transfer time, , spent in the transfer orbit.

• The velocity increments VA and VB.

At A, the required increment VA is the difference between circular velocity Vc1 in the

inner orbit and perihelion velocity VP in the transfer orbit.

VA = VP − Vc1, since VP > Vc1 or

In this case:=GMCentral Body

• At B, the required increment VB is the difference between circular velocity Vc2 in the outer orbit and aphelion velocity VA in the transfer orbit.

VB = Vc2 − VA, since Vc2 > VA

or

Hohmann Transfer(lower to higher orbit)

• Fire motor at perigee to get eccentric orbit with desired orbit altitude equal  to the new apogee.

• Give an apogee kick to recircularize the orbit.

• Most efficient orbital transfer.

Hohmann Transfer(higher to lower orbit)• Retrofire rocket at apogee to make  eccentric 

orbit with lower perigee.

• Retrofire at perigee to recircularize.

• Retrofiring means turning the rocket around and firing against the flight direction.

Orbital Transfer

• Going from a low orbit to higher

– Requires thrust in flight direction

– Satellite speed decreases

• Going from high orbit to lower

– Requires retrofiring

– Satellite speed increases

Rendezvous

• Approaching craft does a Hohmann transfer.

• Since higher orbit is slower, it must start out 

closer to the rendezvous point.

• In rendezvous problem, only the first of the two impulses was required since there was no need to inject the payload into another orbits.

• In the transfer between circular and coplanar orbit, the first burn could be made at any time. But, in the rendezvous problem a time table has to be kept,dictated by the necessity that when the vehicle reaches aphelion in the transfer orbit, the particle moving in the outer circular orbit should also be there.

• Let two particles P1 and P2 revolve in coplanar orbits of radii a1 and a2 about a body of mass M. Let their longitudes, measured from some reference direction be l10 and l20 at time t0.

• The problem is to obtain the time conditions enabling a vehicle to leave particle P1 and arrive at particle P2 by a Hohmann cotangential ellipse.

• The angular velocities of the two particles are n1 and n2, given by:

• Remember that

• The longitudes of the particles at time t are given by:

• If angle BSC= θ,the transfer time is ,given by:

• From previous slide, the transfer time is:

or

So that we have

• Now the longitude of P1 when the vehicle leaves that body is π radians less than the longitude of P2 when the vehicle arrives. Thus, the longitude of the particles at the vehicle departure time must differ by (π − θ) radians, or L12, given by

• The difference in the longitudes of P2 and P1 at any time t is given by

• If the value of L12 is inserted in the left-hand side of equation above, the time t can be calculated.

• For a return of the vehicle from P2 to P1, the transfer time will be the same as on the outward journey and the angle θ between the radius vector of P1 when the vehicle departs and that of the arrival point in P1’s orbit must be given by

• Then the difference in longitudes of P2 and P1 must be L’21 where

• The round trip time or mission time tM is

• If P1 is β radians ‘ahead’ of P2 when a transfer from P1 to P2 has just ended, the first available transfer back from P2 to P1 will begin when P1 is β radians ‘behind’ P2.

• Hence, if S is the synodic period of the two particles,

(If is positive)

• Alternatively, if P1 were β radians ‘behind’ P2 when an outward transfer (P1 to P2) has just ended, the first available return from P2 to P1 can begin when P1 has reached a point β radians ‘ahead’ of P2. And in this case,

(If is negative)

Lunar Trajectory(Apollo)

• Spacecraft must reach escape velocity to  leave Earth’s gravity.

• Spacecraft will fly on a trajectory which will not orbit Earth (it won’t come back!).

Lunar Capture

• Spacecraft falls towards Moon.

• Retrofire is required to be captured by Moon’s  gravity.

• Without retrofire, spacecraft will pass Moon (or crash into it) and orbit the Sun.

Trajectory to Another Planet

• Leaving the Earth

• Orbit the Sun

• Rendezvous with target

• When spacecraft approaches, it starts falling  towards planet.

– Must retrofire to be captured in orbit.

– Better to approach planet from same direction

of travel.

• Hohmann transfer is most efficient but takes a long time and only during “operating windows”.

• Fast transfer is very expensive.