Astronomy Principles and Practice Fourth Edition.pdf

Then V B = V c2 − V A ,sinceV c2 > V A .
Proceeding in the same way, we obtain, finally
( [
)1
µ 2
V B = 1 −
a 1
Interplanetary transfer orbits 199
( 2a1
a 1 + a 2
)1
2
]
. (14.6)
14.8.3 Transfer between particles moving in circular, coplanar orbits
In the first attempts to send packages of scientific instruments from Earth to Venus and Mars, the
transfer orbits closely resembled Hohmann transfers. Only the first of the two impulses was required
since there was no need to inject the payload into either the Venusian or the Martian orbits. Apart
from that difference, there was a very important factor present that was not taken into account in the
discussion of section 14.8.2.
In that problem, the first burn could be made at any time. In the present problem a timetable
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. This problem, often called the
rendezvous problem, arises because departure point and arrival point have their own orbital motions
in the departure and destination orbits. When the transfer burn is initiated, the object in the second
orbit is a little behind the point where it is due to cross the transfer orbit. During the time taken by the
transfer, the object progresses on its orbit to be at the crossing point at the correct time.
Let two particles P 1 and P 2 revolve in coplanar orbits of radii a 1 and a 2 about a body of mass M.
Let their longitudes, measured from some reference direction be l 10 and l 20 at time t 0 . The problem
is to obtain the time conditions enabling a vehicle to leave particle P 1 and arrive at particle P 2 by a
Hohmann cotangential ellipse.
The angular velocities of the two particles are n 1 and n 2 ,givenby
n 1 =
( )1 ( )1
GM
2
GM
2
n 2 =
(14.7)
a1
3 a2
3
since
(
a
3
)1
2
.
n = 2π T
and
T = 2π
GM
The longitudes of the particles at time t are, therefore, given by
l 1 = l 10 + n 1 (t − t 0 ) (14.8)
l 2 = l 20 + n 2 (t − t 0 ). (14.9)
The time spent by the vehicle in the transfer orbit must be the time taken by the particle P 2 to
reach the touching point of transfer and destination orbits. This point C, say, therefore, lies ahead of
the position of P 2 , namely B, when the vehicle leaves P 1 at A (see figure 14.11).
Then, if BSC = θ, the transfer time is τ,givenby
τ = θ n 2
. (14.10)
But, by equation (14.3),
τ = π
(
(a1 + a 2 ) 3
8GM
) 1/2