Astronomy Principles and Practice Fourth Edition.pdf

198 Celestial mechanics: the many-body problem
(i) The semi-major axis, α of the transfer orbit. In figure 14.10 it can be seen that
AB = 2α = a 1 + a 2 .
Hence,
α = a 1 + a 2
. (14.1)
2
(ii) The eccentricity e of the transfer orbit.
SA = a 1 = α(1 − e)
SB = a 2 = α(1 + e).
Hence,
e = a 2 − a 1
. (14.2)
a 2 + a 1
(iii) The transfer time τ spent in the transfer orbit. This is the time interval spent in coasting from
A to B. It must be half the period of revolution T in the transfer orbit. Then by equation (13.31),
τ = T 2 = π (
α
3
or, using equation (14.1),
( ) 1/2
(a1 + a 2 ) 3
τ = π
(14.3)
8GM
since the mass of the rocket is negligible compared to that of the central body.
If distance is measured in astronomical units, time in years and the unit of mass is the Sun’s mass,
then by section 13.7, GM = 4π 2 .
We may write, therefore,
( ) 1/2
(a1 + a 2 ) 3
τ =
. (14.4)
32
(iv) The velocity increments V A and V B .AtA, the required increment V A is the difference
between circular velocity V c1 in the inner orbit and perihelion velocity V P in the transfer orbit. Then
µ
)1
2
V A = V P − V c1 .
By equations (13.26) and (13.30) respectively, we may write
V A =
=
[ ( )] 1 ( )1
µ 1 + e 2 µ 2
−
α 1 − e a 1
( ) µ 1/2
[(1 + e) 1/2 − 1].
a 1
Hence, using equation (14.2) we obtain
V A =
( µ
a 1
)1
2
[ ( 2a2
a 1 + a 2
)1
2
− 1
]
. (14.5)
At B, the required increment V B is the difference between circular velocity V c2 in the outer orbit and
aphelion velocity V A in the transfer orbit.