Astronomy Principles and Practice Fourth Edition.pdf

178 Celestial mechanics: the two-body problem
At every point, however, Kepler’s second law holds, namely that
r 2 ω = h. (13.22)
Hence, at perihelion and aphelion only, wehave
V = h r .
For perihelion,
For aphelion,
so that
Now the energy equation (13.20) is
V P =
V A =
h
a(1 − e) .
h
a(1 + e)
V P
= 1 + e
V A 1 − e . (13.23)
so that at perihelion, we have
1
2 V 2 − µ r = C
1
2 V 2 P − µ
a(1 − e) = C (13.24)
while at aphelion,
1
2 V A 2 − µ
= C. (13.25)
a(1 + e)
Subtracting equation (13.25) from equation (13.24) and using relation (13.23) to eliminate V A ,
we obtain
V 2 P = µ a
( ) 1 + e
. (13.26)
1 − e
In similar fashion, we obtain
Again, equations (13.26) and (13.27) give
V 2 A = µ a
( ) 1 − e
. (13.27)
1 + e
V A V P = µ a . (13.28)
Subtracting equation (13.24) from equation (13.20) and using (13.26) to eliminate V P , we obtain,
after a little reduction, the required relation
( 2
V 2 = µ
r − 1 )
. (13.29)
a