Astronomy Principles and Practice Fourth Edition.pdf

The two-body problem 181
Figure 13.8. Example 13.1—the orbit of an artificial satellite.
The size of the major axis AA ′ of the ellipse is given by
AA ′ = 4023 + 2 × 6372 + 660 = 17 427 km
so that the semi-major axis a 1 is of length 8713·5 km. The period of revolution is T 1 = 134 minutes.
For the Earth orbit,
a = 149·5 × 10 6 km
T = 365·25 × 24 × 60 = 5·2596 × 10 5 minutes.
Letting M and m be the masses of Sun and Earth respectively, we have, using equation (13.35),
(
) ( )
M
m = 134 2 3
149·5 × 10
6
5·2596 × 10 5 8713·5
giving, after a little calculation,
M
= 327 800.
m
Hence, the Earth’s mass is only about three-millionths that of the Sun.
13.6.8 Co-periodic orbits
Rewriting equation (13.31), we have
( )1
a
3 2
T = 2π . (13.31)
µ
Re-arranging equation (13.29), we obtain
/ ( 2
a = 1
r − V )
2
. (13.36)
µ
If we eliminate a between equations (13.31) and (13.36), there results
( 2µ
T = 2πµ
r
− V 2 ) −3/2
. (13.37)
Equations (13.31), (13.36) and (13.37) highlight some interesting properties of elliptical motion.
It is seen that the semi-major axis is a function of the radius vector and the square of the velocity.
If, therefore, a body of mass m 1 is projected at a given distance r from another body of mass m 2 with