Astronomy Principles and Practice Fourth Edition.pdf

The two-body problem 179
13.6.5 The period of revolution of a planet in its orbit
Let us suppose the orbit to be circular so that r = a. Then expression (13.29) becomes
V 2 = µ a . (13.30)
But
V = 2πa
T
where T is the time it takes the planet to describe its circular orbit. Hence,
T = 2π
( )1
a
3 2
. (13.31)
Although it will not be proved here, the expression (13.31) holds even when the orbit is elliptical,
a being the semi-major axis of the orbit.
13.6.6 Newton’s form of Kepler’s third law
Let two planets revolve about the Sun in orbits of semi-major axes a 1 and a 2 , with periods of revolution
T 1 and T 2 . Let the masses of the Sun and the two planets be M, m 1 and m 2 respectively. Then by
equation (13.31),
( )1
a
3 2
T 1 = 2π 1
µ 1
where µ 1 = G(M + m 1 ).Also,
T 2 = 2π
µ
(
a
3
2
µ 2
)1
2
.
Hence,
( ) 2 ( ) 3 ( ) ( ) 3 ( )
T2 a2 µ1 a2 M + m1
=
=
. (13.32)
T 1 a 1 µ 2 a 1 M + m 2
Kepler’s third law would have been written as
( ) 2 ( ) 3 T2 a2
= .
T 1 a 1
The only difference between this last equation and (13.32) is a factor k,where
Dividing top and bottom by M, we obtain
k = M + m 1
M + m 2
.
k = 1 + m 1/M
1 + m 2 /M .
The greatest departure of k from unity arises when we take the two planets to be the most massive
and the least massive in the Solar System. The most massive is Jupiter: in this case m 1 /M = 1/1047·3.
Of the planets known to Newton, the least massive was Mercury, giving m 2 /M = 1/6 200 000. Hence,
to three significant figures, k = 1. Kepler’s third law is, therefore, only an approximation to the truth,
though a very good one. Newton’s form of Kepler’s third law, namely equation (13.32), is much better.