Gravitation - Dynamics and Relativity - page 192

This is clear since the Earth is defined to orbit the Sun such that T = 1 year when

M)

1 year 2 / AU 3 .
Example 9.5.2 The comet Hale-Bopp is (just about) in orbit around the Sun. The
orbit is very eccentric, with ε

1 AU and hence the constant of proportionality 4 π 2 /(G

a = =

0 . 99511 and the distance of closest approach to
the Sun (the perihelion distance) is 0.9141 AU (it was last at perihelion in 1997).
Determine the period of the comet's orbit and its farthest distance from the Sun (the
aphelion distance).

=

Solution 9.5.2 To compute the period of the orbit is a straightforward application
of Kepler's Third Law once we have the distance of the semi-major axis, a.The
perihelion distance is given by

0 . 9141 AU = a( 1 − ε)

which can be solved to give a 187 AU. Since the distance is provided in AU we

need not work too hard, i.e. we can use Eq. (9.70) to get the period:

=

a 3 / 2 T = = 2560 years .

The distance of farthest approach is given by

a( 1 + ε) = 373 AU .

Example 9.5.3 The day before the 1969 moonlanding, the Apollo 11 spacecraft
was put into orbit around the Moon. The spacecraft had a mass of 9970 kg and the
period of the orbit was 119 minutes. In addition, the pericentre and apocentre of
the orbit were 1838 km and 1861 km. Use these data to determine the mass of the
Moon. Also determine the maximum and minimum speeds of Apollo 11 when it was
in lunar orbit.

Solution 9.5.3 Using Eq. (9.69), we can determine the mass of the Moon if we have
the period of the orbit and the length of the semi-major axis a. Since we anticipate
that the mass of Apollo 11 is much smaller than the mass of the Moon we do not need
to worry about the reduced mass. Our task is therefore to deduce a. We know that

a( 1 − ε) = 1838 km

a( 1 + ε) = 1861 km

and adding these two equation together gives

a = 1849 . 5 km.

Re-arranging Eq. (9.69) yields the mass of the Moon:

4 π 2 a 3
GT 2

10 3 ) 3

( 1849 . 5 ×

4 π 2 10 22 kg. M = = = 7 . 35 ×

6 . 67 × 10 − 11 × ( 119 × 60 ) 2

To determine the maximum and minimum speeds we need to compute the speeds
at the pericentre and apocentre of the orbit. We can do this using the conservation