SOLUTIONS TO EXERCISE 32

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a) We can use ratios again.  We know from Part (c) of Exercise 31 that for a
satellite at r = 42300 km, P = 24 hr = 1 day.  With those units, in this orbital
system, K = 1.  Now, 384000 km is 384000/42300 = 9.08 times the
distance of that satellite.  Using Kepler's Third Law, we can say

P^2 = K * r^3 = K * 9.08 * 9.08 * 9.08 = 748 ; so
P   = sqrt(748) = 27.3 days.

Or, using K = 7.6 * 10^(-12) hr^2/km^3 from 3(c), we can compute

P   = sqrt( (7.6 * 10^(-12) hr^2/km^3) * (384000 km)^3 ) = 656 hr.
= 656 / 24 = 27.3 days.

Or, using K = 9.89 * 10^(-14) in MKS units, we can say

P   = sqrt( 9.89 * 10^(-14) * (3.84 * 10^8 m)^3 )
= 2.36 * 10^6 sec = 27.3 days.

(b) According to various reliable sources, the sidereal orbital period
of the Moon is 27.3 days.  Indeed, Part (a) and part (b) are identical to within
the roundoff errors of my calculations.  So the Moon must also be 384,000 km away
from Earth's center!  Earth, this launched satellite, and the Moon will form
a stable triangle in space, rotating around with Earth as its center of rotation.

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