SOLUTIONS TO EXERCISE 31

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(a) Kepler's Third Law holds for all two-body orbits in the
universe!  But the proportionality constant,
K, will be different for each orbiting system; but for any given
system, K will be the same for any object.  As discussed in class,
Newton showed that for nearly circular orbits, K = (4 * pi^2) / (G * M)
where G is the gravitational constant and M is the mass of the primary
object in the orbiting system - that is, the thing around which
everything else orbits.

(b) 24 hours is 22.5 hours longer than 90 minutes.  In terms of ratio,
24 hours is 24/1.5 = 16 times longer than 90 minutes.

(c) We know that a distance of r = 6660 km provides a period of
P = 90 minutes.  In those units, for this orbital system, K = 1.
We want a period 15 times longer than 90 minutes;
since P^2 = K * r^3,

16^2 = K * r^3 = 256, so r needs to be cuberoot(256) = 6.35 times

farther away than 6660 km.  Thus the communications satellite should be
6660 * 6.08 = 42300 km away from Earth's center.

Another way to reach this answer is to compute K for this orbital
system.  This can be achieved using

P^2 = K * r^3  ;  (1.5 hr)^2 = K * (6660 km)^3 ; thus

K = (1.5 * 1.5) / (6660 * 6660 * 6660) = 7.6 * 10^(-12) hr^2/km^3 and

r = cuberoot( (24 hr)^2 / (7.6 * 10^(-12) hr^2/km^3) ) = 42300 km.

Or, remembering that the mass of Earth is 5.98 * 10^24 kg,

K = (4 * 3.14 * 3.14) / (6.67 * 10^(-11) * 5.98 * 10^24)
= 9.89 * 10^(-14) in MKS units; then use distances in mm and
time periods in seconds to get

r = cuberoot( (86400 s)^2 / 9.89 * 10^(-14) ) = 4.23 * 10^7 m.

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