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SOLUTIONS TO EXERCISE 31
(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. |