A11. A space station and a weather satellite are orbiting the Earth. The space s

Question

A11. A space station and a weather satellite are orbiting the Earth. The space station is 8 times as far from the Earth’s center as the weather satellite. If the period of revolution of the weather satellite is 24 hours, what is the period of revolution of the space station in hours? (Hint: use Kepler’s Third Law)

A12. During the lunar mission of Apollo 11, Armstrong and Aldrin descended to the moon's surface in the lunar module while Collins remained in orbit in the command module. If the command module's period was 2 hours and 20 minutes, how high above the moon's surface did it orbit? (mmoon = 7.35 x 10^22 kg, Rmoon = 1.74 x10^6 m, and G = 6.67 x10^–11 N · m2/kg2).

Explanation / Answer

Let the Radius of the space station orbit be Rs And the radius of the weather satellite orbit be Rw Period of revolution of satellite is , Tw= 24 hr s And let the period of revolution of space station be , Ts The condition is  Rs = 8 Rw From Kepler's third law                     T2 R3   Now, (Ts)2 / (Tw)2  =   (Rs)3 /  (Rw)3             (Ts)2 / (24 hrs)^2 = (8Rw)^3 / (Rw)^3                  (Ts)^2 = (8)^3 * (24hrs)^2                      Ts = 543.05 hrs Thus, the time period of space station is 543.05 hrs   Now, (Ts)2 / (Tw)2  =   (Rs)3 /  (Rw)3             (Ts)2 / (24 hrs)^2 = (8Rw)^3 / (Rw)^3                  (Ts)^2 = (8)^3 * (24hrs)^2                      Ts = 543.05 hrs Thus, the time period of space station is 543.05 hrs             (Ts)2 / (24 hrs)^2 = (8Rw)^3 / (Rw)^3                  (Ts)^2 = (8)^3 * (24hrs)^2                      Ts = 543.05 hrs Thus, the time period of space station is 543.05 hrs

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