A satellite orbiting Earth has a time period of 5 hours. If the orbital radius (separation between Earth and satellite) is increased to 4 times its original value, the new time period becomes n times the original. Find the value of n.

Correct answer: 8

Solution

By Kepler's third law, T² proportional to r³. So T2/T1 = (r2/r1)^(3/2) = 4^(3/2) = 8. New period = 5 * 8 = 40 hours. Hence n = 8.

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