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A satellite moves in a circular orbit of radius R around the Earth. Another satellite is placed in a circular orbit of radius 1.01R. By approximately how much is the period of the second satellite greater than that of the first?

1. 0.5%
2. 1.0%
3. 1.5%
4. 3.0%

Correct answer: 1.5%

Solution

By Kepler's third law T proportional to R^(3/2). The ratio of periods is (1.01)^(3/2) ~ 1.015, an increase of about 1.5%.

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