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A geostationary satellite is orbiting the earth at a height of 5R above the surface of the earth, R being the radius of the earth. The time period of another satellite in hours at a height of 2R from the surface of the earth is:

1. 5
2. 10
3. 6 √2
4. 6

Correct answer: 6 √2

Solution

The time period of a satellite is proportional to the square root of the cube of its orbital radius (Kepler's Third Law). For the second satellite at a height of 2R, the orbital radius is 3R. Comparing with the geostationary satellite (orbital radius 6R, time period 24 hours), the time period is T = 24 × (3/6)^(3/2) = 6√2 hours.

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