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Question

A geostationary satellite is orbiting around an arbitrary planet P at a height of 11R above the surface of P, R being the radius of P. The time period of another satellite in hours at a height of 2R from the surface of P is ___. P has the time period of rotation of 24 hours.

A
62
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B
3
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C
62
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D
5
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Solution

The correct option is B 3
From Kepler's law: T2R3where R is distance of satellite from the centre of the planet.

Since time period of a geostationary satellite is equal to the time period of rotation of planet. In this case that is equal to 24 hours.

Using Kepler's law
(24T)2=(12R3R)3

T=3 hour

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