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The time period of a geostationary satellite is 24 h, at a height 6R (R is radius of earth) from surface of earth. The time period of another satellite whose height is 2.5 R from surface will be,

1. 12 / 2.5 h
2. 6 √2 h
3. 12 √2 h
4. 24 / 2.5 h

Correct answer: 12 √2 h

Solution

The time period of a satellite is proportional to the square root of the cube of its orbital radius (T ∝ R^(3/2)). For the geostationary satellite, the orbital radius is 7R (6R + R). For the second satellite, the orbital radius is 3.5R (2.5R + R). Using the ratio of time periods, T2/T1 = (R2/R1)^(3/2), we find T2 = 12√2 h.

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