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Q.

Periodic time of a satellite revolving above Earth’s surface at a height equal to R, radius of Earth, is [g is acceleration due to gravity at Earth’s surface]

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a

2π2Rg

b

42πRg

c

2πRg

d

8πRg

answer is B.

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Detailed Solution

T=2π(R+h)3gR2=2π(2R)3gR2=42πRg
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