MHT CET 2023 Physics question
Periodic time of a satellite revolving above the earth's surface at a height equal to radius of the earth 'R' is [g = acceleration due to gravity]
Answer: (B)
Solution
Orbital radius . For a circular orbit and . Hence \sqrt{\frac{(2R)^3}{gR^2}}\sqrt{\frac{8R}{g}}\sqrt{\frac{2R}{g}}$ =
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