Given radius of Earth ‘R’ and length of a day ‘T’ the height of a geostationary satellite is [G–Gravitational Constant, M–Mass of Earth]

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4π2GMT21/3

4πGMR21/3−R

GMT24π21/3−R

GMT24π21/3+R

answer is C.

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Detailed Solution

T=2πr3GM⇒T2=4π2GM(R+h)3⇒R+h=GMT24π21/3⇒h=GMT24π213−R

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