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Question

A satellite of mass m is orbiting the earth (of radius R) at a height h from its surface. The total energy of the satellite in terms of g0, the value of acceleration due to gravity at the earth's surface is:

A
2mg0R2R+h
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B
2mg0R2R+h
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C
mg0R22(R+h)
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D
mg0R22(R+h)
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Solution

The correct option is D mg0R22(R+h)
We know that the total energy of a satellite is given by TE=12GMm(R+h). Here the symbols have their usual meanings

We also know that the acceleration due to gravity on the earth-surface is g0 given by g0=GMR2
GM=g0R2

TE=12GMm(R+h)

TE=mg0R22(R+h)

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