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Question:

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

mg0R2/(R+h)

mg0R2/2(R+h)

-mg0R2/2(R+h)

2mg0R2/(R+h)

Solution:

Total energy of the satellite T = -GMm/2r where r = R + h
∴ T = -GMm/2(R+h)
Acceleration due to gravity at earth's surface g0 = GM/R2 ⇒ GM = g0R2
⇒ T = -mg0R2/2(R+h)