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 :
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a
2mg0R2R+h
b
−2mg0R2R+h
c
mg0R22(R+h)
d
−mg0R22(R+h)
answer is D.
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Detailed Solution
Total energy =−GMem2(R+h)∵g0=GMeR2⇒Me=g0R2G∴Energy =−mg0R22(R+h)