A body of mass 'm' is moved to height equal to the radius of the earth 'R'. The Increase in its potential energy is 1)2mgR 2)1/2mgR 3)1/4mgR 4)mgR

Asked by  | 9th Feb, 2012, 03:49: PM

Expert Answer:

P.E of the body on surface of the earth = -GMm/R
P.E of body at a height h from the surface of earth =  -GMm/(R+h)
Difference in P.E = -GMm/(R+h) -  -GMm/R
= GMm(1/R - 1/(R+h) )
= mgR^2 h/(R(R+h))
Because, g=GM/R^2
When h=R
Change in P.E = 1/2 (mgR)

Answered by  | 9th Feb, 2012, 07:08: PM

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