A small mass 2kg moved slowly from the surface of earth to a height of 6.4 x 106 m above the earth. The work done [in Mega Joule] is __________.

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answer is 64.

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Detailed Solution

Wext =ΔU=Uf−Ui=−GM(R+h)−−GMRΔU=mgh1+hR=mgR1+Rh=mgR2ΔU=2×10×6.4×1062⇒ΔU=6.4×107 joule ⇒W=ΔU=64 Mega Joule

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