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The change in potential energy of the body when it is taken from the earth's surface to a height above its surface is : |
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Answer» (n-1) mgR `U_(1)=(-GMm)/(R)` At HEIGHT nR from the surface of earth, `U_(2)=(-GMN)/((n+1)R)` `:.` Change in G.P.E is `Delta U=U_(2)-U_(1)` `DeltaU=(-GMn)/((n+1)R)+(GMn)/(R)` `DeltaU=(GMn)/(R)[1-(1)/(n+1)]=(GMn)/(R)((n)/(n+1))` Since`g=(GM)/(R^(2))implies GM=gR^(2)` `DeltaU=(gR^(2)m)/(R)((n)/(n+1))=(n)/(n+1)mgR` |
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