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The change in the gravitational potential energy when a body of mass m is raised to a height nR above the surface of the earth is (here R is the radius of the earth) |
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Answer» `((n)/(n+1))mgR` `U=-(GM m)/(r )` At the surface of earth r=R, `:. U_(S)=-(GM m)/(R )=-mgR""(`:.`g=(GM)/(R^(2))` At the HEIGHT h=nR from the surface of earth `r=R+h=R+nR-R(1+n)` `U_(h)=-(GM m)/(R(1+n))=-(mgR)/((1+n))` `:.` Change in gravitational potential energy is `Delta U=U_(h)-U_(S)=-(mgR)/((1+n))-(-mgR)` `=-(mgR)/(1+n)+mgR=mgR(1-(1)/(1+n))=mgR((n)/(1+n))` |
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