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Work done in taking a body of mass m to a height nR above surface of earth will be :
(R = radius of earth)
- a)mgnR
- b)mgR (n/n + 1)
- c)mgR
- d)
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
Work done in taking a body of mass m to a height nR above surface of e...
Potential energy at surface of earth will be PE1 = (G.M.m)/R
Now, potential energy at height nR from the surface of earth,
PE2 = (G.M.m)/(R+nR)
Here, change in PE = PE1 – PE2
= (G.M.m)/R – (G.M.m)/(n+1)R
= (G.M.m)/R[1 – 1/(n+1)]
= (G.M.m)/R[(n + 1 – 1)/(n+1)]
= [n/(n+1)](G.M.m)/R ….(i)
Also, g = GM/(R^2) ….(ii)
From (i) and (ii),
Change in PE = (n/(n+1))mgR
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Work done in taking a body of mass m to a height nR above surface of e...
Potential energy at surface of earth will be PE1 = (G.M.m)/R
Now, potential energy at height nR from the surface of earth,
PE2 = (G.M.m)/(R+nR)
Here, change in PE = PE1 – PE2
= (G.M.m)/R – (G.M.m)/(n+1)R
= (G.M.m)/R[1 – 1/(n+1)]
= (G.M.m)/R[(n + 1 – 1)/(n+1)]
= [n/(n+1)](G.M.m)/R ….(i)
Also, g = GM/(R^2) ….(ii)
From (i) and (ii),
Change in PE = (n/(n+1))mgR
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Work done in taking a body of mass m to a height nR above surface of earth will be :(R = radius of earth)a)mgnRb)mgR (n/n + 1)c)mgRd)Correct answer is option 'B'. Can you explain this answer?
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