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Question: The change in potential energy when the body of mass \(m\) is raised to height \[nR\] from earth’s s...

The change in potential energy when the body of mass mm is raised to height nRnR from earth’s surface is ( RR=radius of the earth.)

& A.mgR\dfrac{n}{(n-1)} \\\ & B.mgR \\\ & C.mgR\dfrac{n}{(n+1)} \\\ & D.mgR\dfrac{{{n}^{2}}}{({{n}^{2}}+1)} \\\ \end{aligned}$$
Explanation

Solution

Potential energy is the energy stored by the object which is a certain position with respect to another massive body. Gravitational potential energy is the energy due to the energy experienced by one body due the gravity of another massive body.

Formula used:
U=−GMmRU=\dfrac{-GMm}{R}

Complete step-by-step answer:
Consider a body with mass mm at a distance nRnR from the earth, where the RR is the radius of the earth. Let the mass of the earth be denoted by MM. Clearly the mass of earth is greater than the mass of the body or M>>mM >> m.
Then the gravitational potential experienced by the body on the surface earth is given by, U=−GMmRU=\dfrac{-GMm}{R} here the negative sign indicates that the forces are attractive in nature. Which means that the mass mm, is attracted to the earth.
Also if the body is moved to a height hh, then the potential due to the earth at height nRnR above the surface such that h=nR+Rh=nR+R is given as Uh=−GMmnR+RU_{h}=\dfrac{-GMm}{nR+R}
Then we can calculate the change in potential experienced by the body which is given as ΔU=Uh−U\Delta U=U_{h}-U
Then substituting the values we get, ΔU=−GMmnR+R−−GMmR\Delta U=\dfrac{-GMm}{nR+R}-\dfrac{-GMm}{R}
Or, ΔU=−GMm(1nR+R−1R)\Delta U=-GMm\left(\dfrac{1}{nR+R}-\dfrac{1}{R}\right)
Or, ΔU=−GMm(nR(n+1))\Delta U=-GMm\left(\dfrac{n}{R(n+1)}\right)
For simplification, we know that g=−GMR2g=\dfrac{-GM}{R^{2}}
Let us replace the value, then we get ΔU=mgR(nn+1)\Delta U=mgR\left(\dfrac{n}{n+1}\right)
Hence, the answer is C.mgRn(n+1)C.mgR\dfrac{n}{(n+1)}

So, the correct answer is “Option C”.

Note: The gravitational potential energy depends directly on the mass of the object and its distance with respect to the massive body. It is independent of the mass of the massive body, as the mass of the object is almost negligible when compared to the mass of the massive body. Thus we are using gg to show the independence of their massive body.