MHT CET · Physics · Gravitation

If \(g\) is the acceleration due to gravity on earth's surface, the gain of the potential energy of an object of mass \(m\) raised from the surface of the earth to a height equal to the radius \(R\) of the earth is

A \(2 \mathrm{mgR}\)
B \(m g R\)
C \(\frac{1}{2} m g R\)
D \(\frac{1}{4} m g R\)

Verified Solution

Answer & Solution

Correct Answer

(C) \(\frac{1}{2} m g R\)

Step-by-step Solution

Detailed explanation

The potential energy of an object at the surface of the earth
\(U_{1}=-\frac{G M m}{R}\)
The potential energy of the object at a height \(h=R\) from the surface of the earth
\(U_{2}=-\frac{G M m}{R+h}=-\frac{G M m}{R+R}\)
Hence, the gain in potential energy of the object
\(\begin{aligned}\Delta U &=U_{2}-U_{1} \\\Delta U &=-\frac{G M m}{R+R}+\frac{G M m}{R} \\\Delta U &=-\frac{G M m}{2 R}+\frac{G M m}{R} \\\Delta U &=\frac{1}{2} \frac{G M m}{R}\end{aligned}\)
But we know that \(G M=g R^{2}\)
Hence,
\(\Delta U=\frac{1}{2} \frac{g R^{2} m}{R}\)
or
\(\Delta U=\frac{1}{2} g R m\)
or
\(\Delta U=\frac{1}{2} m g R\)

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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