WBJEE · Physics · Gravitation
The change in the gravitational potential energy when a body of mass \(m\) is raised to a height \(n \mathrm{R}\) above the surface of the Earth is (here \(R\) is the radius of the Earth)
- A \(\left(\frac{n}{n+1}\right) m g \mathrm{R}\)
- B \(\left(\frac{n}{n-1}\right) m g \mathrm{R}\)
- C \(n m g R\)
- D \(\frac{m g R}{n}\)
Answer & Solution
Correct Answer
(A) \(\left(\frac{n}{n+1}\right) m g \mathrm{R}\)
Step-by-step Solution
Detailed explanation
Hints: \(\Delta \mathrm{U}=\frac{m g h}{1+\frac{h}{\mathrm{R}}}=\frac{m g \times n \mathrm{R}}{1+\frac{n \mathrm{R}}{\mathrm{R}}}=\frac{n m g \mathrm{R}}{n+1}\)
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