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NEET · Physics · STD 11 - 7. gravitation
Assuming that the gravitational potential energy of an object at inflinity is zero, the change in potential energy (final-initial) of an object of mass \(\mathrm{m}\), when to a height \(h\) from the surface of earth (of radius \(\mathrm{R}\) ), is given
- A \(-\frac{\mathrm{GMm}}{\mathrm{R}+\mathrm{h}}\)
- B \(\frac{\mathrm{GMmh}}{\mathrm{R}(\mathrm{R}+\mathrm{h})}\)
- C \(mgh\)
- D \(\frac{\mathrm{GMm}}{\mathrm{R}+\mathrm{h}}\)
Answer & Solution
Correct Answer
(B) \(\frac{\mathrm{GMmh}}{\mathrm{R}(\mathrm{R}+\mathrm{h})}\)
Step-by-step Solution
Detailed explanation
\(\Delta \mathrm{U}=-\mathrm{GMm}\left[\frac{1}{\mathrm{r}_{\mathrm{f}}}-\frac{1}{\mathrm{r}_{\mathrm{i}}}\right]\)\(=-\mathrm{GMm}\left[\frac{1}{\mathrm{R}+\mathrm{h}}-\frac{1}{\mathrm{R}}\right]\)\(=\frac{\mathrm{GMmh}}{\mathrm{R}(\mathrm{R}+\mathrm{h})}\)
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