A) \[\frac{5GM}{4{{r}^{2}}}\]
B) \[\frac{4GM}{3{{r}^{2}}}\]
C) \[\frac{3GM}{2{{r}^{2}}}\]
D) \[\frac{2GM}{{{r}^{2}}}\]
Correct Answer: B
Solution :
\[{{E}_{g}}=+\,\frac{Gm}{{{r}^{2}}}\,+\frac{Gm}{{{(2r)}^{2}}}..............\infty \] \[=+\,\frac{Gm}{{{r}^{2}}}\,+(1+\frac{1}{4}\,+\frac{1}{16}\,+...........\infty )\] \[=+\,\frac{Gm}{{{r}^{2}}}\,+\left[ \frac{1}{1-\frac{1}{4}} \right]\,=\frac{4Gm}{3{{r}^{2}}}\]You need to login to perform this action.
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