A) \[\propto {{r}^{3}}\]
B) \[\propto {{r}^{7/2}}\]
C) \[\propto {{r}^{9/2}}\]
D) \[\propto {{r}^{3/2}}\]
Correct Answer: B
Solution :
For a satellite rotating around a planet in circular motion Gravitational force = centripetal force \[\frac{GM}{{{r}^{5/2}}}=\frac{m{{v}^{2}}}{r}\] \[\Rightarrow \] \[{{v}^{2}}=\frac{GM}{{{r}^{3/2}}}\] \[\Rightarrow \] \[{{T}^{2}}={{\left( \frac{2\pi r}{v} \right)}^{2}}\] \[=\frac{4{{\pi }^{2}}{{r}^{2}}}{GM/{{r}^{3/2}}}=\frac{4{{\pi }^{2}}{{r}^{7/2}}}{GM}\] \[\Rightarrow \] \[{{T}^{2}}\propto \,{{r}^{7/2}}\]You need to login to perform this action.
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