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(B) For a satellite revolving in a circular orbit of radius $r$ around the Earth of mass $M_e$,the gravitational force provides the necessary centripe

Vedclass · Accepted Answer

$T \propto \sqrt{\frac{r^3}{GM}}$

Vedclass · Answer

(B) For a satellite revolving in a circular orbit of radius $r$ around the Earth of mass $M_e$,the gravitational force provides the necessary centripetal force.$\frac{GM_e M}{r^2} = \frac{M v^2}{r}$Since $v = \frac{2\pi r}{T}$,we substitute this into the equation:$\frac{GM_e}{r^2} = \frac{(2\pi r / T)^2}{r}$$\frac{GM_e}{r^2} = \frac{4\pi^2 r^2}{T^2 r}$$T^2 = \frac{4\pi^2 r^3}{GM_e}$$T = 2\pi \sqrt{\frac{r^3}{GM_e}}$Thus,$T \propto \sqrt{\frac{r^3}{GM_e}}$. Comparing this with the given options,the correct proportionality is $T \propto \sqrt{\frac{r^3}{GM}}$.

$A$ satellite of mass $M$ is revolving in a circular orbit of radius $r$ around the Earth. The time period of revolution of the satellite is:

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