A satellite is orbiting the earth in a circul | vedclass

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Question

$\because$ Satelite is orbiting

$\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\mathrm{r}^{2}}=\frac{\mathrm{mv}^{2}}{\mathrm{r}}$

$\Rightarrow \ma

Vedclass · Accepted Answer

frequency of revolution varies as $\frac{1}{r^{3/2}}$

Vedclass · Answer

$\because$ Satelite is orbiting

$\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\mathrm{r}^{2}}=\frac{\mathrm{mv}^{2}}{\mathrm{r}}$

$\Rightarrow \mathrm{mv}^{2}=\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\mathrm{r}}$

$\mathrm{K} \cdot \mathrm{E}=\frac{1}{2} \mathrm{mv}^{2}=\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{2 \mathrm{r}}$

Angular momentum $=\mathrm{mvr}$

$=\mathrm{m} \sqrt{\frac{\mathrm{GM}}{\mathrm{r}}} \mathrm{r}$

$=\mathrm{L} \propto \mathrm{r}^{1 / 2}$

Momentum $\mathrm{P}=\mathrm{mv}=\mathrm{m} \sqrt{\frac{\mathrm{GM}}{\mathrm{r}}}$

$P$ oc $r^{-1 / 2}$

$\mathrm{T}^{2} \propto \mathrm{r}^{3} \Rightarrow \mathrm{T} \propto \mathrm{r}^{3 / 2} \quad \mathrm{f}=\frac{1}{\mathrm{T}}$

$\Rightarrow \mathrm{f} \propto \mathrm{r}^{-3 / 2}$

A satellite is orbiting the earth in a circular orbit of radius $r.$ Its

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