Suppose the gravitational force varies invers | vedclass

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Question

$\mathrm{T}=\frac{2 \pi \mathrm{r}}{\mathrm{v}}$

and $\quad \frac{\mathrm{mv}^{2}}{\mathrm{r}}=\frac{\mathrm{k}}{\mathrm{r}^{\mathrm{n}}} \Rightarr

Vedclass · Accepted Answer

${R^{\left( {n + 1} \right)/2}}$

Vedclass · Answer

$\mathrm{T}=\frac{2 \pi \mathrm{r}}{\mathrm{v}}$

and $\quad \frac{\mathrm{mv}^{2}}{\mathrm{r}}=\frac{\mathrm{k}}{\mathrm{r}^{\mathrm{n}}} \Rightarrow \mathrm{v}=\frac{\mathrm{A}}{\mathrm{r}^{\frac{\mathrm{n}-1}{2}}}$

$	herefore \mathrm{T}=\frac{2 \pi \mathrm{r}}{\mathrm{A}} \mathrm{r}^{\frac{\mathrm{n}-1}{2}}=\frac{2 \pi}{\mathrm{A}} \mathrm{r}^{\frac{\mathrm{n}+1}{2}} \Rightarrow \mathrm{T} \propto \mathrm{r}^{\frac{\mathrm{n}+1}{2}}$

Suppose the gravitational force varies inversely as the $n^{th}$ power of the distance. Then, the time period of a planet in circular orbit of radius $R$ around the sun will be proportional to

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