A planet is revolving around the sun in a cir | vedclass

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Question

As, force $=\frac{G M m}{r^{3 / 2}}=m \omega^{2} r$

$=\frac{G M m}{r^{3 / 2}}=\frac{4 \pi^{2} m r}{T^{2}}\left[\because T=\frac{2 \pi}{\omega}\righ

Vedclass · Accepted Answer

$r^{5 / 2}$

Vedclass · Answer

As, force $=\frac{G M m}{r^{3 / 2}}=m \omega^{2} r$

$=\frac{G M m}{r^{3 / 2}}=\frac{4 \pi^{2} m r}{T^{2}}\left[\because T=\frac{2 \pi}{\omega}\right]$

$\Rightarrow T^{2}=\left(\frac{\Delta \pi^{2}}{G M}\right) \cdot r^{5 / 2}$

$\Rightarrow T^{2} \propto r^{5 / 2}$

A planet is revolving around the sun in a circular orbit with a radius $r$. The time period is $T$. If the force between the planet and star is proportional to $r^{-3 / 2}$, then the square of time period is proportional to

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