Question 3: Gravitation

Question

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

Accepted Answer

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

Answer

$$r^{3 / 2}$$

Answer

$$r^2$$

Answer

$$r$$

Answer

$$r^{5 / 2}$$

Gravitation

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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