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Question

(A) For the planet to orbit around the star,the centripetal force must be provided by the gravitational force. Let the gravitational force be $F_g \pr

Vedclass · Accepted Answer

$r^{9/2}$

Vedclass · Answer

(A) For the planet to orbit around the star,the centripetal force must be provided by the gravitational force. Let the gravitational force be $F_g \propto r^{-7/2}$. The centripetal force required for circular motion is $F_c = m \omega^2 r$,where $\omega = \frac{2\pi}{T}$. Equating the forces,we have $m \omega^2 r \propto r^{-7/2}$. Since $m$ is constant,$\omega^2 r \propto r^{-7/2}$. Dividing by $r$,we get $\omega^2 \propto r^{-9/2}$. Substituting $\omega = \frac{2\pi}{T}$,we get $\left(\frac{2\pi}{T}\right)^2 \propto r^{-9/2}$. This implies $\frac{1}{T^2} \propto r^{-9/2}$. Therefore,$T^2 \propto r^{9/2}$.

Consider a light planet revolving around a massive star in a circular orbit of radius $r$ with time period $T$. If the gravitational force of attraction between the planet and the star is proportional to $r^{-7/2}$,then $T^2$ is proportional to:

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