Gujarati
Hindi
7.Gravitation
normal

The radius of a planet is $R$. A satellite revolves around it in a circle of radius $r$ with angular velocity $\omega _0.$ The acceleration due to the gravity on planet’s surface is

A

$\frac{r^3\omega _0}{ R}$

B

$\frac{r^3\omega _0^3}{ R}$

C

$\frac{r^3\omega _0^2}{ R}$

D

$\frac{r^3\omega _0^2}{ R^2}$

Solution

As, $F=m a$ $\frac{G M m}{r^{2}}=m \omega_{0}^{2} r$

$G M=\omega_{0}^{2} r^{3}$

$F=m a$

$a=\frac{F}{m}$

$a=\frac{G M}{R^{2}}=\frac{\omega_{0}^{2} r^{3}}{R^{2}}$

Standard 11
Physics

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