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
$\frac{r^3\omega _0}{ R}$
$\frac{r^3\omega _0^3}{ R}$
$\frac{r^3\omega _0^2}{ R}$
$\frac{r^3\omega _0^2}{ R^2}$
The dependence of acceleration due to gravity $'g'$ on the distance $'r'$ from the centre of the earth, assumed to be a sphere of radius $R$ of uniform density is as shown in figure below
Suppose the gravitational force varies inversely as the $n^{th}$ power of distance. Then the time period of a planet in circular orbit of radius $R$ around the sun will be proportional to
Imagine a light planet revolving around a very massive star in a circular orbit of radius $R$ with a period of revolution $T$. If the gravitational force of attraction between the planet and the star is proportional to $R^{-5/2}$, then,
A satellite is launched into a circular orbit of radius $R$ around earth, while a second satellite is launched into a circular orbit of radius $1.02\, {R}$. The percentage difference in the time periods of the two satellites is -
The period of a satellite, in a circular orbit near an equatorial plane, will not depend on