If $g \propto \frac{1}{R^3}$ (instead of $\frac{1}{R^2}$),then the relation between the time period $T$ of a satellite near the Earth's surface and the radius $R$ will be:

The distance between two stars of masses $3 M_S$ and $6 M_S$ is $9 R$. Here $R$ is the mean distance between the centers of the Earth and the Sun,and $M_S$ is the mass of the Sun. The two stars orbit around their common center of mass in circular orbits with period $n T$,where $T$ is the period of Earth's revolution around the Sun. The value of $n$ is. . . . .

$A$ body is projected with a velocity greater than orbital velocity but less than escape velocity. Its path is

$A$ 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 the star is proportional to $r^{-3/2}$,then the square of the time period is proportional to

What is the orbital velocity of a satellite? Derive its equation.

$A$ satellite of mass $m$ is circulating around the Earth with constant angular velocity. If the radius of the orbit is $R_0$ and the mass of the Earth is $M$,the angular momentum about the centre of the Earth is: