The orbital angular momentum of a satellite is $L$,when it is revolving in a circular orbit at height $h$ from the earth's surface. If the distance of the satellite from the earth's centre is increased by eight times its initial value,then the new angular momentum will be $............\,L$.

Two satellites of mass $m$ and $9m$ are orbiting a planet in orbits of radius $R$. Their periods of revolution will be in the ratio of

$A$ satellite revolving around the earth at a certain height experiences acceleration due to gravity equal to $\frac{16}{49} g_0$,where $g_0$ is the acceleration due to gravity on the earth's surface. If $R$ is the radius of earth,then the square of time period of the satellite's revolution is equal to $K\left[\frac{\pi^2 R^3}{G M}\right]$. The value of $K$ is

Consider a star of mass $m_2 \ kg$ revolving in a circular orbit around another star of mass $m_1 \ kg$ with $m_1 \gg m_2$. The heavier star slowly acquires mass from the lighter star at a constant rate of $\gamma \ kg/s$. In this transfer process,there is no other loss of mass. If the separation between the centers of the stars is $r$,then its relative rate of change $\frac{1}{r} \frac{dr}{dt} \ (\text{in } s^{-1})$ is given by:

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 ${T^2}$ is proportional to:

The time period of a satellite,revolving above earth's surface at a height equal to $R$ will be (Given $g = \pi^2 \ m/s^2$,$R =$ radius of earth).