The radius of the orbit of an Earth satellite is $R$. Its kinetic energy is proportional to:

Two satellites $A$ and $B$ having masses in the ratio $4: 3$ are revolving in circular orbits of radii $3r$ and $4r$ respectively around the earth. The ratio of total mechanical energy of $A$ to $B$ is.

The energy required to take a satellite to a height $h$ above the Earth's surface (radius of Earth $R = 6.4 \times 10^3 \, km$) is $E_1$,and the kinetic energy required for the satellite to be in a circular orbit at this height is $E_2$. The value of $h$ for which $E_1 = E_2$ is:

When a satellite going round the Earth in a circular orbit of radius $r$ and speed $v$ loses some of its energy,then $r$ and $v$ change as:

In a satellite,if the time of revolution is $T$,then the kinetic energy $(K.E.)$ is proportional to:

$A$ satellite of earth of mass $m$ is moved from an orbital radius of $2R$ to $3R$. Calculate the minimum work done.