The additional kinetic energy to be provided to a satellite of mass $m$ revolving around a planet of mass $M$, to transfer it from a circular orbit of radius $R_1$ to another of radius $R_2\,(R_2 > R_1)$ is
$GmM\,\left( {\frac{1}{{{R_1}}} - \frac{1}{{{R_2}}}} \right)$
$2GmM\,\left( {\frac{1}{{{R_1}}} - \frac{1}{{{R_2}}}} \right)$
$\frac{1}{2}GmM\,\left( {\frac{1}{{{R_1}}} - \frac{1}{{{R_2}}}} \right)$
$GmM\,\left( {\frac{1}{{R_1^2}} - \frac{1}{{R_2^2}}} \right)$
A body of mass $m$ is lifted up from the surface of the earth to a height three times the radius of the earth. The change in potential energy of the body is
where $g$ is acceleration due to gravity at the surface of earth.
A satellite in force free space sweeps stationary interplanetary dust at a rate of $\frac{d M}{d t}=\alpha v$ where $M$ is mass and $v$ is the speed of satellite and $\alpha$ is a constant. The acceleration of satellite is
Assume that a tunnel is dug through earth from North pole to south pole and that the earth is a non-rotating, uniform sphere of density $\rho $. The gravitational force on a particle of mass $m$ dropped into the tunnel when it reaches a distance $r$ from the centre of earth is
The Earth is assumed to be a sphere of radius $R$. A platform is arranged at a height $R$ from the surface of the Earth. The escape velocity of a body from this platform is $fv$, where $v$ is its escape velocity from the surface of the Earth. the value of $f$ is
A body of mass $m$ falls from a height $R$ above the surface of the earth, where $R$ is the radius of the earth. What is the velocity attained by the body on reaching the ground? (Acceleration due to gravity on the surface of the earth is $g$)