If potential energy of a body of mass $m$ on the surface of earth is taken as zero then its potential energy at height $h$ above the surface of earth is [ $R$ is radius of earth and $M$ is mass of earth]
$\frac{-G M m}{R+h}$
$\frac{-G M m}{h}$
$\frac{G M m h}{R(R+h)}$
$\frac{G M m h}{h+2 R}$
Asatellite is launched into a circular orbit of radius $R$ around the earth. A second satellite is launched into an orbit of radius $1.02\,R.$ The period of second satellite is larger than the first one by approximately ........ $\%$
$Assertion$ : The escape speed does not depend on the direction in which the projectile is fired.
$Reason$ : Attaining the escape speed is easier if a projectile is fired in the direction the launch site is moving as the earth rotates about its axis.
The force of gravitation is
The value of escape velocity on a certain planet is $2\, km/s$ . Then the value of orbital speed for a satellite orbiting close to its surface is
In a certain region of space, the gravitational field is given by $-k/r$ , where $r$ is the distance and $k$ is a constant. If the gravitational potential at $r = r_0$ be $V_0$ , then what is the expression for the gravitational potential $(V)$ ?