The mean radius of earth is $R$, and its angular speed on its axis is $\omega$. What will be the radius of orbit of a geostationary satellite?
$\left(\frac{R g}{\omega^2}\right)^{1 / 3}$
$\left(\frac{R^2 g}{\omega^2}\right)^{1 / 3}$
$\left(\frac{R^2 g}{\omega}\right)^{1 / 3}$
$\left(\frac{R^2 \omega^2}{g}\right)^{1 / 3}$
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]
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 $(V)$ ?
If $g$ is the acceleration due to gravity on the earth's surface, the gain in the potential energy of an object of mass $m$ raised from the surface of the earth to a height equal to the radius $R$ of the earth, is
An object is taken to height $2 R$ above the surface of earth, the increase in potential energy is $[R$ is radius of earth]
Two identical spheres are placed in contact with each other. The force of gravitation between the spheres will be proportional to ($R =$ radius of each sphere)