At surface of earth potential energy of a particle is $\mathrm{U}$ and potential is $\mathrm{V}$. Change in potential energy and potential at height $\mathrm{h}=\mathrm{R}$ are suppose $∆\mathrm{U}$ and $∆\mathrm{V}$. Then.

If one wants to remove all the mass of the earth to infinity in order to break it up completely. The amount of energy that needs to be supplied will be $\frac{x}{5}\frac{G{M}^2}{R}$, where $x$ is ________.

(Round off to the Nearest Integer)

($M$ is the mass of earth, $R$ is the radius of earth, $G$ is the gravitational constant)

A spherically symmetric gravitational system of particles has mass density $\rho =\left\{\begin{array}{l}{\rho }_0\text{ for }r\le R\\ 0\text{ for }r>R\end{array}\right.$ where, ${\rho }_0$ is a constant. A test mass can undergo circular motion under the influence of the gravitational field of particles. Its speed $v$ as a function of distance $r(0<r<\infty )$ from the centre of the system is represented by

A body of mass $\left(2M\right)$ splits into four masses $\{m,m,M-m,M-m\},$ which are rearranged to form a square as shown in the figure. The ratio of $\frac{M}{m}$ for which, the gravitational potential energy of the system becomes maximum is $x : 1.$ The value of $x$ is ________.

Four spheres each of mass $m$ form a square of side $d$ (as shown in figure). A fifth sphere of mass $M$ is situated at the centre of square. The total gravitational potential energy of the system is