If the gravitational potential on the surface of the earth is $V_0$,then the potential at a point at a height equal to half of the radius of the earth is ..........

The gravitational potential energy of a body on the surface of the earth is $E$. If the body is taken from the surface of the earth to a height equal to $150 \%$ of the radius of the earth,its gravitational potential energy is (in $E$)

Three masses $200 \ kg$,$300 \ kg$,and $400 \ kg$ are placed at the vertices of an equilateral triangle with sides $20 \ m$. They are rearranged on the vertices of a bigger triangle of side $25 \ m$ with the same center. The work done in this process is . . . . . . $J$.

$A$ body of mass $m$ is taken from the Earth's surface to a height $h$ equal to twice the radius of the Earth $(R_e)$. The increase in potential energy will be: (where $g$ is the acceleration due to gravity on the surface of the Earth)

The gravitational field in a region is given by the equation $E = (5\hat{i} + 12\hat{j}) \text{ N/kg}$. If a particle of mass $2 \text{ kg}$ is moved from the origin $(0, 0)$ to the point $(12 \text{ m}, 5 \text{ m})$ in this region, the change in gravitational potential energy is: (in $\text{ J}$)

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