If the gravitational field in space is given as $E = -\frac{K}{r^2}$. Taking the reference point at $r = 2\,cm$ with gravitational potential $V = 10\,J/kg$. Find the gravitational potential at $r = 3\,cm$ in $SI$ units. (Given: $K = 6\,J\cdot cm/kg$)

If three particles each of mass $M$ are placed at the corners of an equilateral triangle of side $a$,the potential energy of the system and the work done if the side of the triangle is changed from $a$ to $2a$,are

An infinite number of bodies,each of mass $2 \, kg$,are situated on the $x-$axis at distances $1 \, m, 2 \, m, 4 \, m, 8 \, m, \dots$ respectively,from the origin. The resulting gravitational potential due to this system at the origin will be:

On the $x$-axis and at a distance $x$ from the origin,the gravitational field due to a mass distribution is given by $\frac{Ax}{(x^2+a^2)^{3/2}}$ in the $x$-direction. The magnitude of gravitational potential on the $x$-axis at a distance $x$,taking its value to be zero at infinity,is

$A$ uniform solid sphere of mass $M$ and radius $a$ is surrounded by a concentric uniform thin spherical shell of mass $0.5 M$ and radius $1.5 a$. The gravitational potential energy of a unit mass kept at a distance of $2.5 a$ from the center is

$A$ body of mass $m$ is raised through a height $h$ above the Earth's surface such that the increase in potential energy is $\frac{mgR}{5}$. The height to which the body is raised is ($R=$ radius of Earth,$g=$ acceleration due to gravity).