Four particles,each of mass $M$ and equidistant from each other,move along a circle of radius $R$ under the action of their mutual gravitational attraction. The speed of each particle is

The dependence of the intensity of the gravitational field $(E)$ of the Earth on the distance $(r)$ from the center of the Earth is correctly represented by:

Three particles of mass $1 \, kg$ each are placed at $(0, 0)$,$(0, 0.2 \, m)$,and $(0.2 \, m, 0)$. What is the net gravitational force on the particle placed at the origin?

Three particles,each of mass $M$,are situated at the vertices of an equilateral triangle of side length $L$. The only forces acting on the particles are their mutual gravitational forces. It is desired that each particle moves in a circle while maintaining the original separation $L$. The initial speed that should be given to each particle is

The orbital velocity of a body near the surface of a planet $A$ is equal to the escape velocity of a body from the planet $B$. If the masses of planets $A$ and $B$ are the same,the ratio of their radii is

$A$ large spherical mass $M$ is fixed at one position and two identical point masses $m$ are kept on a line passing through the centre of $M$ (see figure). The point masses are connected by a rigid massless rod of length $\ell$ and this assembly is free to move along the line connecting them. All three masses interact only through their mutual gravitational interaction. When the point mass nearer to $M$ is at a distance $r = 3\ell$ from $M$,the tension in the rod is zero for $m = k\left(\frac{M}{288}\right)$. The value of $k$ is