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

$A$ body projected vertically from the earth reaches a height equal to earth's radius before returning to the earth. The power exerted by the gravitational force is greatest

Take the mean distance of the moon and the sun from the earth to be $0.4 \times 10^6 \, km$ and $150 \times 10^6 \, km$ respectively. Their masses are $8 \times 10^{22} \, kg$ and $2 \times 10^{30} \, kg$ respectively. The radius of the earth is $6400 \, km$. Let $\Delta F_1$ be the difference in the forces exerted by the moon at the nearest and farthest points on the earth and $\Delta F_2$ be the difference in the force exerted by the sun at the nearest and farthest points on the earth. Then,the number closest to $\frac{\Delta F_1}{\Delta F_2}$ is

Three identical spheres of mass $m$ are placed at the vertices of an equilateral triangle of side length $a$. When released,they interact only through gravitational force and collide after a time $T = 4 \text{ s}$. If the sides of the triangle are increased to length $2a$ and the masses of the spheres are made $2m$,then they will collide after . . . . . . seconds.

The magnitudes of the gravitational field at distances $r_1$ and $r_2$ from the centre of a uniform sphere of radius $R$ and mass $M$ are $F_1$ and $F_2$ respectively. Then-

The distance between the Sun and the Earth is $1.6 \times 10^{11} \,m$ and the radius of the Earth is $6.4 \times 10^6 \,m$. The ratio of the angular momentum of the Earth around the Sun to the angular momentum around its own axis is approximately (Assume the Earth as a solid sphere with uniform mass density and that it rotates around the Sun in a circular path.)