$A$ particle is dropped on Earth from a height $R$ (radius of Earth) and it bounces back to a height $R/2$. The coefficient of restitution for the collision is (ignore air resistance and rotation of Earth).

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.)

Weighing the Earth: You are given the following data: $g = 9.81 \; m/s^2$, $R_E = 6.37 \times 10^6 \; m$, the distance to the moon $R = 3.84 \times 10^8 \; m$, and the time period of the moon's revolution is $27.3$ days. Obtain the mass of the Earth $M_E$ in two different ways.

$A$ very long (length $L$) cylindrical galaxy is made of uniformly distributed mass and has radius $R$ $(R << L)$. $A$ star outside the galaxy is orbiting the galaxy in a plane perpendicular to the galaxy and passing through its centre. If the time period of the star is $T$ and its distance from the galaxy's axis is $r$,then:

$A$ star of mass $M$ and radius $R$ is made up of gases. The average gravitational pressure compressing the star due to the gravitational pull of the gases making up the star depends on $R$ as

Two spherical bodies of mass $M$ and $5M$ and radii $R$ and $2R$ are released in free space with initial separation between their centres equal to $12R.$ If they attract each other due to gravitational force only,then the distance covered by the smaller body before collision is (in $R$)