A meteor of mass ' $m$ ' having a speed ' $V$ ' at infinity reaches the surface of the Earth with a speed of ( $v_c$ is escape speed from the Earth's surface)

If the orbital speed of a body revolving in a circular path near the surface of the Earth is $8 \mathrm{kms}^{-1}$, then the orbital speed of a body revolving around the Earth in a circular orbit at height of $19,200 \mathrm{~km}$ from the surface of Earth is (Radius of the Earth $=6400 \mathrm{~km}$ )

A body is projected from the Earth's surface with a speed $\sqrt{5}$ times the escape speed $\left(V_e\right)$. The speed of the body when it escapes from the gravitational influence of the Earth is

The ratio of the time periods of a simple pendulum at heights $2 R_E$ and $3 R_E$ from the surface of the Earth is ( $R_E$ is radius of the Earth)