The density of a new planet is twice that of earth. The acceleration due to gravity at the surface of the planet is equal to that at the surface of earth. If $R$ is the radius of earth,then the radius of the planet would be

The time period of a simple pendulum on the surface of the earth is $T$. The height above the surface of the earth at which the time period of the pendulum becomes $2T$ is (Radius of the earth $= 6400 \text{ km}$) (in $\text{ km}$)

Two satellites $P$ and $Q$ are moving in different circular orbits around the Earth (radius $R$). The heights of $P$ and $Q$ from the Earth's surface are $h_p$ and $h_Q$,respectively,where $h_p = R / 3$. The accelerations of $P$ and $Q$ due to Earth's gravity are $g_p$ and $g_Q$,respectively. If $g_p / g_Q = 36 / 25$,what is the value of $h_Q$?

The gravitational potential at a point above the surface of the Earth is $-5.12 \times 10^7 \,J/kg$ and the acceleration due to gravity at that point is $6.4 \,m/s^2$. Assume that the mean radius of the Earth is $6400 \,km$. The height of this point above the Earth's surface is: (in $\,km$)

If the value of $g$ (acceleration due to gravity) at the Earth's surface is $10 \ m/s^2$,what is its value in $m/s^2$ at the center of the Earth,assuming the Earth is a sphere of radius $R$ meters with uniform mass density?

The rotation of the earth having radius $R$ about its axis speeds up to a value such that a man at latitude angle $60^{\circ}$ feels weightless. The duration of the day in such case will be: