At what depth $d$ below the surface of the Earth does the acceleration due to gravity become $\frac{1}{n}$ times its value at the surface? ($R$ = radius of the Earth)

The density of a newly discovered 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 the Earth. If the radius of the Earth is $R$,the radius of the planet would be

The gravitational pull of the moon is $(1/6)^{\text{th}}$ of the earth and the mass of the moon is $(1/8)^{\text{th}}$ of the earth. This implies that the:

The height $h$ above the earth's surface at which the value of acceleration due to gravity $g$ becomes $\frac{g}{3}$ is ($R=$ radius of the earth).

What is the time period of a seconds pendulum on a planet whose mass and radius are twice that of the Earth?

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 $\,km$)