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?

Consider two spherical planets of same average density. The second planet is $8$ times as massive as the first planet. The ratio of the acceleration due to gravity of the second planet to that of the first planet is

$A$ $90 \,kg$ body placed at $2R$ distance from the surface of the Earth experiences a gravitational pull of: ($R=$ Radius of Earth,$g=10 \,ms^{-2}$) (in $\,N$)

$A$ simple pendulum performing small oscillations at a height $R$ above the Earth's surface has a time period of $T_1 = 4 \ s$. What would be its time period $T_2$ if it is brought to a point at a height $2R$ from the Earth's surface? Choose the correct relation ($R =$ radius of Earth).

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 length of a seconds pendulum at a height $h=2R$ from the earth's surface will be. (Given: $R =$ radius of earth and acceleration due to gravity at the surface of earth $g = \pi^{2} \ m/s^{2}$)