Calculate the difference in the value of $g$ at the equator and at the poles due to the rotation of the Earth.

State whether the following statements are true or false:
$(a)$ $G$ is a vector quantity.
$(b)$ The relation $g = \frac{GM}{r^2}$ holds well for all celestial bodies.
$(c)$ If the Earth suddenly stops rotating,the value of acceleration due to gravity at the equator will decrease.

$T$ is the time period of a simple pendulum on the Earth's surface. Its time period becomes $xT$ when taken to a height $R$ (equal to the Earth's radius) above the Earth's surface. Then,the value of $x$ will be:

$A$ body weighs $300 \ N$ on the surface of the earth. How much will it weigh at a distance $\frac{R}{2}$ below the surface of the earth (in $N$)? ($R$ is the radius of the earth.)

The lengths of seconds pendulums on the surface of the earth and at an altitude '$h$' from the surface of the earth are $l_s$ and $l_h$ respectively. The radius of the earth is

$A$ ball is launched from the top of Mt. Everest,which is at an elevation of $9000 \, m$. The ball moves in a circular orbit around the Earth. Acceleration due to gravity near the Earth's surface is $g$. The magnitude of the ball's acceleration while in orbit is