What is the acceleration due to gravity at a distance of $2R$ from the surface of the Earth,where $R$ is the radius of the Earth?

If a man at the equator would weigh $(3/5)^{th}$ of his weight,the angular speed of the earth is

If a planet has a mass and radius both half that of the Earth,the acceleration due to gravity at its surface would be ......... $m/s^2$ ($g$ on Earth $= 9.8\, m/s^2$)

The weight of a body at the centre of the earth is

The depth $d$ at which the value of acceleration due to gravity becomes $\frac{1}{n-1}$ times the value at the earth's surface is ($R=$ radius of the earth).

Let $g$ be the acceleration due to gravity at the Earth's surface and $K$ be the rotational kinetic energy of the Earth. Suppose the Earth's radius decreases by $2 \%$ keeping its mass the same,then: