When a body is taken from the pole to the equator,its weight:

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 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 $\text{ km}$)

At what height from the surface of the earth will the value of acceleration due to gravity fall to half of its value on the surface of the earth?

If the density of the Earth increases $4$ times and its radius becomes half of its present value,our weight will

The mass of a planet and its diameter are three times those of the Earth. Then the acceleration due to gravity on the surface of the planet is ....... $m/s^2$.