The radius of the Earth is $6400 \,km$ and the acceleration due to gravity is $g=10 \,ms^{-2}$. For the weight of a body of mass $5 \,kg$ to be zero at the equator,the rotational angular velocity of the Earth must be (in $rad/s$):

The value of acceleration due to gravity at a height of $4 R_E$ from the surface of the Earth is (where $R_E$ is the radius of the Earth and acceleration due to gravity on the surface of the Earth $g = 10 \,ms^{-2}$): (in $\,ms^{-2}$)

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)

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

At what height over the earth's pole,the free fall acceleration decreases by one percent? (Assume the radius of earth to be $6400 \, km$)

Imagine Earth to be a solid sphere of mass $M$ and radius $R$. If the value of acceleration due to gravity at a depth $d$ below Earth's surface is the same as its value at a height $h$ above its surface and equal to $\frac{g}{4}$ (where $g$ is the value of acceleration due to gravity on the surface of Earth),the ratio of $\frac{h}{d}$ will be