At what height above the surface of the Earth does the value of $g$ decrease by $2 \%$? [Radius of the Earth is $6400 \, km$]

If it is assumed that the spinning motion of the Earth increases,then the weight of a body on the equator

Assuming Earth to be a sphere of uniform density,what is the value of gravitational acceleration in a mine $100 \, km$ below the Earth's surface? (Given $R = 6400 \, km$,$g = 9.8 \, m/s^2$)

The gravitational force of the Earth on a rocket at a height $h$ from the surface of the Earth is $\frac{1}{3}$ of the gravitational force on the surface of the Earth. Find the relationship between $h$ and $R_e$ (the radius of the Earth).

The depth from the surface of the earth of radius $R$,at which the acceleration due to gravity will be $60 \%$ of its value on the earth's surface,is:

$A$ box weighs $196 \; N$ on a spring balance at the North Pole. Its weight recorded on the same balance if it is shifted to the equator is close to ....... $N$ (Take $g = 10 \; m/s^2$ at the North Pole and the radius of the Earth $R = 6400 \; km$).