At what height above the earth’s surface does the acceleration due to gravity fall to $1\%$ of its value at the earth’s surface ?

The acceleration due to gravity on the earth's surface at the poles is $g$ and angular velocity of the earth about the axis passing through the pole is $\omega .$ An object is weighed at the equator and at a height $h$ above the poles by using a spring balance. If the weights are found to be same, then $h$ is $:( h << R ,$ where $R$ is the radius of the earth)

Assuming the earth to be a sphere of uniform density the acceleration due to gravity

Weight of $1 \,kg$ becomes $1/6$ on moon. If radius of moon is $1.768 \times {10^6}\,m$, then the mass of moon will be

Mass density of a solid sphere is $\rho $ . Radius of the sphere is $R$. The gravitational field at a distance $r$ from the centre of the sphere inside it is

A uniform spherical planet (Radius $R$) has acceleration due to gravity at its surface $g.$ Points $P$ and $Q$ located inside and outside the planet have acceleration due to gravity $\frac{g}{4} .$ Maximum possible separation between $P$ and $Q$ is