What should be the angular speed of the Earth so that a body lying on the equator may appear weightless? $(g = 10\,m/s^2, R = 6400\,km)$

$A$ 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 $g/4$. The maximum possible separation between $P$ and $Q$ is:

The height in terms of radius of the earth $(R)$,at which the acceleration due to gravity becomes $g/9$,where $g$ is acceleration due to gravity on earth's surface,is . . . . . . .

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:

If a hole is bored along the diameter of the earth and a stone is dropped into the hole,what happens to the stone?

At what distance above and below the surface of the earth will a body have the same weight? (Take the radius of the earth as $R$.)