The acceleration of a body due to the attraction of the earth (radius $R$) at a distance $2R$ from the surface of the earth is ($g =$ acceleration due to gravity at the surface of the earth).

The value of acceleration due to gravity at Earth's surface is $9.8\, m\,s^{-2}$. The altitude above its surface at which the acceleration due to gravity decreases to $4.9\, m\,s^{-2}$,is close to: (Radius of Earth $= 6.4 \times 10^6\, m$)

At what altitude in metres will the acceleration due to gravity be $25\%$ of that at the earth's surface? (Radius of earth $= R \ m$)

The ratio of the radii of a planet and the earth is $1: 2$, and the ratio of their mean densities is $4: 1$. If the acceleration due to gravity on the surface of the earth is $9.8 \,ms^{-2}$, then the acceleration due to gravity on the surface of the planet is: (in $\,ms^{-2}$)

$A$ spherical planet has a mass $M$ and diameter $D$. $A$ particle of mass $m$ falling freely near the surface of this planet will experience an acceleration due to gravity equal to:

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