The height of any point $P$ above the surface of earth is equal to diameter of earth. The value of acceleration due to gravity at point $P$ will be : (Given $g=$ acceleration due to gravity at the surface of earth)

If the radius of the earth were to shrink by $1\%$ its mass remaining the same, the acceleration due to gravity on the earth's surface would

A clock $S$ is based on oscillation of a spring and a clock $ P$ is based on pendulum motion. Both clocks run at the same rate on earth. On a planet having the same density as earth but twice the radius

Figure shows variation of acceleration due to gravity with distance from centre of a  uniform spherical planet, Radius of planet is $R$. What is $r_2 -r_1.$

The height at which the acceleration due to gravity becomes $\frac{g}{9}$ (where $g$ = the acceleration due to gravity on the surface of the earth) in terms of $R$, the radius of the earth, is

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 ?