At what depth below the surface of the earth,will the acceleration due to gravity $g$ be half of its value at $1600 \, km$ above the surface of the earth?

$A$ man can jump to a height of $1.5 \, m$ on a planet $A$. What is the height he may be able to jump on another planet whose density and radius are,respectively,one-quarter and one-third that of planet $A$?

The figure shows the variation of acceleration due to gravity with distance from the center of a uniform spherical planet of radius $R$. What is $r_2 - r_1$?

$A$ certain planet completes one rotation about its axis in time $T$. The weight of an object placed at the equator on the planet's surface is a fraction $f$ ($f$ is close to unity) of its weight recorded at a latitude of $60^{\circ}$. The density of the planet (assumed to be a uniform perfect sphere) is given by

$A$ body weighs $W$ newton on the surface of the earth. Its weight at a height equal to half the radius of the earth will be:

The ratio of gravitational acceleration at height $3R$ to that at height $4R$ from the surface of the earth is (where $R$ is the radius of the earth):