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?

If the radius of the Earth becomes double while its mass remains unchanged,what will be the change in the weight of an object of mass $m$ on the surface of the Earth?

$A$ 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

The magnitudes of the gravitational field at distances $r_1$ and $r_2$ from the center of a uniform sphere of radius $R$ and mass $M$ are $F_1$ and $F_2$ respectively. Then:

The angular speed of rotation of the Earth about its axis,at which the weight of a man standing on the equator becomes half of his weight at the poles,is given by:

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).