Assuming the Earth to be a sphere of uniform mass density,the weight of a body at a depth $d = R/2$ from the surface of the Earth,if its weight on the surface of the Earth is $200 \, N$,will be $........... \, N$ (Given $R =$ Radius of Earth).

Imagine Earth to be a solid sphere of mass $M$ and radius $R$. If the value of acceleration due to gravity at a depth $d$ below Earth's surface is the same as its value at a height $h$ above its surface and equal to $\frac{g}{4}$ (where $g$ is the value of acceleration due to gravity on the surface of Earth),the ratio of $\frac{h}{d}$ will be

The percentage decrease in the weight of a rocket,when taken to a height of $32 \ km$ above the surface of the Earth,will be $.....\%$
(Radius of Earth $= 6400 \ km$)

The maximum vertical distance through which a fully dressed astronaut can jump on the earth is $0.5\, m$. If the mean density of the moon is two-thirds that of the earth and the radius is one-quarter that of the earth,what is the maximum vertical distance through which he can jump on the moon and the ratio of the duration of the jump on the moon to that on 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

Earth is assumed to be a sphere of radius $R$. If $g_{\phi}$ is the value of effective acceleration due to gravity at latitude $30^{\circ}$ and $g$ is the value at the equator,then the value of $|g - g_{\phi}|$ is ($\omega$ is the angular velocity of rotation of the Earth,$\cos 30^{\circ} = \frac{\sqrt{3}}{2}$).