The value of $g$ at the surface of the earth is $9.8 \, m/s^2$. Then the value of $g$ at a place $480 \, km$ above the surface of the earth will be nearly .......... $m/s^2$ (radius of the earth is $6400 \, km$).

In order to make the effective acceleration due to gravity equal to zero at the equator,the angular velocity of rotation of the earth about its axis should be $(g = 10\,m/s^2$ and radius of earth is $6400\,km)$.

The acceleration due to gravity at a height $h$ above the surface of the $Earth$ is $g_h$. At a depth $d = 90 \text{ km}$ below the $Earth$'s surface, the acceleration due to gravity is also $g_h$. The value of $h$ is: (in $\text{ km}$)

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

Define acceleration due to gravity. Give the magnitude of $g$ on the surface of earth.

$A$ body weighs $72 \ N$ on the surface of the earth. What is the gravitational force on it at a height equal to half the radius of the earth (in $N$)?