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 $6400 \, km$).

The acceleration due to gravity on the Moon is $\frac{1}{6}$ times that of the Earth. If the ratio of the density of the Earth $(\rho_e)$ to the Moon $(\rho_m)$ is $\frac{\rho_e}{\rho_m} = \frac{5}{3}$,what is the radius of the Moon $(R_m)$ in terms of the radius of the Earth $(R_e)$?

Assertion $(A)$: $A$ particle of mass $m$ dropped into a hole made along the diameter of the Earth from one end to the other possesses simple harmonic motion.
Reason $(R)$: Gravitational force between any two particles is inversely proportional to the square of the distance between them.

$A$ body is taken to a height of $n R$ from the surface of the earth. The ratio of the acceleration due to gravity on the surface to that at the altitude is

If the acceleration due to gravity experienced by a point mass at a height $h$ above the surface of the Earth is the same as the acceleration due to gravity at a depth $d = \alpha h$ $(h \ll R_{e})$ from the Earth's surface,then the value of $\alpha$ will be: (use $R_{e} = 6400 \ km$)

How does the acceleration due to gravity $(g)$ at a location on Earth change with latitude?