Find the gravitational field at a distance of $2000\, km$ from the center of the Earth. (in $m/s^2$)
(Given: $R_{\text{earth}} = 6400\, km$,$r = 2000\, km$,$M_{\text{earth}} = 6 \times 10^{24}\, kg$)

$R$ is the radius of the Earth,$\omega$ is its angular velocity,and $g_p$ is the value of acceleration due to gravity at the poles. The effective value of $g$ at latitude $\lambda = 60^\circ$ will be equal to:

If the radius of the Earth shrinks by $1.5 \%$ (mass remaining same),then the value of gravitational acceleration changes by ......... $\%$

The density of a planet is $3$ times that of Earth and its radius is $2.5$ times that of Earth. If $g_p$ and $g_e$ represent the acceleration due to gravity on the planet and Earth respectively, then the ratio of $g_p$ to $g_e$ is:

An iron ball and a wooden ball of the same radius are released from the same height in vacuum. They take the same time to reach the ground. The reason for this is

As we go from the equator of the earth to the pole of the earth,the value of acceleration due to gravity