The value of $g$ on the Earth's surface is $980 \, cm/s^2$. Its value at a height of $64 \, km$ from the Earth's surface is ........ $cm/s^2$ (Radius of the Earth $R = 6400 \, km$).

Two particles are projected vertically upwards with the same velocity on two different planets with accelerations due to gravity $g_1$ and $g_2$ respectively. If they fall back to their initial points of projection after lapse of times $t_1$ and $t_2$ respectively,then

If two planets have their radii in the ratio $x: y$ and densities in the ratio $m: n$,then the ratio of the acceleration due to gravity on them is

The acceleration due to gravity at a height $(1/20)^{th}$ of the radius of the Earth above the Earth's surface is $9 m s^{-2}$. Its value at an equal depth below the surface of the Earth is: (in $m s^{-2}$)

If the earth stops rotating,the value of $g$ at the equator will

$T$ is the time period of a simple pendulum on the Earth's surface. Its time period becomes $xT$ when taken to a height $R$ (equal to the Earth's radius) above the Earth's surface. Then,the value of $x$ will be: