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

The acceleration due to gravity on the planet $A$ is $9$ times the acceleration due to gravity on planet $B$. $A$ man jumps to a height of $2 \ m$ on the surface of $A$. What is the height of jump by the same person on the planet $B$?

$A$ uniform spherical planet (Radius $R$) has acceleration due to gravity at its surface $g$. Points $P$ and $Q$ located inside and outside the planet have acceleration due to gravity $g/4$. The maximum possible separation between $P$ and $Q$ is:

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 $6400 \, km$).

If $M$ is the mass of the Earth and $R$ is its radius,the ratio of the gravitational acceleration $g$ to the gravitational constant $G$ is:

If $g_E$ and $g_M$ are the accelerations due to gravity on the surfaces of the earth and the moon respectively and if Millikan's oil drop experiment could be performed on the two surfaces,one will find the ratio (electronic charge on the moon / electronic charge on the earth) to be