The height at which the weight of a body becomes $\left(\frac{1}{9}\right)^{\text{th}}$ of its weight on the surface of the Earth is $(R = \text{radius of Earth})$: (in $R$)

The height at which the weight of a body becomes $\frac{1}{9}$ of its weight on the surface of the earth (radius of the earth is $R$) is:

$A$ hole is drilled halfway to the center of the Earth. $A$ body weighs $300 \ N$ on the surface of the Earth. How much will it weigh at the bottom of the hole (in $N$)?

The angular velocity of the earth with which it has to rotate so that acceleration due to gravity on $60^o$ latitude becomes zero is (Radius of earth $= 6400 \, km$. At the poles $g = 10 \, m/s^2$)

The acceleration due to gravity on the surface of Earth is $g$. If the diameter of Earth reduces to half of its original value and mass remains constant,then the acceleration due to gravity on the surface of Earth would be:

The mass and diameter of a planet have twice the value of the corresponding parameters of Earth. Acceleration due to gravity on the surface of the planet is ........ $m/s^2$.