If a planet has a mass and radius both half that of the Earth,the acceleration due to gravity at its surface would be ......... $m/s^2$ ($g$ on Earth $= 9.8\, m/s^2$)

If the radius of the earth contracts by $2\%$ and its mass remains the same,then the weight of a body at the earth's surface:

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:

$A$ simple pendulum has a time period $T_1$ when on the earth's surface and $T_2$ when taken to a height $R$ above the earth's surface,where $R$ is the radius of the earth. The value of $T_2/T_1$ 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 $m s^{-2}$)

Suppose that the angular velocity of rotation of the earth is increased. Then,as a consequence: