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 $m s^{-2}$)

The gravitational potential at a point above the surface of the Earth is $-5.12 \times 10^7 \,J/kg$ and the acceleration due to gravity at that point is $6.4 \,m/s^2$. Assume that the mean radius of the Earth is $6400 \,km$. The height of this point above the Earth's surface is: (in $\,km$)

The height at which the weight of a body becomes $\frac{1}{16}^{th}$ of its weight on the surface of the Earth (radius $R$) is: (in $R$)

$T$ 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:

The weight of a body of mass $m$ decreases by $1\%$ when it is raised to a height $h$ above the Earth's surface. If the body is taken to a depth $h$ in a mine,the change in its weight is:

$A$ body has a weight of $90 \ kgf$ on the earth's surface. The mass of the moon is $1/9$ of the earth's mass and its radius is $1/2$ of the earth's radius. On the moon,the weight of the body is .......... $kgf$.