The weight of a body of mass $m$ decreases by $1\%$ when it is taken to a height $h$ above the surface of the Earth. What is the percentage decrease in its weight when it is taken to a depth $h$ below the surface of the Earth?

The acceleration due to gravity at a height $h$ above the surface of the $Earth$ is $g_h$. At a depth $d = 90 \text{ km}$ below the $Earth$'s surface, the acceleration due to gravity is also $g_h$. The value of $h$ is: (in $\text{ km}$)

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 $= 9.8\, m/s^2$)

The acceleration due to gravity on the moon is $\frac{1}{6}$ times the acceleration due to gravity on the earth. If the ratio of the density of the earth $\rho_e$ to the density of the moon $\rho_m$ is $\frac{5}{3}$,then the radius of the moon $R_m$ in terms of the radius of the earth $R_e$ is:

The depth $d$ at which the value of acceleration due to gravity becomes $\frac{1}{n-1}$ times the value at the earth's surface is ($R=$ radius of the earth).

If the radius of the earth were to shrink by $1\%$ with its mass remaining the same,the acceleration due to gravity on the earth's surface would