The height of the point vertically above the earth's surface,at which the acceleration due to gravity becomes $1\%$ of its value at the surface is (Radius of the earth $= R$). (in $,R$)

$A$ $90 \,kg$ body placed at $2R$ distance from the surface of the Earth experiences a gravitational pull of: ($R=$ Radius of Earth,$g=10 \,ms^{-2}$) (in $\,N$)

The mass of the earth is $81$ times that of the moon and the radius of the earth is $3.5$ times that of the moon. The ratio of the acceleration due to gravity at the surface of the moon to that at the surface of the earth is

At what depth $d$ below the surface of the Earth does the acceleration due to gravity become $\frac{1}{n}$ times its value at the surface? ($R$ = radius of the Earth)

The value of acceleration due to gravity at a height of $10 \,km$ from the surface of the Earth is $x$. At what depth inside the Earth is the value of the acceleration due to gravity the same value $x$ (in $\,km$)?

The height $h$ above the earth's surface at which the value of acceleration due to gravity $g$ becomes $\frac{g}{3}$ is ($R=$ radius of the earth).