The approximate height from the surface of earth at which the weight of the body becomes $\frac{1}{3}$ of its weight on the surface of earth is $.......... \, km$ : [Radius of earth $R = 6400 \, km$ and $\sqrt{3} = 1.732$]

$A$ 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 $N$)?

$A$ planet of radius $R = \frac{1}{10} \times$ (radius of Earth) has the same mass density as Earth. Scientists dig a well of depth $\frac{R}{5}$ on it and lower a wire of the same length and of linear mass density $\lambda = 10^{-3} \ kg \ m^{-1}$ into it. If the wire is not touching anywhere,the force applied at the top of the wire by a person holding it in place is (take the radius of Earth $= 6 \times 10^6 \ m$ and the acceleration due to gravity on Earth $g_e = 10 \ m \ s^{-2}$) (in $N$)

At a height $R$ above the earth's surface,the gravitational acceleration is ($R$ = radius of earth,$g$ = acceleration due to gravity on earth's surface).

Acceleration due to gravity on the surface of Earth is $g$. If the diameter of Earth is reduced to one-third of its original value and mass remains unchanged,then the acceleration due to gravity on the surface of the Earth is . . . . . . $g$.

If the change in the value of $g$ at a height $h$ above the surface of the earth is the same as at a depth $x$ below it,then (both $x$ and $h$ being much smaller than the radius of the earth)