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).

Weight of a body decreases by $1.5 \%$ when it is raised to a height $h$ above the surface of the Earth. When the same body is taken to the same depth $h$ in a mine,its weight will show ........

$A$ 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 $N$)

The acceleration due to gravity on the Moon is $\frac{1}{6}$ times that of the Earth. If the ratio of the density of the Earth $(\rho_e)$ to the Moon $(\rho_m)$ is $\frac{\rho_e}{\rho_m} = \frac{5}{3}$,what is the radius of the Moon $(R_m)$ in terms of the radius of the Earth $(R_e)$?

Write the equation of gravitational acceleration which is used for any height $h$ from the surface of the Earth.

What should be the angular velocity of the Earth due to rotation about its own axis so that the weight at the equator becomes $\left(\frac{3}{5}\right)$ of its initial value? (Radius of Earth at the equator $R = 6400 \ km$,$g = 10 \ m/s^2$,$\cos 0^{\circ} = 1$)