If a man at the equator would weigh $(3/5)^{th}$ of his weight,the angular speed of the earth is

The radius of the Moon is $\frac{1}{4}$ times that of the Earth,and its mass is $\frac{1}{80}$ times that of the Earth. If the acceleration due to gravity on Earth is $g$,what is the acceleration due to gravity on the Moon?

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 value of acceleration due to gravity at Earth's surface is $9.8\, m\,s^{-2}$. The altitude above its surface at which the acceleration due to gravity decreases to $4.9\, m\,s^{-2}$,is close to: (Radius of Earth $= 6.4 \times 10^6\, m$)

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

If the earth were to cease rotating about its own axis,the increase in the value of $g$ in the $C.G.S.$ system at a place of latitude of $45^{\circ}$ will be ........ $cm/sec^{2}$.