The speed with which the earth would have to rotate about its axis so that a person on the equator would weigh $\frac{3}{5}$ th as much as at present weight is ($g=$ gravitational acceleration,$R=$ equatorial radius of the earth).

$A$ body weighs $500 \,N$ on the surface of the earth. At what distance below the surface of the earth will it weigh $250 \,N$ (in $\,km$)? (Radius of earth, $R = 6400 \,km$)

The weight of $1 \, kg$ becomes $1/6$ on the moon. If the radius of the moon is $1.768 \times 10^6 \, m$,then the mass of the moon will be:

If the radius of the Earth becomes double while its mass remains unchanged,what will be the change in the weight of an object of mass $m$ on the surface of the Earth?

The acceleration due to gravity on the earth's surface at the poles is $g$ and the angular velocity of the earth about the axis passing through the poles is $\omega$. An object is weighed at the equator and at a height $h$ above the poles by using a spring balance. If the weights are found to be the same,then $h$ is: ($h << R$,where $R$ is the radius of the earth)

The period of a seconds pendulum on a planet,whose mass and radius are three times that of Earth,is