If the Earth were to shrink to half its radius without changing its mass, the value of 'g' on its surface would become:

Determine the speed with which the earth would have to rotate on its axis so that a person on the equator would weigh $3 / 5$th as much as at present. Take the equatorial radius as $6400 \mathrm{~km}.$

The height from surface of earth at which value of $g$ becomes one fourth of that on earth's surface will be ( $R$ is radius of earth)

A box weighs $196\,N$ on a spring balance at the north pole. Its weight recorded on the same balance, if it is shifted to the equator is close to (Take, $g = 10\,m{s^{ - 2}}$ at the north pole and the radius of the earth = 6400 km)

A particle does uniform circular motion in a horizontal plane. The radius of the circle is 20 cm. The centripetal force acting on the particle is 10 N. It's kinetic energy is

Acceleration due to gravity at earth's surface is $10 \mathrm{~ms}^{-2}$. The value of acceleration due to gravity at the surface of a planet of mass $\frac{1}{5}$ th and radius $\frac{1}{2}$ of the earth is -

If the earth stops rotating in its orbit about the sun, the value of $g$ changes at

The acceleration due to gravity at the poles and the equator is $g_p$ and $g_e$ respectively. If the earth is a sphere of radius ${Re}$ and rotation about its axis with angular speed $\omega$, then $g_p-g_e$ is given by :-

Statement-1: There is almost no effect of rotation of earth at poles.Statement-2: Because rotation of earth is about polar axis.

A man goes to a height equal to the radius of the earth from its surface. The weight of the person at that height relative to his weight on the surface of earth is:

In order to make the effective acceleration due to gravity equal to zero at the equator, the angular velocity of rotation of the earth about its axis should be $\left( {g = 10\,m{s^{ - 2}}\,\,{ext{and radius of earth is 64000 km}}} \right)$