If acceleration due to gravity at distance $d < R$ from the centre of the Earth is $\beta$,then its value at distance $d$ above the surface of the Earth will be [where $R$ is the radius of the Earth].

If the radius of the earth shrinks by $1 \%$,its mass remaining the same,then the acceleration due to gravity on the earth's surface would

There is a planet which is $8$ times more massive and $27$ times denser than the Earth. If $g^{\prime}$ and $g$ are the accelerations due to gravity on the surfaces of the planet and the Earth respectively,then:

Find the angular speed of rotation of the Earth,so that the apparent $g$ at the equator becomes $(1/6)$th of its original value. $(R = 6.4 \times 10^6 \ m)$

The mass and the diameter of a planet are three times the respective values for the Earth. The period of oscillation of a simple pendulum on the Earth is $2 \ s$. The period of oscillation of the same pendulum on the planet would be

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)