The height above the surface of the earth where acceleration due to gravity becomes $\frac{g}{9}$ is ( $R$ is the radius of the earth,$g$ is the acceleration due to gravity at the surface).

If the radius of the Earth decreases without any change in its mass,will the value of $g$ on its surface increase or decrease?

If all objects on the equator of the earth feel weightless,then the duration of the day will nearly become ....... $hr$.

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 radius of the Earth is $6400 \,km$ and the acceleration due to gravity is $g=10 \,ms^{-2}$. For the weight of a body of mass $5 \,kg$ to be zero at the equator,the rotational angular velocity of the Earth must be (in $rad/s$):

$A$ pendulum is oscillating with frequency $n$ on the surface of the Earth. If it is taken to a depth $d = \frac{R}{2}$ below the surface of the Earth,where $R$ is the radius of the Earth,what is the new frequency of oscillations at this depth?