The time period of a simple pendulum on the surface of the earth is $T$. If the pendulum is taken to a height equal to half of the radius of the earth,then its time period 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 height at which the weight of a body becomes $\frac{1}{16}^{th}$ of its weight on the surface of the Earth (radius $R$) is: (in $R$)

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:

Obtain an expression for the variation in effective gravitational acceleration $g'$ with latitude due to the Earth's rotation.

$A$ 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?