If the value of acceleration due to gravity on the surface of the Earth is $g$,what will be the value of $g$ at a height equal to the radius of the Earth from the surface?

The value of the acceleration due to gravity is $g_{1}$ at a height $h = \frac{R}{2}$ ($R$ = radius of the earth) from the surface of the earth. It is again equal to $g_{1}$ at a depth $d$ below the surface of the earth. The ratio $\left(\frac{d}{R}\right)$ equals

The linear speed of a particle at the equator of the earth due to its spin motion is $V$. The linear speed of the particle at latitude $30^{\circ}$ is:

The depth $d$ at which the value of acceleration due to gravity becomes $\frac{1}{n-1}$ times the value at the earth's surface is ($R=$ radius of the earth).

$A$ uniform spherical planet (Radius $R$) has acceleration due to gravity at its surface $g$. Points $P$ and $Q$ located inside and outside the planet have acceleration due to gravity $g/4$. The maximum possible separation between $P$ and $Q$ is:

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