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 value of acceleration due to gravity at a depth $d$ from the surface of the Earth and at an altitude $h$ from the surface of the Earth are in the ratio:

The magnitudes of the gravitational force at distances $r_1$ and $r_2$ from the centre of a uniform sphere of radius $R$ and mass $M$ are $F_1$ and $F_2$ respectively. Then

The length of a seconds pendulum at a height $h=2R$ from the earth's surface will be. (Given: $R =$ radius of earth and acceleration due to gravity at the surface of earth $g = \pi^{2} \ m/s^{2}$)

$A$ body weighs $72 \ N$ on the surface of the earth. What is the gravitational force on it at a height equal to half the radius of the earth (in $N$)?

The value of acceleration due to gravity at a height of $4 R_E$ from the surface of the Earth is (where $R_E$ is the radius of the Earth and acceleration due to gravity on the surface of the Earth $g = 10 \,ms^{-2}$): (in $\,ms^{-2}$)