The acceleration due to gravity becomes $\left(\frac{g}{2}\right)$ ($g =$ acceleration due to gravity on the surface of the earth) at a height equal to

What is the height from the surface of earth,where acceleration due to gravity will be $\frac{1}{4}$ of that of the earth (in $km$)? $(R_E = 6400 \ km)$

If a body takes time $t$ to reach the ground when dropped from a height $h$ on Earth,how much time will it take to reach the ground when dropped from the same height $h$ on the Moon?

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 $g = \pi^{2} \ m/s^{2}$)

Assuming the density of the Earth is constant, which graph correctly represents the variation of acceleration due to gravity $(g)$ with the distance $(r)$ from the center of the Earth (radius of the Earth $= R$)?

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