The weight of an object on the surface of the Earth is $72 \, N$. What will be the weight of the object at a height of $R/2$ from the surface of the Earth? ($R$ = radius of the Earth)

How does the acceleration due to gravity change with altitude?

The acceleration due to gravity at a height of $(\sqrt{2}-1) R$ from the surface of the earth is (Acceleration due to gravity on the surface of the earth $= 10 \ m \ s^{-2}$ and $R$ is radius of the earth). (in $m \ s^{-2}$)

The height $h$ at which the weight of a body will be the same as that at the same depth $h$ from the surface of the earth is (Radius of the earth is $R$ and effect of the rotation of the earth is neglected):

The percentage decrease in the weight of a rocket,when taken to a height of $32 \ km$ above the surface of the Earth,will be $.....\%$
(Radius of Earth $= 6400 \ km$)

The height $h$ above the earth's surface at which the value of acceleration due to gravity $g$ becomes $\frac{g}{3}$ is ($R=$ radius of the earth).