Question

A body is fired from the surface of the earth. It goes to a maximum height $R$ (radius of the earth) from the surface of the earth. Find the initial velocity given.

• A. $6.9km{s}^{-1}$
• B. $7.4km{s}^{-1}$
• C. $7.9km{s}^{-1}$
• D. $8.4km{s}^{-1}$

Answer 1

Answer: (C)

$7.9km{s}^{-1}$

The potential energy at earth surface is $-GMm/R$,

where $M$ is mass of the earth.

$R$ is the radius of earth.

$G$ is the gravitationsl constant

The potential energy at height $R$, is $-GMm/2R$,

So change in potential energy to reach height $R$ is $-GMm/2R-(-GMm/R)=GMm(1/R-1/2R)=GMm/R$

By conservation of energy, change in potential energy $=$ Kinetic energy.

So, $m{v}^2/2=GMm/2R\Rightarrow v=\sqrt{GM/R}=\sqrt{6.67\times {10}^{-11}\times 6\times {10}^{24}/6400\times {10}^3}=7.9km{s}^{-1}$