Question

A rocket is launched vertically from the surface of a planet with radius $R$. Rocket has an initial speed $v$. If atmospheric resistance is neglected, the maximum height attained by the rocket is given by________.
Take $g=\frac{GM}{R^2}$.

• A. $h=\frac{R}{(\frac{2gR}{v^2}-1)}$
• B. $h=\frac{R}{(\frac{2gR}{v^2}+1)}$
• C. $h=R(\frac{2gR}{v^2}-1)$
• D. $h=R(\frac{2gR}{v^2}+1)$

Answer 1

The correct option is A

$h=\frac{R}{(\frac{2gR}{v^2}-1)}$

On the surface of the earth, the total energy is
$K{E}_i+P{E}_i=\frac{1}{2}m{v}^2-\frac{GmM}{R}$
where:
m: Mass of the rocket
M : Mass of the earth.

At the highest point, speed is 0 and the energy is entirely potential.
$P{E}_f=-\frac{GmM}{(R+h)}$
where:
h: Maximum height attained.

From the law of conservation of energy, we have
$\frac{1}{2}m{v}^2-\frac{GmM}{R}=-\frac{GmM}{(R+h)}$
$\frac{GmMh}{R(R+h)}=\frac{1}{2}m{v}^2$
$\Rightarrow \frac{R+h}{h}=\frac{2gR}{v^2}(\text{Given:}g=\frac{GM}{R^2})$
$\Rightarrow h=\frac{R}{(\frac{2gR}{v^2}-1)}$
Hence, the correct choice is (a).