Question

A particle is fired vertically from the surface of the earth with a velocity $k{\upsilon }_e$, where ${\upsilon }_e$ is the escape velocity and k < 1. Neglecting air resistance and assuming earth's radius as $R_e$. The height to which it will rise from the surface of the earth is

• A. $\frac{R_{e}k}{1-k^2}$
• B. $\frac{R_e}{1-k^2}$
• C. $\frac{R_e{k}^2}{1-k^2}$
• D. $\frac{R_e(1+k)}{1-k}$

Answer 1

The correct option is C $\frac{R_e{k}^2}{1-k^2}$

Equating initial and final energy:
$-\frac{GMm}{R}+\frac{1}{2}m{\left(k{v}_e\right)}^2=-\frac{GMm}{r}$
$-\frac{GMm}{R}+\frac{1}{2}m{\left(k\sqrt{\frac{2GM}{R}}\right)}^2=-\frac{GMm}{r}$
$(1-k^2)\frac{GMm}{R}=\frac{GMm}{r}$
$r=\frac{R}{1-k^2}$
Since $r$ is the distance from the centre, thus distance from surface is:
$r-R=\frac{R{k}^2}{1-k^2}$