Question

A particle of mass $m$ is projected with a velocity $v=k{v}_e(k<1)$ from the surface of the earth. $(v_e=$ escape velocity$)$. The maximum height above the surface reached by the particle is -

$R$ is the radius of earth.

• A. $R{\left(\frac{k}{1-k}\right)}^2$
• B. $R{\left(\frac{k}{1+k}\right)}^2$
• C. $\frac{R^{2}k}{1+k}$
• D. $\frac{R{k}^2}{1-k^2}$

Answer 1

The correct option is D $\frac{R{k}^2}{1-k^2}$

From conservation of mechanical energy,

$\frac{1}{2}m{v}^2-\frac{GmM}{R}=-\frac{GmM}{(R+h)}$

$\Rightarrow \frac{v^2}{2}=\frac{GM}{R}-\frac{GM}{(R+h)}=\frac{GMh}{R(R+h)}$

$\Rightarrow \frac{1}{2}{k}^2{v}_e^2=\frac{GMh}{R(R+h)}$

We know that, $v_e=\sqrt{\frac{2GM}{R}}$

$\frac{1}{2}{k}^2\left(\frac{2GM}{R}\right)=\frac{GMh}{R(R+h)}$

$\Rightarrow k^2=\frac{h}{R+h}$

$\Rightarrow R{k}^2+h{k}^2=h$

$\Rightarrow R{k}^2=h(1-k^2)$

$\Rightarrow h=\frac{R{k}^2}{1-k^2}$

Hence, option $\left(A\right)$ is the correct answer.