Question

A body is projected horizontally from the surface of the Earth $(radius=R)$ with a velocity equal to $n$ times the escape velocity. Neglect rotational effects of the earth. The maximum height attained by the body from the earth's surface is $R/2$. Then $n$ must be

• A. $\sqrt{0.6}$
• B. $\left(\sqrt{3}\right)/2$
• C. $\sqrt{0.4}$
• D. $1/2$

Answer 1

The correct option is A $\sqrt{0.6}$

let escape velocity be $v_e$.

We have: $v_e=\sqrt{\frac{2GM}{R}}$

By angular momentum conservation:

$m\left(n{v}_e\right)R=\frac{3}{2}mvR\Rightarrow v=\frac{2}{3}n{v}_e$ where $v$ is the velocity at the highest point.

By energy conservation:

$\frac{1}{2}m{n}^2{v}_e^2-\frac{GMm}{R}=\frac{1}{2}m{n}^2{v}_e^2\times \frac{4}{9}-\frac{2GMm}{3R}\Rightarrow \frac{5}{18}m{n}^2{v}_e^2=\frac{GMm}{3R}\Rightarrow n{v}_e=\sqrt{\frac{6GM}{5R}}$

$\Rightarrow n=\sqrt{\frac{3}{5}}=\sqrt{0.6}$