Question

The escape velocity of a body on the earth's surface is $V_e$. A body is thrown vertically up with a speed $\left(k{V}_e\right)(k<1)$. The maximum height reached by the body above the earth is

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

Answer 1

The correct option is A $R\left(\frac{k^2}{1-k^2}\right)$

Given $V_e$ is the escape velocity of a body on the earth's surface.

Let us assume the body reaches a maximum height $h$ above earth's surface when thrown vertically up with speed $k{V}_e$.

As there are no other forces than the gravitational foce in action, we can conserve the total mechanical energy

(Kinetic energy + Gravitational Potential energy) of the system.

If mass of the particle is $m$, mass of earth is $M$, radius of earth is $R$.

From prior knowledge we know that

$V_e=\sqrt{\frac{2GM}{R}}$

Initially the kinetic energy is $\frac{1}{2}m(k{V}_e{)}^2$

At the maximum height, Kinetic energy of the particle is $0$.

Initial potential energy on the surface of earth is $-\frac{GMm}{R}$

Final potential energy at a height $h$ from the surface of earth is $-\frac{GMm}{R+h}$

$\therefore \frac{1}{2}m(k{V}_e{)}^2-\frac{GMm}{R}=0-\frac{GMm}{R+h}$

Substituting $V_e$ from above formula,

$\therefore \frac{1}{2}m{k}^2\left(\frac{2GM}{R}\right)-\frac{GMm}{R}=-\frac{GMm}{R+h}$

$\therefore \frac{GMm}{R}{k}^2-\frac{GMm}{R}=-\frac{GMm}{R+h}$

$\therefore \frac{1}{R}(1-k^2)=\frac{1}{R+h}$

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

$\therefore h=R\left(\frac{1}{1-k^2}\right)-R$

$\therefore h=R\left(\frac{k^2}{1-k^2}\right)$

The maximum height reached by the body is $R\left(\frac{k^2}{1-k^2}\right)$