Escape Velocity - Practice Questions & MCQ

Escape velocity is defined as the minimum velocity an object must have in order to escape from the planets gravitational pull.

•  Escape velocity ( in terms of the radius of the earth)

To escape a body from earth surface means to displace it from the surface of the earth to infinity.

The work done to displace a body from the surface of the earth (r = R) to infinity ( r =  ) is 

$
W=\int_R^{\infty} \frac{G M m}{x^2} d x=\frac{G M m}{R}
$

So if we provide kinetic energy equal to W to body at the surface of the earth then it will be able to escape from the earth's gravitational pull.

$
\mathrm{So} K E=\frac{G M m}{R}
$

And Kinetic energy can be written as $K E=\frac{1}{2} m V_e^2$
Where $V_e$ is the required escape velocity.
By comparing we get

$
\begin{aligned}
& \frac{1}{2} m V_e^2=\frac{G M m}{R} \\
\Rightarrow & V_e=\sqrt{\frac{2 G M}{R}}
\end{aligned}
$

Using $G M=g R^2$
We get $V_e=\sqrt{2 g R}$
$V_e \rightarrow$ Escape velocity
$R \rightarrow$ Radius of earth
And using $g=\frac{4}{3} \pi \rho G R$

$
V_e=R \sqrt{\frac{8}{3} \pi G \rho}
$
 

For the earth

$
V_e=11.2 \mathrm{Km} / \mathrm{s}
$

- Escape velocity is independent of the mass of the body.
- Escape velocity is independent of the direction of projection of the body.
- Escape velocity depends on the mass and radius of the earth/planet.
I.e Greater the value of $\frac{M}{R}$ or $(g R)$ of the planet greater will be the escape velocity
- If the body projected with velocity less than escape velocity $\left(V<V_e\right.$ )

In this case, the first body will reach a certain maximum height ( $H_{\text {max }}$ )
And after that, it may either move in an orbit around the earth/planet or may fall back down towards the earth/planet.

Let's find Maximum height attained by the body
At maximum height, the velocity of the particle is zero
So at $\mathrm{h}=H_{\max }$ it's Kinetic energy $=0$
By the law of conservation of energy
Total energy at surface $=$ Total energy at the height $H_{\max }$

$
\frac{-G M m}{R}+\frac{1}{2} m V^2=\frac{-G M m}{H_{\max }}+0
$

And using

$
V_e=\sqrt{\frac{2 G M}{R}}
$
 

            We get  

$
H_{\max }=R\left[\frac{V^2}{V_e^2-V^2}\right]
$

$V_e \rightarrow$ escape velocity
$V \rightarrow$ Projection velocity of the body
$R \rightarrow$ Radius of planet
- If a body is projected with a velocity greater than escape velocity $\left(V>V_e\right)$

Then By the law of conservation of energy
Total energy at surface $=$ Total energy at infinity

$
\begin{gathered}
\frac{-G M m}{R}+\frac{1}{2} m V^2=0+\frac{1}{2} m\left(V^{\prime}\right)^2 \\
\text { And using } V_e=\sqrt{\frac{2 G M}{R}}
\end{gathered}
$

We get

$
V^{\prime}=\sqrt{V^2-V_e^2}
$

new velocity of the body at infinity= $V^{\prime}$
$V \rightarrow$ projection velocity
$V_e \rightarrow$ Escape velocity

• Escape energy

  Energy to be given to an object on the surface of the earth so that it's total energy is 0

  $\frac{G M m}{R}=$ Escape Energy
  $M \rightarrow$ Mass of planet
  $m \rightarrow$ mass of the body
  $G \rightarrow$ Gravitational constant