Escape Velocity MCQ - Practice Questions with Answers

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

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

To escape a body from the earth's 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 $(r=\infty)$ 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 the 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 a velocity less than escape velocity ( $V<V_e$ ) In this case, the first body will reach a certain maximum height ( $H_{\max }$ )

After that, it may either move in an orbit around the earth/planet or may fall back down towards the earth/planet.

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 the Maximum height attained by the body

At maximum height, the velocity of the particle is zero

So at $\mathrm{h}=H_{\text {max }}$ it's Kinetic energy $=0$
By the law of conservation of energy
Total energy at surface $=$ Total energy at the height $H_{\text {max }}$

$$
\begin{aligned}
& \frac{-G M m}{R}+\frac{1}{2} m V^2=\frac{-G M m}{H_{\max }}+0 \\
& \qquad V_e=\sqrt{\frac{2 G M}{R}} \\
& \text { And using }
\end{aligned}
$$

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 ( $V>V_e$ )

  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

  $$
  \frac{G M m}{R}=\text { Escape Energy }
  $$

  $M \rightarrow$ Mass of planet
  $m \rightarrow$ mass of the body
  $G \rightarrow$ Gravitational constant