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 $(r=\mathrm{\infty })$ is

$$W={\int }_R^{\mathrm{\infty }}\frac{GMm}{x^2}dx=\frac{GMm}{R}$$

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.

$$\mathrm{so}KE=\frac{GMm}{R}$$

And Kinetic energy can be written as

$$KE=\frac{1}{2}m{V}_e^2$$

Where $V_e$ is the required escape velocity.
By comparing we get

$$\begin{array}{rl}& \frac{1}{2}m{V}_e^2=\frac{GMm}{R}\\ \Rightarrow & V_e=\sqrt{\frac{2GM}{R}}\end{array}$$

Using $GM=g{R}^2$
We get $V_e=\sqrt{2gR}$
$V_e\to$ Escape velocity
$R\to$ Radius of earth

And using $g=\frac{4}{3}\pi \rho GR$

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

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 $(gR)$ 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 $H_{\text{max }}$

$$\begin{array}{rl}& \frac{-GMm}{R}+\frac{1}{2}m{V}^2=\frac{-GMm}{H_{max}}+0\\ & V_e=\sqrt{\frac{2GM}{R}}\\ & \text{ And using }\end{array}$$

We get

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

$V_e\to$ escape velocity
$V\to$ Projection velocity of the body
$R\to$ 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

  $$\begin{array}{c}\frac{-GMm}{R}+\frac{1}{2}m{V}^2=0+\frac{1}{2}m{\left(V^{\prime }\right)}^2\\ \text{ And using }{V}_e=\sqrt{\frac{2GM}{R}}\end{array}$$

  
  We get

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

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

  $$\frac{GMm}{R}=\text{ Escape Energy }$$

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