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# Escape Velocity - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

## Quick Facts

• Escape Velocity is considered one the most difficult concept.

• 25 Questions around this concept.

## Solve by difficulty

A planet in a distant solar system is 10 times more  massive than the Earth and its radius is  10 times smaller Given that the escape velocity from the earth is  11 km s-1, the escape velocity ( in km s-1) from the  surface of the planet would be :

The kinetic energy needed to project a body of  mass m from the earth's surface (radius R) to infinity is :

The escape velocity of a body depends upon mass  as :

The escape velocity for a body projected vertically upwards from the surface of the earth is 11 Km/s. If the body is projected at an angle of 45o with the vertical, the escape velocity will be :

## Concepts Covered - 1

Escape Velocity

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 =$\infty$  ) is

$W=\int_{R}^{\infty}\frac{GMm}{x^2}dx=\frac{GMm}{R}$

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.

So $KE=\frac{GMm}{R}$

And Kinetic energy can be written as $KE= \frac{1}{2}mV_e^2$

Where $V_e$ is the required escape velocity.

By comparing we get

$\frac{1}{2}mV_e^2=\frac{GMm}{R}\\ \Rightarrow V_e=\sqrt{\frac{2GM}{R}}$

Using $GM=gR^2$

We get $V_{e}=\sqrt{2gR}$

$V_{e} \rightarrow$ Escape velocity

$R \rightarrow$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

$V_{e}=11.2Km/s$

• 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 $\left ( gR \right )$ of the planet greater will be the escape velocity

• If the body projected with velocity less than escape velocity  ($V< V_{e}$)

In this case, the first body will reach a certain maximum height ($H_{max}$)

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 h=$H_{max}$ it's Kinetic energy=0

By the law of conservation of energy

Total energy at surface = Total energy at the height $H_{max}$

$\frac{-GMm}{R}+\frac{1}{2}mV^2=\frac{-GMm}{H_{max}}+0$

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

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 ($V>V_e$)

Then By the law of conservation of energy

Total energy at surface = Total energy at infinity

$\frac{-GMm}{R}+\frac{1}{2}mV^2= 0+ \frac{1}{2}m(V')^2$

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

We get

${V}'=\sqrt{V^{2}-V_{e}^{2}}$

new velocity of the body at infinity= ${V}'$

${V}\rightarrow$ 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

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

$M\rightarrow$  Mass of planet

$m\rightarrow$ mass of the body

$G\rightarrow$ Gravitational constant

## Study it with Videos

Escape Velocity

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