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Escape Velocity - Practice Questions & MCQ

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

Quick Facts

  • Escape Velocity is considered one the most difficult concept.

  • 35 Questions around this concept.

Solve by difficulty

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 :

The ratio of escape velocity of earth \left( {\upsilon _\text{e} } \right)  to the escape velocity at a planet \left( {\upsilon _\text{p} } \right) whose radius and mean density are twice as that of earth is:

A particle of mass 'm' is kept at rest at a height 3 R from the surface of earth, where 'R' is radius of earth and 'M' is mass of earth. The minimum speed with which it should be projected, so that it does not return back, is (g is acceleration due to gravity on the surface of the earth)

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 :

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 \rightarrowRadius 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

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