Escape Velocity

This article explains the concept of escape velocity and how to calculate it for different celestial bodies. It discusses the formula for escape velocity and provides an example calculation for Earth. The article also explores the interpretation and implications of escape velocity, including the concept of conservation of energy. References to relevant physics textbooks are provided.

1Escape Velocity
ESCAPE VELOCITY

2Escape Velocity
Introduction
According to Verma(1999), When a stone (projectile) is thrown up from the earth
surface, it goes up to a maximum height and then returns due to Gravitational Force
of Attraction of the earth. This formula for calculating this force was given by Sir
Isaac Newton which is
F = GMm
r2 ȓ
where,
G is the Universal Gravitational Constant. Its value is 6.67 x 10-11 Nm2kg-2
M and m are the masses in kg
r is the distance between the 2 masses
Now, as the stone continues to go up, its kinetic energy becomes zero and it starts to
return back from there to the surface.
In order to remove the stone completely from the influence of Earth’s Gravitational
Potential, the work done on the stone is provided by the Initial Kinetic Energy from
which we can estimate the Escape Velocity for the given planet (Earth). Here, we are
neglecting Air resistance (atmospheric friction) and gravitational pull due to other
celestial bodies.
Analysis
As discussed in Arora (2018), let an object be thrown vertically upward from a Planet
of mass (M) and Radius (R) as shown in Fig.-1. Let the mass of the projectile be
(m) and initial velocity when thrown from the earth surface be (ve ).

3Escape Velocity
∞
dx ȓ
x
R
Earth
Figure-1
Then, the force acting on the object-
F = - GMm
x2 ȓ , where ȓ is the unit vector in the direction away from the centre of
earth
Now, according to Morin (208), Newton’s 2nd Law states that-
F = ma
- GMm
x2 ȓ = ma ȓ
- GM
x2 = v dv
dx (Since acceleration can be written as v dv
dx ,as stated in Giri (2015) )
vdv = - GM
x2 dx
Integrating both sides within the limits, velocity going from ve to 0 and distance
going from R to ∞ where R is the Radius of the Planet (Earth)

4Escape Velocity
∫
ve
0
vdv = - ∫
R
∞
GM
x2 dx
[ 0 - ve
2
2 ] = [ GM
∞ − GM
R ]
ve
2 = 2GM
R (1)
ve = √ 2 GM
R is the required expression for calculation of the Escape velocity of the
object projected vertically from a planet.
In vector form, it is : ve = √ 2 GM
R ȓ
Solution
Calculating Escape velocity for the earth, we need to know its mass (M) and its
Radius (R) which is 5.972 x 1024 kg and 6378100 m respectively as stated in Mathur
and Kumar (2017). Also, the value of universal gravitational constant (G) is 6.67 x 10−11
N m2 kg -2. Putting these values in equation (1),
ve = √ 2 GM
R = √ 2 x 6.67 x 10−11 x 5.972 x 1024
6378100
= √ 79.67 x 1013
6378100
= √ 124.8867 x 103
= 11.175 x 103
ve = 11175 m s -1 = 11.175 km s -1
This is a very high speed. It means we have to throw the object at a speed such that
it covers 11.175 km in just 1 second !
We can calculate Escape Velocity for any planet or Celestial body in space provided
it is spherically symmetric in shape and we know its mass and radius. For example,
Escape Velocity of Moon whose radius is 1740 km and mass is 7.4 x 1022 kg comes
out to be 2.4 km s -1 using the formula : ve = √ 2 GM
R .

5Escape Velocity
Interpretation
According to Halliday, Resnick & Walker (2015), this velocity can be implied as that
if we throw a projectile vertically upward from the earth surface with a velocity
greater than or equal to 11.175 km s -1, then, the object will not return to the surface
and it will escape the gravitational field of the earth. If the initial velocity is less
than 11.175 km s -1 even by a small amount, the object will retrace its path and will
come back to the surface of earth. Once the object is thrown with a velocity greater
than or equal to escape velocity, it will move away from the body and its velocity
slows down with time but never reaches to zero unless it comes under the influence
of some other massive body.
Also, the Law of Conservation of energy is fulfilled here. As the object moves away,
its Kinetic Energy starts decreasing while its gravitational potential energy starts
increasing. We can also calculate the escape velocity using Concept of Conservation
of energy.
Discussion
Here, as stated in Gomber & Gogia (2018), we notice that the escape velocity for a
particular planet is constant (since G, M and R are fixed) and is independent of the
mass and size of the object. This means whether we throw a 10gm metal ball or a
10kg metal ball from the surface, it will escape the planet only when thrown with a
velocity greater than or equal to √ 2 GM
R The thing that make the difference is the
amount of energy required to throw an object with the velocity equal to Escape
Velocity.

6Escape Velocity
References
Arora, S.L. (2018). New Simplified Physics. 9th ed. New Delhi : Dhanpat Rai Publications.
Giri, G. (2015). Conceptual Physics. Delhi : Prachi India Pvt . Ltd.
Gomber, K.L. and Gogia K.L. (2018). Fundamental Physics. 27th ed. Delhi : Pradeep
Publications.
Halliday, D., Resnick, R. and Walker, J. (2015). Principles of Physics. Noida : 10th ed.
Wiley India Pvt. Ltd.
Mathur, S.B. and Kumar, A. (2017). Core Physics. New Delhi : Bharti Bhawan
Publications.
Morin, D. (2009). Introduction to Classical Mechanics. New Delhi : Cambridge University
Press
Verma, H.C. (1999). Concepts of Physics. New Delhi : Bharti Bhawan Publications