Escape Velocity

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)