[Physics Class Notes] on Derivation of Escape Velocity Pdf for Exam

There are several instances in our daily life when we look up to the sky and wonder if we can go out there, or maybe we wonder how the astronauts reach space. You might be thinking of how difficult it is to launch a massive object into space. How much velocity does it require to complete this phenomenon? However, all objects, be it a rocket or a baseball moving deep into space, require the same speed. Notably, this velocity or acceleration is known as escape velocity.

What is Escape Velocity?

Escape velocity in Physics is the speed that an object requires to escape from the gravitational force of the Earth. However, it must not accelerate further. It is the minimum speed with which an object must be launched so that it is able to overpower the gravitational pull of the Earth and hence is able to escape to space. This means the escape velocity should remain constant throughout the way out and must not change while the object is under the influence of the gravitational force of the Earth.

 

The gravitational pull of the Earth is dependent on the size and the mass of the Earth and the escape velocity is dependent on the strength of the gravitational force of the Earth. This means that the escape velocity is dependent on the size and mass of the Earth.

Also, the escape speed is dependent on several factors. It is determined by scientists that the escape rate of an enormous body like a star, or a planet is evaluated using the following escape velocity equation:

[V_{e} = frac{sqrt{2GM}}{R}]

The expression for escape velocity is derivable by taking the initial kinetic energy of a body and initial gravitational potential energy at a certain height.

[PE_{i} = frac{-GMm}{R_{i}}]

Where, [PE_{i} ]: initial gravitational potential energy in kg-[kg-m^{2}/s^{2}].

G: Universal Gravitational Constant = [6.674*10 – 20 km^{3}/kg-s^{2}]

m: mass of the attracting body (kg)

M: mass of the escaping body (kg)

Note: M is much greater than m (M >> m)

[R_{i}]: initial separation between the centers of the bodies (km)

[KE_{i} = frac{mv{_{e}}^{2}}{2}]

Where [KE_{i}]: initial kinetic energy ([kg-m^{2}/s^{2}])

M: mass of the object (kg)

[V_{e}]: initial velocity—and thus “escape velocity” (km/s)

Hence, the sum of kinetic and potential energies equals to total initial energy:

[ TE_{i} = KE_{i} = PE_{i}]

[ TE_{i} = frac mv{_{e}}^{2}{2} – frac{GMm}{R_{i}}]

Final energy

Also, gravitational fields are assumed to reach infinity. Hence, a body will move to infinity if the initial speed is strong enough. As a result, it “escapes” gravitational pull.

[PE infty = frac{-GMm}{R infty}]

Where, [PE infty]: gravitational potential energy at infinity

[ R infty ]: infinite separation between the centres of the objects

As, [ R infty ] = ∞ (infinity), thus [PE infty = 0]

[KE infty = frac{mv{_{infty}}^{2}}{2}]

Where, [KE_{infty}]: final kinetic energy

[v_{infty}]: final velocity at infinity

However, at infinity, the velocity of the body is zero: [v_{infty}] = 0

Therefore, [KE_{infty} = 0]

As, kinetic energy is acting upward and potential energy is moving downward, total energy at the initial location:

[ TE_{infty} = KE_{infty} + PE_{infty} ]

Therefore, [ TE_{infty} = 0 + 0 ]

According to the Law of Conservation of Energy, this total energy remains constant for a closed system. So, in this scenario, the closed system contains two bodies with a gravitational pull between each other. Moreover, no outside force or energy is acting on any of the masses.

Hence, [TE_{i} = TE_{infty} ]

[ KE_{i} + PE_{i} = 0]

By substituting values:

[frac{mv{_{e}}^{2}}{2} – frac{GMm}{R_{i}} = 0]

By adding [frac{GMm}{R_{i}}] to both sides:

Therefore, [frac{mv{_{e}}^{2}}{2} = frac{GMm}{R_{i}}]

Now for [v{_{e}}^{2}] solution:

[v{_{e}}^{2} = frac{2GM}{Ri}]

By applying square root on both sides:

[v_{e} = pm sqrt{frac{2GM}{Ri}}]

However, in this case, the gravitational convention is away from the other body, so escape velocity expression is negative:

[v_{e} = -sqrt{frac{2GM}{Ri}}]

Note: Although with regards to this convention, the negative version is right; in maximum escape velocity books, it is considered to be positive.

Important Facts About the Escape Velocity

The escape velocity of an object from the earth is an important factor that is taken into consideration while launching satellites and other space probes, scientists very precisely calibrate the satellites and other space probes so as to ensure that at no point of time does the velocity of the satellite is higher or lower than the required escape velocity of 11.18 Km/s. Even when the satellites are launched, continuous monitoring is done to ensure the satellites are having the parameters which are necessary for a successful launch. 

The launch sites are always designed at places near to the equator because at the equator the gravitational pull of the Earth is the weakest. At the two poles, the gravitational pull of the earth is very high because along with the gravitational force the magnetic forces are also very crucial in the successful launching of the satellites.  

Escape velocity is always calculated when there is a very huge difference between the size and mass of the two bodies. That is, one of the bodies has a very high mass and size as compared to the body that is trying to escape. Since the escape velocity is dependent on the size and mass of the body whose gravitational pull is to escape, the higher is the difference between the weights and sizes of the two bodies, the higher would be the escape velocity.  There are certain space bodies that have a very huge mass and a  small size e.g black holes. In the case of black holes, the density is so high that the escape velocity is more than the speed of light and as a result, even the light can’t escape the black holes. 

Do It Yourself

(i) Gas escapes from the surface of a planet as it acquires an escape speed. Determine the factors escape velocity depends on:

1. Mass of the planet 

2. Mass of the element escaping 

3. The radius of the planet 

4. Temperature of planet

(ii) Choose the correction combination of answers from the following options:

(a) 1 and 2 are correct 

(b) 1 and 3 are correct 

(c) 1,2 and 3 are correct 

(d) 1 and 4 are correct