Derivation of Escape Velocity: Definition, Formula and Derivation

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Derivation of Escape Velocity of an entity can be obtained by comparing kinetic and potential energy values at a specific point. Furthermore, it is calculated with the application of the Law of Conservation of Energy which states that energy can neither be created nor destroyed. It can be transformed from one form to another e.g potential energy of a ball at a certain height can be converted to its kinetic energy if the ball falls from a height.

Key Terms: Escape Velocity, Gravitation, Escape Velocity of Earth, Escape Velocity of Moon, Escape Speed


What is Escape Velocity?

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Escape velocity can be defined as the minimum velocity value that an object should acquire to be launched into space so that it overcomes the gravitational force of earth or any other planet. The conditions that we need to be mindful of for this mechanism are that the object needs to reach an infinite distance from the planet and that the object should be free.

The general formula for escape velocity is:

ve= √2GM/R

Escape Velocity

Escape Velocity


Important Formulas

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Derivation of escape velocity involves the relationship between an object and its velocity. It also includes kinetic energy and gravitational potential energy. We use the initial kinetic and initial gravitational energy for a particular height to find escape velocity. Let's see how we can find the final formula of the escape velocity.

Initial Kinetic Energy

We know that kinetic energy is defined as the energy of an object in its motion. We are using the kinetic energy value where the initiation of the movement happened. The formula for initial kinetic energy is:

KEi = mve2/2

Where,

  • KEi is the initial kinetic energy
  • m is mass of the body
  • ve is the initial velocity which is taken in km/s and will eventually give us escape velocity

Initial Potential Energy

Potential energy is the amount of energy stored in an object when it is in a stationary state. The formula for initial potential energy is:

PEi = -GMm/Ri

Here,

  • PE denotes the initial gravitational potential energy which will be taken in kg-km2/s2
  • G is a gravitational constant which has a value of 6.674 X 10-20 km3/kg-s2
  • M is the mass of the bigger body which is exerting gravitational force
  • m is taken as the mass value for the escaping body
  • R is the value for the initial distance between both objects

This distance will be measured from the center of the body in km. 

Remember that M has a larger value than m.

Taking the sum of initial kinetic and potential energy values, we will get:

TEi = KEi + PEi

TEi = mve2/2 - GMm/Ri

Now, we will calculate both of these energies at infinity.

Potential Energy at Infinity

The formula for potential energy at infinity is:

PEn = -GMm/Rn

Here,

PEn is the gravitational potential energy at infinity distance

R­n is the infinite distance between both objects which is taken from the center of the objects.

From the infinite distance, the value of PEn is 0.

Kinetic Energy at Infinity

The formula for kinetic energy at infinity distance is:

KE­­n = mvn2/2

Here, KEn is the kinetic energy at infinity distance,

m is the mass of the object, and

vn is the final velocity at infinity distance. 

Now the value of the KEn is 0 because vn is also 0.

Whole Final Energy

As the potential energy acts downward and the kinetic energy moves upward, the whole final energy at the initial location from all the above formulas will be:

TEn = KEn+ PEn

TEn = 0 + 0 

= 0


Derivation of Escape Velocity

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Considering the law of conservation of energy, if we take a closed system then all the energy remains conserved. Let's assume that the closed system encompasses two objects. Both of them are attracting each other through gravitational force. So in that case,

TEi = TE

KEi+ PEi = 0

If we put the values in the above equation then,

mve2/2 - GMm/Ri = 0

Now, if we open the ve2= 2GM/Ri

Applying square root on both sides then, the equation will become

ve= √(2GM/Ri)

Now as the gravitation convention is away from another body, the velocity expression will be negative-

ve = –√(2GM/Ri)

As per the convention, the escape velocity is negative. However, the escape velocity is generally taken as positive to get the maximum value.

Also check: Universal Gravitation Formula


Escape Velocity of Earth

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ve = √(2GM/Ri)

Putting g = GM

ve = √(2gRi)

Hence, the escape velocity of a planet depends on the radius and mass of the planet and not on the mass of the object.

Now, for Earth

g = 9.8 m

R = 6378100 m

Hence, the escape velocity for Earth becomes:

ve = √2 x 9.8 x 6378100

= 11.2 km/s

Escape Velocity of Earth
Escape Velocity of Earth

Things to Remember based on Derivation of Escape Velocity

  • Escape velocity can be defined as the minimum velocity value that an object should acquire to be launched into space so that it overcomes the gravitational force of earth or any other planet.
  • Derivation of escape velocity involves the relationship between an object and its velocity.
  • Initial Kinetic Energy: KEi = mve2/2
  • Initial Potential Energy: PEi = -GMm/Ri
  • Potential Energy at Infinity: PEn = -GMm/Rn
  • Kinetic Energy at Infinity: KE­­n = mvn2/2
  • Escape Velocity Formula: ve = √(2GM/Ri)
  • Escape Velocity of Earth: 11.2 km/s

Also check:


Sample Questions based on Derivation of Escape Velocity

Ques. Which law is used for deriving escape velocity? (2 Marks)

Ans. The law of conservation of energy is used for deriving the escape velocity formula. It signifies that the amount of energy always remains conserved and that’s why the value of escape velocity remains the same for each object.

Ques. What is the minimum value of escape velocity? (2 Marks)

Ans. The minimum value of escape velocity is 11.186 km/s. This signifies that the object needs to be launched with a speed of a minimum of 11.186 km/s so that it can overcome the gravitational force of the planet or the attracting body.

Ques. Give the value of escape velocity of a planet if its radius is given as 7000km and mass is 107 kg. (3 Marks)

Ans. The formual for minimum ve is √(2GM/Ri)1/2

So, by putting values G = 6.67X10-11 N kg-2m2

M= 107kg

R= 7000 X 1000 = 7 X 106

Now putting these values in the formula-

(2 X 6.67X10-11 X 107 / 7 X 106)½

= 0.21 km/s

Ques. Discuss the dependency of escape velocity on mass. (2 Marks)

Ans. The escape velocity of an object doesn’t depend on its own mass. But it depends on the mass of the object which is exerting force or from which the other object is trying to escape. The size of the attractive body also matters in such a case.

Ques. State the escape velocity of the sun and moon. (2 Marks)

Ans. The escape velocity of the sun and moon is 618 km/s and 2.38 km/s respectively.

Ques. What is the formula for: (5 Marks)
i. Initial Kinetic energy
ii. Initial Potential Energy
iii. Potential energy at infinity
iv. Kinetic energy at infinity 
v. Escape velocity, ve

Ans. The formulas are as follows:

  • Initial kinetic energy, KEi = mve2/2
  • Initial potential energy, PEi = -GMm/Ri
  • Potential energy for infinity, PEn = -GMm/Rn
  • Kinetic energy for infinity, KE­­n = mvn2/2
  • The formula of escape velocity, ve = √(2GM/Ri)

Also check:

Check out:

CBSE CLASS XII Related Questions

  • 1.
    In a region, the electric potential varies as \(V = 10 - 50x\), where \(V\) is in volts and \(x\) is in meters. The electric field in the region is

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    • 2.
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        • 3.

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            • 4.
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                • \(\frac{1}{4\pi\epsilon_0} \frac{\sqrt{2}Q}{d^2} (-\hat{i} - \hat{j})\)
                • \(\frac{1}{4\pi\epsilon_0} \frac{Q}{d^2} (\hat{i} - \hat{j})\)
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              • 5.

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                      CBSE CLASS XII Previous Year Papers

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