Escape Velocity - Formula, Derivation, Escape Velocity of Earth, Moon, Solved Questions

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Escape velocity can be defined as the minimum speed or velocity required by a mass that must be projected from the earth's surface to escape the gravitational pull of the planet. In simpler terms, this is the speed required by the object to escape from the gravitational force applied on the object by our planet. This is called escape velocity (Ve).

Escape Speed is an essential topic covered in CBSE Class 11 Physics Unit VI Chapter VIII Gravitation. As per the syllabus released by CBSE, Units IV-VI will carry a total weightage of 17 marks in Class 11 Examinations. Also Check: CBSE Class 12 Physics

Check : CBSE Class 11 Chapter 8 Important Questions free pdf download


What is Escape Velocity?

[Click Here for Sample Questions]

Gravitational Pull prevents any object from escaping into space. This raises the question, how can space shuttles, rovers and satellites leave the atmosphere and reach space. 

Space shuttles require an immense amount of energy to break through the gravitational pull of the earth. This so-called speed is what is referred to as Escape Speed or Escape Velocity, a necessity a space shuttle must achieve to escape the velocity of celestial bodies such as planets. 

Escape speed can be defined as the minimum speed or velocity required by a mass that must be projected from the earth's surface to escape the gravitational pull of the planet. 

In simpler terms, this is the speed required by the object to escape from the gravitational force applied on the object by our planet. This is called escape velocity (Ve).

If we view our planet as the massive body exerting a gravitational pull on every object, the escape velocity is the minimum speed or velocity which is required by the object to overcome this gravitational field and travel into space without falling or being pulled back.

Unit of Escape Velocity

The unit of escape speed or velocity is meter per second (m.s-1). This is also termed the SI unit of escape velocity. Its dimension is MT-1

Factors Affecting Value of Escape Velocity

Escape velocity depends on several factors other than just the gravitational pull of the earth.

Escape Velocity doesn't just depend on the mass of the object. This escape velocity depends on the mass and radius of the massive object which we assume is our planet from which our object is being released.

The more the mass of the massive body and closer the object to our planet, the higher will be the escape velocity. Similarly, if the mass of the massive body isn't as large and it is further away from the massive body, the escape velocity required isn't as great.

This has two consequences –

  1. Different planets therefore will have different escape velocities.
  2. The escape velocity will remain the same for a rocket as compared to an atom trying to escape the massive object's gravitational pull.

In correspondence to the principle of conservation of energy;

(K + Ug)i = (K + Ug)f

Here,

K = ½ mv2

U = GMm/r

Here, Ugf will be considered zero as the distance is infinity and Kf will also become zero as the final velocity equates to zero.

Therefore, the minimum velocity required to escape the gravitational pull of earth or any massive body remains;

Ve = √2gr

Where,

g = GM/r2

Dimensional Formulas of Escape Velocity

The dimensional formula of the earth’s mass is M1L0T0.

Whereas the dimensional formula of the universal gravitational constant is M-1L3T-2.

The dimensional formula of the centre of the earth to the distance covered is M0L1T0.

Therefore, after substituting the following equations, the dimensional formula of escape velocity is M0L1T-1.


Sample Questions

Question. What is the Escape Velocity formula? (1 mark)

Ans. The escape velocity formula is expressed as;

Ve = √2GM/r

Where,

Ve – The escape speed or velocity

G – Newton’s universal constant gravitational force (6.67 x 10-11 meters3/(kg)(second)2)

M – Mass of the planet

r – Radius of the planet (since in this case we consider the radius of planet earth, we take the value which is approximately 5.97 x 1024 kg.

Question. What does Escape velocity depend on? (2 Marks)

Ans. The escape speed or velocity depends on only the mass and size of the massive body from which the object is trying to escape. The mass of the actual object doesn't matter, therefore the escape speed required for a space shuttle and a ball to leave the earth's atmosphere would be the same.

Question. What is Escape Velocity Derivation? (2 Marks)

Ans: It is easy to calculate the minimum velocity of an object required to overcome a particular planet’s or object’s gravitational pull. Here, the derivation of escape velocity is outlined in a straightforward and easy to understand manner that will help to learn the concept without any hassles.

To derive an expression for escape velocity, the formula is written is as follows:

Ve = √2gR

CBSE CLASS XII Related Questions

  • 1.
    A beam of light coming from a distant source is refracted by a spherical glass ball (refractive index 1.5) of radius 15 cm. Draw the ray diagram and obtain the position of the final image formed.


      • 2.
        Two wires made of the same material have the same length \( l \) but different cross-sectional areas \( A_1 \) and \( A_2 \). They are connected together with a cell of voltage \( V \). Find the ratio of the drift velocities of free electrons in the two wires when they are joined in: (i) series, and (ii) parallel.


          • 3.
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            potential difference across capacitor C


              • 4.
                A conductor of length \( l \) is connected across an ideal cell of emf E. Keeping the cell connected, the length of the conductor is increased to \( 2l \) by gradually stretching it. If R and \( R' \) are initial and final values of resistance and \( v_d \) and \( v_d' \) are initial and final values of drift velocity, find the relation between:
                \( R' \) and \( R \)
                \( R' = 4R \)


                  • 5.
                    You are required to design an air-filled solenoid of inductance 0.016 H having a length 0.81 m and radius 0.02 m. The number of turns in the solenoid should be:

                      • 2592
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                      • 2976
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                    • 6.
                      A square loop MNPK of side \( l \) carrying a current \( I_2 \) is kept close to a long straight wire in the same plane and the wire carries a steady current \( I_1 \) as shown in the figure. Obtain the magnitude of magnetic force exerted by the wire on the loop.
                      square loop MNPK of side l carrying a current I2

                        CBSE CLASS XII Previous Year Papers

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