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.
    An astronomical telescope consists of two converging lenses. One of them of large aperture and large focal length is called objective lens and the other one, of smaller focal length and smaller aperture is called the eyepiece. It is used to see distant objects which are not seen clearly with naked eyes. The image formed by the objective lens acts as an object for the eyepiece and the final image produced by the eyepiece is magnified.


      • 2.
        Two air-filled capacitors of capacitances $C_1$ and $C_2$ are connected in parallel with a dc battery. After the capacitors are fully charged, a slab of dielectric constant K is inserted between the plates of each capacitor. How will the (i) charge on each capacitor and (ii) energy stored in the capacitor affected after the slab is introduced.


          • 3.
            A student sets up the circuit as shown in the figure to find the value of unknown resistance X and records a set of readings of the voltmeter and the ammeter by using the rheostat.


              • 4.
                In a telescope the objective has much larger aperture than the eye piece. Why ?


                  • 5.
                    This ‘average velocity’ is found be few mm/s for currents in range of a few amperes. How then is current established almost the instant a circuit is closed ?


                      • 6.
                        Two metal spheres of radii $r_1$ and $r_2$ ($> r_1$) having charges $q_1$ and $q_2$ respectively kept in air, are brought in contact. Which of the following statements is not correct ?

                          • The total charge of the two spheres is conserved.
                          • Both spheres attain the same potential.
                          • The final potential of the system equals $\frac{1}{4\pi\epsilon_0} \frac{(q_1 + q_2)}{(r_1 + r_2)}$
                          • The final potential of the system equals $\frac{1}{4\pi\epsilon_0} \frac{(q_1 + q_2) (r_1 + r_2)}{r_1 r_2}$
                        CBSE CLASS XII Previous Year Papers

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