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Escape Velocity

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Escape velocity as the name suggests, is the velocity required by an object to escape from the gravitational barrier of any celestial object. “What happens when you throw a stone upward in the air?” The stone comes back to the Earth’s surface. If we throw the stone with a much higher force still it comes back to the Earth’s surface because of the gravitational force of Earth. 

What happens when we keep increasing the force the initial velocity of the stone keeps getting higher till it reaches the point when it never comes back to the Earth’s surface. The velocity at this point is called Escape velocity. Let’s learn more about Escape Velocity and others in detail in this article.

Escape Velocity Definition

When an object reaches the speed that it is capable of breaking the Earth’s gravitational barrier then we say that it is travelling at escape velocity. i.e. Escape velocity is the minimum velocity that when an object travels at that speed can break the gravitational barrier of that celestial body. 

The image added below shows a rocket escaping the Earth’s gravitational field, the velocity must be greater than the escape velocity of the Earth for the rocket to leave Earth’s gravitational pull. The satellite revolving around the Earth has a specific velocity that is less than the Escape velocity of the Earth but high enough to provide the satellite with enough energy to revolve around the Earth. If the velocity of the object is less than the orbital velocity of the planet then the object thrown in the atmosphere returns back to the Earth’s surface as the ball thrown by the boy does.

Escape Velocity and other of Earth

 

Note: The escape velocity at the Earth’s Surface is 11.2 km/sec.

Escape Velocity Equation

The escape velocity equation is obtained by equating the kinetic energy of an object with mass m and travelling with a velocity of v and the gravitational potential energy of the same object. When the Kinetic energy of the object equals the gravitational energy of the celestial object the equation so obtained is called the Escape Velocity Equation. 

For Earth let the escape velocity is vc then the kinetic energy is 1/2 m(vc)2 if m is the mass of the object and now the gravitational potential energy is equal to the kinetic energy and the gravitational potential energy of the object is GMm/r where G is the universal gravitational constant and M is the mass of the Earth then,

1/2 m(vc)2 = GMm/r 

Thus, the equation of Escape Velocity is given by

\bold{v_c = \sqrt{\frac{2GM}{r}}}

Where,

  • vc is the Escape Velocity of the object, 
  • G is the Universal Gravitational Constant, 
  • M is the Mass of the Celestial Object whose gravitational force is to break, and 
  • r is the distance from the object to the centre of mass of the body which has to be escaped.

Escape Velocity of Earth

The velocity required by an object to escape from Earth’s Gravitational Force is called the Escape Velocity of Earth. Now,

Let the escape velocity of the Earth be vc then,

vc = √(2GM/r)

We know that,

GM/r2 = g (where g is Earth’s gravity)

vc = √(2gr)

We know, Radius of Earth, r = 6.4 × 106 m, and

g = 9.8 m/s2

Now the escape velocity of the Earth, 

vc = √(2 × 9.8 × 6.4 × 106)

⇒ vc = 11.2 × 103 m/s 

⇒ vc = 11.2 km/s

Thus, the escape velocity of the Earth is 11.2 km/s.

Unit and Dimension of Escape Velocity

Escape velocity is the velocity that is required to break the gravitational barrier of any celestial object so the unit of escape velocity is similar to the unit of velocity.

The Escape Velocity is measured in Km/sec, m/sec, or in the other units of velocity and the dimensional formula of Escape Velocity is [LT-2].

Escape Velocity on Different Celestial Objects

Different Celestial Objects in our solar system have different escape velocities due to different gravitation forces of attraction (as they have different masses and different radii). The Escape velocity of different objects in our solar system is discussed in the table below,

Celestial Body

 Escape Velocity (in km/s)

Sun

618 km/s

Moon

2.38 km/s

Earth

11.2 km/s

Jupiter

59.5 km/s

Mercury

4.25 km/s

Mars

5.03 km/s

Venus

10.36 km/s

Solved Examples on Escape Velocity

Example 1: Find the escape velocity for a planet whose mass is 7.35 × 1022 Kg and radius is 1.5 × 106m. 

Solution: 

The formula for escape velocity is given by, 

V_i = \sqrt{\frac{2GM}{R}}

Given,

M = 7.35 × 1022 Kg

R = 1.5 × 106m

G = 6.6 × 10-11

plugging the values into the equation, 

V_i = \sqrt{\frac{2GM}{R}}

V_i = \sqrt{\frac{2 \times 6.6 \times 10^{-11} \times 7.35 \times 10^{22}}{1.5 \times 10^6}}

Vi = 7.6 × 105 m/s

Example 2: Find the escape velocity for a planet whose mass is 14 × 1022 Kg and radius is 2 × 106m. 

Solution: 

The formula for escape velocity is given by, 

V_i = \sqrt{\frac{2GM}{R}}     …(1)

Given,

M = 14.7 × 1022 Kg

R = 3 × 106m

G = 6.6 × 10-11

Putting the values in equation (1)

V_i = \sqrt{\frac{2GM}{R}}

V_i = \sqrt{\frac{2 \times 6.6 \times 10^{-11} \times 14.7 \times 10^{22}}{3 \times 10^6}}      

Vi = 3.04 × 103 m/s

Escape Velocity of the planet is .04 × 103 m/s

Example 3: Find the mass of a planet whose escape velocity is 2 × 104 m/s and radius is 2 × 105m. 

Solution: 

The formula for escape velocity is given by, 

V_i = \sqrt{\frac{2GM}{R}}     …(1)

Given,

R = 2 × 105m

G = 6.6 × 10-11

Vi = 2 × 104 m/s

Putting the values in equation (1)

V_i = \sqrt{\frac{2GM}{R}}

 2 \times 10^4 = \sqrt{\frac{2 \times 6.6 \times 10^{-11} \times M}{2 \times 10^5}}

8 × 1013 = 2 × 6.6 × 10-11 × M 

4 × 0.15 × 1024 = M 

6 × 1023 Kg = M

The mass of the planet is 6 × 1023 Kg

Example 4: Find the escape velocity if the radius of the Earth is increased by 4 times. 

Solution: 

The formula for escape velocity is given by, 

V_i = \sqrt{\frac{2GM}{R}}

In Earth’s case it becomes, 

(V_{min}) = \sqrt{2gR_e}

Now,

V_{min} \propto  \sqrt{R_e}

If the radius is made four times, this means that the velocity must be doubled. 

Escape Velocity of Earth = 11.2 Km/s. 

Escape Velocity of Earth with with 4 times radius = 11.2 × 2 = 22.4 Km/s. 

FAQs on Escape Velocity

Q1: What is Escape Velocity?

Answer:

The velocity required to break free the gravity barrier of any celestial object is called the escape velocity of that object. Escape velocity of Earth is 11.2 km/sec

Q2: What is the Escape Velocity Formula?

Answer:

The formula which is used to calculate the escape velocity of any object is,

vc = √(2GM/r)

Where,

  • vc is the Escape Velocity of the object, 
  • G is the Universal Gravitational Constant, 
  • M is the Mass of the Celestial Object whose gravitational force is to break, and 
  • r is the distance from the object to the centre of mass of the body which has to be escaped.

Q3: What is the difference between orbital velocity and escape velocity?

Answer:

Orbital Speed: The speed required to reach the orbit of a planet or any celestial object is called as the orbital speed of that object. Orbital speed of the Earth is 7.34 km/sec.

Escape Velocity: The velocity required to break the gravitational barrier of the celestial object is called the escape velocity of that object. Escape velocity of Earth is 11.2 km/sec.

Q4: Which planet has the highest escape velocity in our solar system?

Answer:

Jupiter has the highest escape velocity among all the planets in our solar system. The escape velocity of Jupiter is 59.5 km/sec.

Q5: Which planet has the lowest escape velocity in our solar system?

Answer:

Mercury has the lowest escape velocity among all the planets in our solar system. The escape velocity of Mercury is 4.2.5 km/sec.



Last Updated : 29 May, 2023
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