Escape Velocity of Earth: Definition, Formula and Derivation

The escape velocity of the earth is defined as the velocity with which an object is projected from the surface of the earth such that the object escaped from its gravitational pull. 

• When we kick a soccer ball into the air, the force of gravity pulls it back down to the ground. 
• This is because the ball does not have enough speed or energy to overcome the gravitational pull of the Earth. 
• However, a rocket is designed to escape the gravitational pull of Earth and reach outer space.
• To escape the Earth's gravitational pull, a rocket needs to reach a speed known as the escape velocity. 
• The escape velocity of Earth is about 11.2 kilometers per second, which means that a rocket needs to reach this speed in order to break free from Earth's gravitational pull and move into space.
• Escape velocity is independent of mass and direction and of projection of the body.

Key Terms: Escape velocity, Gravitational force, Acceleration due to gravity, Value of G, Mass of the earth, Radius of the earth.

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Escape Velocity

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The escape velocity or escape speed is defined as the minimum velocity with which the body has to be projected vertically upwards from the surface of a massive body so that it just crosses the gravitational field of the massive body and does not return on its own.

• It is independent of the mass of the object to be thrown but increases with the increase in mass of the massive body.
• It decreases with the increases in the distance of the object to be thrown from the massive body.
• SI unit of escape velocity is meters per second (m/s or ms^-1).
• The dimensional formula of escape velocity is [M⁰ L¹ T^-1]

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Escape Velocity Formula

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The formula for Escape vlocity is given as

v_e = \(\sqrt{\frac{2GM}{R}}\)

Where

• v_e = escape velocity
• G = Universal Gravitational Constant
• R = distance of the object from the center of mass of massive body

If a body is projected from height h, then escape velocity is given by

v_e = \(\sqrt{\frac{2GM}{R + h}}\)

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Escape Velocity of Earth

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The escape velocity of the earth is the velocity with which an object is thrown vertically upward from the surface of the earth such that the object escaped from its gravitational pull.

We have,

• Mass of the earth, M = 6 x 10²⁴ kg
• Distance of object to be thrown from the center of mass of the earth = Radius of the earth = 6.4 x 10⁶ m.
• G = 6.67 x 10^-11 N m² kg^-2

Substituting the values in the formula of escape velocity, we will get the escape velocity of the earth.

Escape velocity of Earth, v_e = \(\sqrt{\frac{2GM}{R}}\)

⇒ v_e = \(\sqrt{\frac{26.67 \times 10^{-11} \times 6 \times 10^{24}}{6.4\times 10^6}}\)

⇒ v_e = 11.2 x 10³ m/s= 11.2 km/s = 40320 km/h

Hence, the escape velocity of Earth is approximately 11.2 kilometers per second (or about 40,320 kilometers per hour). This means that any object that is launched from the Earth's surface with a velocity of 11.2 km/s or greater will escape the Earth's gravitational pull and move away into space.

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Derivation of Escape Velocity

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Let a body of mass m is projected with velocity equal to escape velocity v_e from a massive body of mass M. Let the distance between the body to be thrown and center of mass of the massive body is R. Then

Initial mechanical energy of the system,

E_i = K_i + P_i = 1/2 mv_e² + ( – GMm/R)

⇒ E_i = 1/2 mv_e² – GMm/R

Where

• K_i = initial kinetic energy
• P_i = initial potential energy

Final mechanical energy of the system at infinity,

E_f = K_f + P_f = 0

Where

• K_f = kinetic energy at infinity
• P_f = potential energy at infinity

As per the principle of conservation of energy, we can write

E_i = E_f

⇒ 1/2 mv_e² - GMm/R = 0

⇒ 1/2 mv_e² = GMm/R

v_e = \(\sqrt{\frac{2GM}{R}}\)

Above equation is valid for all plates.

Also, By multiplying numerator and denominator of RHS by R, we get

v_e = \(\sqrt{\frac{2GMR}{R^2}}\)

But GM/R² = g (acceleration due to gravity)

v_e = \(\sqrt{2gR}\)

If a body is projected from height h, then escape velocity is given by

v_e = \(\sqrt{\frac{2GM}{R+h}}\)

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Escape Velocity of Various Objects

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The following table is the escape velocity of different celestial bodies:

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Things to Remember

• The formula for escape velocity is v_e = √(2GM/R).
• The mass of the object attempting to escape has no effect on the escape velocity.
• The escape velocity of Earth is approximately 11.2 km/s or 40,270 km/h.
• In addition to escape velocity, there is also something called the "gravitational assist," which can be used to help spacecraft achieve greater velocities and escape the gravitational pull of planets or other objects.
• Escape velocity depends on the mass and radius of the planet or object. 
• Generally, larger and more massive objects have a stronger gravitational pull and therefore require higher escape velocities.
• Escape velocity is the minimum velocity an object needs to achieve to escape the gravitational pull of a planet or other celestial body.
• The formula for escape velocity is v_e = √(2GM/R), where G is the gravitational constant, M is the mass of the planet or object, and R is the distance from the center of the planet or object to the object.