Escape Velocity Formula with Solved Examples

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An object requires an escape velocity to overcome earth’s gravitational pull and escape out without falling back. For instance, when a stone is hurled from the ground, it flies into the air before crashing back to earth. The stones can be flung at even higher speeds when using machines. The question then arises as to how fast a person must throw a stone into the sky for it to beat the gravitational pull and fly out of the earth into space. This is the concept behind the term "escape velocity," which is used when discussing outer space and space exploration missions. To comprehend and grasp how much work and effort is put into designing these missions, it is necessary to study the concepts behind this word. Let's take a closer look at these ideas.

Read Also: Relation Between G and g

Table of Content

What is Escape Velocity
The formula of Escape Velocity
Things To Remember
Sample Questions

Key Terms: Gravitational pull, Velocity, Speed, Escape Velocity, Potential Energy, Kinetic Energy

What is Escape Velocity?

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Escape velocity is the minimum velocity that any item requires to escape the earth's gravitational pull and travel into space without falling back on earth. Here, it's important to consider how gravity affects and works on the items. When an object is flung into space, gravitational pull works on it. As a result of this work, the object's kinetic energy is reduced and transformed to potential energy. The object comes to a halt and falls back to the earth's surface due to the decrease in kinetic energy.

In this scenario, the notion of energy conservation comes to our aid. Assume that an object is taken from the earth and reaches beyond its gravitational field, at a velocity of Vf. The total energy of the object is the sum of its potential and kinetic energy.

Read More:

Neptune

Value of Gravitational Constant

Central Force

The formula of Escape Velocity

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The formula for escape velocity is presented.

Where,

V is the escape velocity
G is the gravitational constant is 6.67408 × 10-11 m³ kg^-1 s^-2
M is the mass of the planet
R is the radius from the center of gravity.

An alternate expression for escape velocity that is particularly useful at the body's surface is

Read More: Derivation of Escape Velocity

Things To Remember

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The square root of the mass (M) of the planet (earth) around which the satellite orbits determines the escape velocity of a body.
The square root of the planet's radius is inversely proportional to a body's escape velocity.
The term m' is missing from the equation, indicating that the critical velocity is independent of the satellite's mass.
The direction of projection has no bearing on a body's escape velocity.

Read More:

Value of g on Moon

Gravitational Force Formula

Specific Gravity Formula

Sample Questions

Ques. Determine the escape velocity of Jupiter if its radius is 7149 Km and mass is 1.898 × 10²⁷ Kg.

Ans. Given: Mass M = 1.898 × 10²⁷ Kg,

Radius R = 7149 Km

Gravitational Constant G = 6.67408 × 10-11 m³ kg^-1 s^-2

Escape Velocity is given as

Vesc = √2GM / R

=√2 x 6.67408 × 10-11 x 1.898 ×10²⁷/ 7149

50.3 km/s

Ques. Determine the escape velocity of the moon if Mass is 7.35 × 1022 Kg and the radius is 1.5 × 10⁶ m.

Ans. Given

M = 7.35 × 10²² Kg,

R = 1.5 × 10⁶ m

Escape Velocity formula is given by

Vesc = √2GMR

= √2×6.673×10^-11×7.35×10²² / 1.5×10⁶

= 7.59 × 10⁵m/s

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

Ans. The formula for calculating escape velocity is as follows:

Escape Velocity

Given: M = 7.35 × 10²² Kg

R = 1.5 × 10⁶m

G = 6.6 × 10^-11

plugging the values into the equation,

Vi = 7.6 × 10⁵ m/s.

Ques. Find the escape velocity for a planet whose mass is 14.7 × 10²² Kg and radius is 3 × 10⁶m.

Ans. The formula for escape velocity is given by,

Escape Velocity

Given: M = 14.7 × 10²² Kg

R = 3 × 10⁶m

G = 6.6 × 10^-11

plugging the values into the equation,

Given: R = 2 × 10⁵m

G = 6.6 × 10^-11

Vi = 10⁴ m/s

plugging the values into the equation,

⇒

⇒ 2 × 10¹³ = 2 × 6.6 × 10^-11 × M

⇒ 0.15 × 10²⁴ = M

⇒ 1.5 × 10²³ = M

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

Ans. The formula for escape velocity is given by,

Given: R = 2 × 10⁵m

G = 6.6 × 10^-11

Vi = 104 m/s

plugging the values into the equation,

⇒

⇒ 2 × 10¹³ = 2 × 6.6 × 10^-11 × M

⇒ 0.15 × 10²⁴ = M

⇒ 1.5 × 10²³ = M

Ques. Find the escape velocity if the radius of the earth is increased by 4 times.

Ans. The formula for escape velocity in earth’s case becomes,

Notice that,

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

The escape velocity of the earth is 11.2 Km/s.

The escape velocity of the earth with an enlarged radius will be 22.4 Km/s.

Ques. To leave the moon, the Apollo astronauts had to take off in the lunar module, and reach the escape velocity of the moon. The radius of the moon is 1.74x10⁶ m, and the mass of the moon is 7.35x10²² kg. What velocity did Neil Armstrong and Buzz Aldrin in the lunar module have to reach to leave the moon?

Ans. The escape velocity from the moon can be found using the formula:

Escape Velocity

V_escape= 2374 m/s

The escape velocity from the moon is 2374 m/s or approximately 2.37 km/s.

Ques. The radius of Earth is 6.38x10⁶ m, and the mass of the Earth is 5.98x10²⁴ kg. What is the escape velocity from Earth?

Ans. The escape velocity from Earth can be found using the formula:

Escape Velocity

V_escape= 11184 m/s

The escape velocity from Earth is 11 184 m/s or approximately 11.2km/s.

Physics Related Links:

Orbital Velocity Formula	Earth Satellites	Escape Speed
Gravitational Potential Energy	Weightlessness	Motion of Celestial Bodies in Space

Escape Velocity Formula with Solved Examples

What is Escape Velocity?

The formula of Escape Velocity

Things To Remember

Sample Questions

CBSE CLASS XII Related Questions

2.
The figure shows three point charges kept at the vertices of triangle ABC. The net electric field, due to this system of charges, at the midpoint M of base BC will be:

3.
Two small identical metallic balls having charges \( q \) and \( -2q \) are kept far at a separation \( r \). They are brought in contact and then separated at distance \( \frac{r}{2} \). Compared to the initial force \( F \), they will now:

4.
Write the expression for the magnetic field due to a current element in vector form. Consider a 1 cm segment of a wire, centered at the origin, carrying a current of 10 A in positive x-direction. Calculate the magnetic field \( \mathbf{B} \) at a point \( (1 \, \text{m}, 1 \, \text{m}, 0) \).

5.
Two thin lenses of focal length \( f_1 \) and \( f_2 \) are placed in contact with each other coaxially. Prove that the focal length \( f \) of the combination is given by \[ f = \frac{f_1 f_2}{f_1 + f_2}. \]

6.
Photoemission of electrons occurs from a metal (\( \phi_0 = 1.96 \, \text{eV} \)) when light of frequency \( 6.4 \times 10^{14} \, \text{Hz} \) is incident on it. Calculate: Energy of a photon in the incident light, The maximum kinetic energy of the emitted electrons, and The stopping potential.

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Escape Velocity Formula with Solved Examples

What is Escape Velocity?

The formula of Escape Velocity

Things To Remember

Sample Questions

CBSE CLASS XII Related Questions

2.The figure shows three point charges kept at the vertices of triangle ABC. The net electric field, due to this system of charges, at the midpoint M of base BC will be:

3.Two small identical metallic balls having charges \( q \) and \( -2q \) are kept far at a separation \( r \). They are brought in contact and then separated at distance \( \frac{r}{2} \). Compared to the initial force \( F \), they will now:

4.Write the expression for the magnetic field due to a current element in vector form. Consider a 1 cm segment of a wire, centered at the origin, carrying a current of 10 A in positive x-direction. Calculate the magnetic field \( \mathbf{B} \) at a point \( (1 \, \text{m}, 1 \, \text{m}, 0) \).

5.Two thin lenses of focal length \( f_1 \) and \( f_2 \) are placed in contact with each other coaxially. Prove that the focal length \( f \) of the combination is given by \[ f = \frac{f_1 f_2}{f_1 + f_2}. \]

6.Photoemission of electrons occurs from a metal (\( \phi_0 = 1.96 \, \text{eV} \)) when light of frequency \( 6.4 \times 10^{14} \, \text{Hz} \) is incident on it. Calculate: Energy of a photon in the incident light, The maximum kinetic energy of the emitted electrons, and The stopping potential.

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2.
The figure shows three point charges kept at the vertices of triangle ABC. The net electric field, due to this system of charges, at the midpoint M of base BC will be:

3.
Two small identical metallic balls having charges \( q \) and \( -2q \) are kept far at a separation \( r \). They are brought in contact and then separated at distance \( \frac{r}{2} \). Compared to the initial force \( F \), they will now:

4.
Write the expression for the magnetic field due to a current element in vector form. Consider a 1 cm segment of a wire, centered at the origin, carrying a current of 10 A in positive x-direction. Calculate the magnetic field \( \mathbf{B} \) at a point \( (1 \, \text{m}, 1 \, \text{m}, 0) \).

5.
Two thin lenses of focal length \( f_1 \) and \( f_2 \) are placed in contact with each other coaxially. Prove that the focal length \( f \) of the combination is given by \[ f = \frac{f_1 f_2}{f_1 + f_2}. \]

6.
Photoemission of electrons occurs from a metal (\( \phi_0 = 1.96 \, \text{eV} \)) when light of frequency \( 6.4 \times 10^{14} \, \text{Hz} \) is incident on it. Calculate: Energy of a photon in the incident light, The maximum kinetic energy of the emitted electrons, and The stopping potential.