Gravitational Force Escape Velocity

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Gravitational Force Escape Velocity

1.0What is Gravitational Force?

Gravitational force is the universal attractive force between any two masses. According to Newton’s Universal Law of Gravitation: the force is directly proportional to the product of the masses and inversely proportional to the square of the distance between their centers. This fundamental concept is essential for JEE-level physics understanding.

2.0Newton’s Law of Gravitation

Sir Isaac Newton formulated the law of gravitation in the 17th century, which states:

Every particle of matter in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

This law provided the foundation to explain Kepler’s laws of planetary motion and laid the groundwork for space science.

3.0Gravitational Force Formula and Its Importance

The gravitational force formula is given by: $F = G \frac{m _{1} m _{2}}{r ^{2}}$

Where:

F = gravitational force between the two masses
$m_{1} an d m_{2}$ = interacting masses
r = distance between the centers of the masses
G = universal gravitational constant = $6.67 \times 1 0^{- 11} N m^{2} / k g^{2}$

Newton’s Universal Law of Gravitation explains that every particle in the universe attracts every other particle with a force that is:

Directly proportional to the product of their masses.
Inversely proportional to the square of the distance between them.

$F = G \frac{m _{1} m _{2}}{r ^{2}}$

This law not only explains the falling of objects on Earth but also the motion of planets around the Sun, the Moon around the Earth, and the structure of galaxies.

4.0Escape Velocity ( $V_{e}$ )

It is the minimum velocity required for an object located at the planet's surface so that it just escapes the planet's gravitational field.

Consider a projectile of mass m, leaving the surface of a planet (or some other astronomical body or system), of radius R and mass M with escape speed.

When the projectile just escapes to infinity, it has neither kinetic energy nor potential energy.

From conservation of mechanical energy

$\frac{1}{2} m v_{e}^{2} + (\frac{- GM m}{R}) = 0 + 0 \Rightarrow v_{e} = \frac{2 GM}{R}$

The escape velocity of a body from a location which is at height 'h' above the surface of planet, we can use :-

$v_{e} = \frac{2 GM}{r} = \frac{2 GM}{R + h}$ ${∵ r = R + h}$

Where, r = Distance from the centre of the planet, h = Height above the surface of the planet

Escape speed depends on:

(i) Mass (M) and radius (R) of the planet

(ii) Position from where the particle is projected

Escape speed does not depend on:

(i) Mass (m) of the body which is projected

(ii) Angle of projection

If a body is thrown from the Earth's surface with escape speed, it goes out of earth's gravitational field and never returns back to the earth's surface.

For Earth,

$v_{e} = \frac{2 GM}{R _{e}}$

Other forms, $g_{s} = \frac{GM}{R _{e}^{2}} \Rightarrow v_{e} = 2 g_{s} R_{e}$

On putting values : $v_{e} = 11.2 km/s, g_{s} = 9.81 m/s, R_{e} = 6400 km$

The escape velocity formula is:

$v_{esc} = \frac{2 GM}{R}$

Where:

$v_{esc}$ = escape velocity
GG = gravitational constant
MM = mass of the planet
RR = radius of the planet

For the Earth, substituting values,

$M = 5.972 \times 1 0^{24} k g$

$R = 6.371 \times 1 0^{6} m$

$G = 6.67 \times 1 0^{- 11} N m^{2} / k g^{2}$

$v_{esc} = \frac{2 \times 6.67 \times 1 0 ^{- 11} \times 5.972 \times 1 0 ^{24}}{6.371 \times 1 0 ^{6}}$

$v_{esc} \approx 11.2 km / s$

Thus, the escape velocity of Earth is approximately 11.2 km/s. This means that if a body is projected with this speed (without air resistance), it will leave Earth’s gravitational field.

5.0Escape velocity from a point other than surface

Keep following points in mind !!

Total energy is zero at any point when particle start moving with escape velocity, TE = 0
If given point is at distance r (> R) from center of Earth, $P E_{o u t} = - G M_{e} m / r$
If given point is at distance r (< R) from center of Earth, $P E_{in} = - GM (3 R^{2} - r^{2}) / (2 R^{3}) \times m$

6.0Escape Energy-Binding Energy

Minimum energy given to a particle in the form of kinetic energy so that it can just escape the Earth's gravitational field.

Magnitude of escape energy = GMm / R

(−ve of PE on the Earth's surface)
Escape energy = Kinetic Energy corresponding to the escape velocity ⇒

GMm / R = 1/2 mvₑ²

Note: In the above discussion it can be any planet for that matter

KE < Escape Energy v₀ < vₑ Body returns to Earth surface
KE = Escape Energy v₀ = vₑ Body comes to rest at infinity
KE > Escape Energy v₀ > vₑ Body has residual velocity at infinity

Binding energy

Total energy of a particle near Earth.

BE < 0 Particle cannot escape the gravitational field of Earth
BE ≥ 0 Particle can escape the gravitational field of Earth

Now that we understand gravitational attraction, a natural question arises:

How fast must an object be launched from the surface of a planet so that it escapes the planet’s gravitational pull without further propulsion?

This minimum speed is called the escape velocity.

If an object has speed less than escape velocity → it will eventually fall back.
If its speed equals or exceeds escape velocity → it will move away indefinitely and not return.

Escape velocity is independent of the mass of the object being launched and depends only on the planet’s mass and radius.

7.0Relation Between Gravitational Force and Escape Velocity

The connection between gravitational force and escape velocity lies in the work required to overcome gravitational potential energy.

Gravitational force holds objects close to the Earth by continuously pulling them inward.
Escape velocity is the exact speed required to provide a body with enough kinetic energy to overcome the gravitational potential energy binding it to the planet.

The stronger the gravitational force (i.e., the larger the mass of the planet and the smaller its radius), the higher the escape velocity. For massive planets like Jupiter, escape velocity is much higher than Earth’s, while for smaller bodies like the Moon, it is much lower.

8.0Factors Affecting Escape Velocity

The escape velocity of a planet depends only on its mass and radius, not on the mass of the escaping object.

From the formula: $v_{esc} = \frac{2 GM}{R}$

A larger planetary mass (M) → stronger gravitational pull → higher escape velocity.
A larger planetary radius (R) → weaker gravitational pull at the surface → lower escape velocity.

For example:

Escape velocity on Moon ≈ 2.38 km/s (much smaller than Earth due to smaller mass and radius).
Escape velocity on Jupiter ≈ 59.5 km/s (much larger because of massive size).

9.0Applications of escape velocity

The concept of escape velocity is central in astrophysics, space exploration, and exam-oriented problem-solving.

Space Missions: To launch satellites or spacecraft, rockets must reach at least escape velocity to overcome Earth’s gravitational pull.
Planetary Atmospheres: Planets with low escape velocities (like Mercury) cannot retain lighter gases in their atmosphere, explaining why they lack a thick atmosphere.
Astrophysics: Understanding black holes relies on escape velocity. At the event horizon, the escape velocity equals the speed of light.

10.0Solved Problems on Gravitational Force

Question 1: Two masses of 5 kg and 10 kg are placed 2 m apart in vacuum. Find the gravitational force between them.

Solution:

Using Newton’s law of gravitation:

$F = G \frac{m _{1} m _{2}}{r ^{2}}$

Substitute values:

$F = \frac{6.67 \times 1 0 ^{- 11} \times 5 \times 10}{( 2 ) ^{2}}$

$F = \frac{3.335 \times 1 0 ^{- 9}}{4}$

$F = 8.34 \times 1 0^{- 10} N$

Answer: The gravitational force is $8.34 \times 1 0^{- 10} N$

Question 2: The gravitational force between two masses is $2.67 \times 1 0^{- 9} N$ If one mass is 1 kg, the other is 2 kg, find the separation between them.

Solution:

$F = G \frac{m _{1} m _{2}}{r ^{2}}$

$r^{2} = G \frac{m _{1} m _{2}}{F}$

$r^{2} = \frac{6.67 \times 1 0 ^{- 11} \times 1 \times 2}{2.67 \times 1 0 ^{- 9}}$

$r^{2} = \frac{1.334 \times 1 0 ^{- 10}}{2.67 \times 1 0 ^{- 9}} \approx 0.05$

$r = 0.05 \approx 0.223 m$

Answer: The separation is approximately 0.22 m.

Question: A body of mass 2 kg is placed on the Earth’s surface. If Earth’s mass is $6 \times 1 0^{24} k g$ and its radius is $6.4 \times 1 0^{6} m$ calculate the gravitational force acting on it.

Solution:

$F = G \frac{m _{1} m _{2}}{r ^{2}}$

$F = \frac{6.67 \times 1 0 ^{- 11} \times 2 \times 6 \times 1 0 ^{24}}{( 6.4 \times 1 0 ^{6} ) ^{2}}$

$F = \frac{8.004 \times 1 0 ^{14}}{4.096 \times 1 0 ^{13}}$

$F \approx 19.55 N$

Answer: The gravitational force is approximately 19.6 N, which equals the weight of the body.

Frequently Asked Questions

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(Session 2026 - 27)

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Class

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Preferred Programs

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Gravitational Force Escape Velocity

1.0What is Gravitational Force?

Gravitational force is the universal attractive force between any two masses. According to Newton’s Universal Law of Gravitation: the force is directly proportional to the product of the masses and inversely proportional to the square of the distance between their centers. This fundamental concept is essential for JEE-level physics understanding.

2.0Newton’s Law of Gravitation

Sir Isaac Newton formulated the law of gravitation in the 17th century, which states:

Every particle of matter in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

This law provided the foundation to explain Kepler’s laws of planetary motion and laid the groundwork for space science.

3.0Gravitational Force Formula and Its Importance

The gravitational force formula is given by: $F = G \frac{m _{1} m _{2}}{r ^{2}}$

Where:

F = gravitational force between the two masses
$m_{1} an d m_{2}$ = interacting masses
r = distance between the centers of the masses
G = universal gravitational constant = $6.67 \times 1 0^{- 11} N m^{2} / k g^{2}$

Newton’s Universal Law of Gravitation explains that every particle in the universe attracts every other particle with a force that is:

Directly proportional to the product of their masses.
Inversely proportional to the square of the distance between them.

$F = G \frac{m _{1} m _{2}}{r ^{2}}$

This law not only explains the falling of objects on Earth but also the motion of planets around the Sun, the Moon around the Earth, and the structure of galaxies.

4.0Escape Velocity ( $V_{e}$ )

It is the minimum velocity required for an object located at the planet's surface so that it just escapes the planet's gravitational field.

Consider a projectile of mass m, leaving the surface of a planet (or some other astronomical body or system), of radius R and mass M with escape speed.

When the projectile just escapes to infinity, it has neither kinetic energy nor potential energy.

From conservation of mechanical energy

$\frac{1}{2} m v_{e}^{2} + (\frac{- GM m}{R}) = 0 + 0 \Rightarrow v_{e} = \frac{2 GM}{R}$

The escape velocity of a body from a location which is at height 'h' above the surface of planet, we can use :-

$v_{e} = \frac{2 GM}{r} = \frac{2 GM}{R + h}$ ${∵ r = R + h}$

Where, r = Distance from the centre of the planet, h = Height above the surface of the planet

Escape speed depends on:

(i) Mass (M) and radius (R) of the planet

(ii) Position from where the particle is projected

Escape speed does not depend on:

(i) Mass (m) of the body which is projected

(ii) Angle of projection

If a body is thrown from the Earth's surface with escape speed, it goes out of earth's gravitational field and never returns back to the earth's surface.

For Earth,

$v_{e} = \frac{2 GM}{R _{e}}$

Other forms, $g_{s} = \frac{GM}{R _{e}^{2}} \Rightarrow v_{e} = 2 g_{s} R_{e}$

On putting values : $v_{e} = 11.2 km/s, g_{s} = 9.81 m/s, R_{e} = 6400 km$

The escape velocity formula is:

$v_{esc} = \frac{2 GM}{R}$

Where:

$v_{esc}$ = escape velocity
GG = gravitational constant
MM = mass of the planet
RR = radius of the planet

For the Earth, substituting values,

$M = 5.972 \times 1 0^{24} k g$

$R = 6.371 \times 1 0^{6} m$

$G = 6.67 \times 1 0^{- 11} N m^{2} / k g^{2}$

$v_{esc} = \frac{2 \times 6.67 \times 1 0 ^{- 11} \times 5.972 \times 1 0 ^{24}}{6.371 \times 1 0 ^{6}}$

$v_{esc} \approx 11.2 km / s$

Thus, the escape velocity of Earth is approximately 11.2 km/s. This means that if a body is projected with this speed (without air resistance), it will leave Earth’s gravitational field.

5.0Escape velocity from a point other than surface

Keep following points in mind !!

Total energy is zero at any point when particle start moving with escape velocity, TE = 0
If given point is at distance r (> R) from center of Earth, $P E_{o u t} = - G M_{e} m / r$
If given point is at distance r (< R) from center of Earth, $P E_{in} = - GM (3 R^{2} - r^{2}) / (2 R^{3}) \times m$

6.0Escape Energy-Binding Energy

Minimum energy given to a particle in the form of kinetic energy so that it can just escape the Earth's gravitational field.

Magnitude of escape energy = GMm / R

(−ve of PE on the Earth's surface)
Escape energy = Kinetic Energy corresponding to the escape velocity ⇒

GMm / R = 1/2 mvₑ²

Note: In the above discussion it can be any planet for that matter

KE < Escape Energy v₀ < vₑ Body returns to Earth surface
KE = Escape Energy v₀ = vₑ Body comes to rest at infinity
KE > Escape Energy v₀ > vₑ Body has residual velocity at infinity

Binding energy

Total energy of a particle near Earth.

BE < 0 Particle cannot escape the gravitational field of Earth
BE ≥ 0 Particle can escape the gravitational field of Earth

Now that we understand gravitational attraction, a natural question arises:

How fast must an object be launched from the surface of a planet so that it escapes the planet’s gravitational pull without further propulsion?

This minimum speed is called the escape velocity.

If an object has speed less than escape velocity → it will eventually fall back.
If its speed equals or exceeds escape velocity → it will move away indefinitely and not return.

Escape velocity is independent of the mass of the object being launched and depends only on the planet’s mass and radius.

7.0Relation Between Gravitational Force and Escape Velocity

The connection between gravitational force and escape velocity lies in the work required to overcome gravitational potential energy.

Gravitational force holds objects close to the Earth by continuously pulling them inward.
Escape velocity is the exact speed required to provide a body with enough kinetic energy to overcome the gravitational potential energy binding it to the planet.

The stronger the gravitational force (i.e., the larger the mass of the planet and the smaller its radius), the higher the escape velocity. For massive planets like Jupiter, escape velocity is much higher than Earth’s, while for smaller bodies like the Moon, it is much lower.

8.0Factors Affecting Escape Velocity

The escape velocity of a planet depends only on its mass and radius, not on the mass of the escaping object.

From the formula: $v_{esc} = \frac{2 GM}{R}$

A larger planetary mass (M) → stronger gravitational pull → higher escape velocity.
A larger planetary radius (R) → weaker gravitational pull at the surface → lower escape velocity.

For example:

Escape velocity on Moon ≈ 2.38 km/s (much smaller than Earth due to smaller mass and radius).
Escape velocity on Jupiter ≈ 59.5 km/s (much larger because of massive size).

9.0Applications of escape velocity

The concept of escape velocity is central in astrophysics, space exploration, and exam-oriented problem-solving.

Space Missions: To launch satellites or spacecraft, rockets must reach at least escape velocity to overcome Earth’s gravitational pull.
Planetary Atmospheres: Planets with low escape velocities (like Mercury) cannot retain lighter gases in their atmosphere, explaining why they lack a thick atmosphere.
Astrophysics: Understanding black holes relies on escape velocity. At the event horizon, the escape velocity equals the speed of light.

10.0Solved Problems on Gravitational Force

Question 1: Two masses of 5 kg and 10 kg are placed 2 m apart in vacuum. Find the gravitational force between them.

Solution:

Using Newton’s law of gravitation:

$F = G \frac{m _{1} m _{2}}{r ^{2}}$

Substitute values:

$F = \frac{6.67 \times 1 0 ^{- 11} \times 5 \times 10}{( 2 ) ^{2}}$

$F = \frac{3.335 \times 1 0 ^{- 9}}{4}$

$F = 8.34 \times 1 0^{- 10} N$

Answer: The gravitational force is $8.34 \times 1 0^{- 10} N$

Question 2: The gravitational force between two masses is $2.67 \times 1 0^{- 9} N$ If one mass is 1 kg, the other is 2 kg, find the separation between them.

Solution:

$F = G \frac{m _{1} m _{2}}{r ^{2}}$

$r^{2} = G \frac{m _{1} m _{2}}{F}$

$r^{2} = \frac{6.67 \times 1 0 ^{- 11} \times 1 \times 2}{2.67 \times 1 0 ^{- 9}}$

$r^{2} = \frac{1.334 \times 1 0 ^{- 10}}{2.67 \times 1 0 ^{- 9}} \approx 0.05$

$r = 0.05 \approx 0.223 m$

Answer: The separation is approximately 0.22 m.

Question: A body of mass 2 kg is placed on the Earth’s surface. If Earth’s mass is $6 \times 1 0^{24} k g$ and its radius is $6.4 \times 1 0^{6} m$ calculate the gravitational force acting on it.

Solution:

$F = G \frac{m _{1} m _{2}}{r ^{2}}$

$F = \frac{6.67 \times 1 0 ^{- 11} \times 2 \times 6 \times 1 0 ^{24}}{( 6.4 \times 1 0 ^{6} ) ^{2}}$

$F = \frac{8.004 \times 1 0 ^{14}}{4.096 \times 1 0 ^{13}}$

$F \approx 19.55 N$

Answer: The gravitational force is approximately 19.6 N, which equals the weight of the body.

Frequently Asked Questions

What is gravitational force in simple words?

What is the escape velocity definition?

What is the relation between escape velocity and orbital velocity?

Why is Earth’s escape velocity 11.2 km/s?

Does the escape velocity depend on the mass of the object?

Join ALLEN!

Gravitational Force Escape Velocity

1.0What is Gravitational Force?

2.0Newton’s Law of Gravitation

3.0Gravitational Force Formula and Its Importance

4.0Escape Velocity ( $V_{e}$ )

5.0Escape velocity from a point other than surface

6.0Escape Energy-Binding Energy

7.0Relation Between Gravitational Force and Escape Velocity

8.0Factors Affecting Escape Velocity

9.0Applications of escape velocity

10.0Solved Problems on Gravitational Force

Table of Contents

Frequently Asked Questions

What is gravitational force in simple words?

What is the escape velocity definition?

What is the relation between escape velocity and orbital velocity?

Why is Earth’s escape velocity 11.2 km/s?

Does the escape velocity depend on the mass of the object?

Join ALLEN!

Gravitational Force Escape Velocity

1.0What is Gravitational Force?

2.0Newton’s Law of Gravitation

3.0Gravitational Force Formula and Its Importance

4.0Escape Velocity ( $V_{e}$ )

5.0Escape velocity from a point other than surface

6.0Escape Energy-Binding Energy

7.0Relation Between Gravitational Force and Escape Velocity

8.0Factors Affecting Escape Velocity

9.0Applications of escape velocity

10.0Solved Problems on Gravitational Force

Table of Contents

Frequently Asked Questions

What is gravitational force in simple words?

What is the escape velocity definition?

What is the relation between escape velocity and orbital velocity?

Why is Earth’s escape velocity 11.2 km/s?

Does the escape velocity depend on the mass of the object?

Join ALLEN!

Gravitational Force Escape Velocity

1.0What is Gravitational Force?

2.0Newton’s Law of Gravitation

3.0Gravitational Force Formula and Its Importance

4.0Escape Velocity (Ve​)

5.0Escape velocity from a point other than surface

6.0Escape Energy-Binding Energy

7.0Relation Between Gravitational Force and Escape Velocity

8.0Factors Affecting Escape Velocity

9.0Applications of escape velocity

10.0Solved Problems on Gravitational Force

Table of Contents

Frequently Asked Questions

What is gravitational force in simple words?

What is the escape velocity definition?

What is the relation between escape velocity and orbital velocity?

Why is Earth’s escape velocity 11.2 km/s?

Does the escape velocity depend on the mass of the object?

Join ALLEN!

Gravitational Force Escape Velocity

1.0What is Gravitational Force?

2.0Newton’s Law of Gravitation

3.0Gravitational Force Formula and Its Importance

4.0Escape Velocity (Ve​)

5.0Escape velocity from a point other than surface

6.0Escape Energy-Binding Energy

7.0Relation Between Gravitational Force and Escape Velocity

8.0Factors Affecting Escape Velocity

9.0Applications of escape velocity

10.0Solved Problems on Gravitational Force

Table of Contents

4.0Escape Velocity ( $V_{e}$ )

4.0Escape Velocity ( $V_{e}$ )