Gravitational Potential Energy: Fields, Potential and Escape Velocity

Published on: March 25, 2025in

Understanding gravitational potential energy and escape velocity, including gravitational fields and example problems.

Estimated reading time: 11 minutes

Gravity is one of the four fundamental forces of nature, governing the motion of celestial bodies and objects on Earth. Moreover, Newton’s Law of Universal Gravitation provides a mathematical framework for understanding gravitational interactions. In this article, we will explore Gravitational Potential Energy in depth, covering key concepts such as gravitational fields, escape velocity, and the energy of orbiting satellites.

Contents

Newton’s Law of Universal Gravitation

Newton’s Law of Universal Gravitation states that every point mass in the universe attracts every other point mass with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them:

$F = G \frac{m_1 m_2}{r^2}$

where:

$F$ is the gravitational force,
$G$ is the gravitational constant ( $6.674 \times 10^{-11} \; \text{Nm}^2\text{/kg}^2$ ),
$m_1, m_2$ are the masses of the objects,
$r$ is the distance between the centers of the masses.

Consequently, this law explains planetary motion, tides, and the motion of satellites around Earth.

What is a Gravitational Field?

A gravitational field is a region in space where a mass experiences a force due to gravity. Hence, it describes how the gravitational force is distributed around a massive object.

Earth Moon Gravity — **Fig 1. The Law of Gravity**

Definition of Gravitational Field

To clarify, the gravitational field strength at any point is defined as the force per unit mass experienced by a small test mass placed at that point. Hence, mathematically given by:

$g = \frac{F}{m}$

where:

$g$ : the gravitational field strength (measured in N/kg\text{N/kg}N/kg or m/s2\text{m/s}^2m/s2),
$F$ : the gravitational force,
$m$ : the mass experiencing the force.

Gravitational Field Due to a Point Mass

Hence, for a mass $M$ , the gravitational field at a distance $r$ from its center is given by:

$g = \frac{G M}{r^2}$

where:

$G$ is the gravitational constant ( $6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2$ ),
$r$ is the distance from the mass $M$ .

Therefore, this equation shows that the gravitational field decreases as the distance from the mass increases.

Gravitational Field Near Earth’s Surface

Near the Earth’s surface, the gravitational field strength is approximately:

$g \approx 9.8 \text{ m/s}^2$

Consequently, this means that any object in Earth’s gravitational field experiences an acceleration of 9.8 m/s² toward the Earth’s center.

Direction of the Gravitational Field

The gravitational field is always attractive, pointing toward the center of the mass creating the field.

Gravitational Potential Energy

Gravitational potential energy is the energy stored in an object due to its position in a gravitational field. That is, it represents the work done to bring an object from infinity to a given point against gravitational attraction.

Formula for Gravitational Potential Energy

For an object of mass $m$ at a distance $r$ from a mass $M$ , the gravitational potential energy is given by:

$U = -\frac{G M m}{r}$

where:

$U$ = Gravitational potential energy (Joules, $J$ )
$G$ = Gravitational constant ( $6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2$ )
$M$ = Mass of the larger body (kg)
$m$ = Mass of the object (kg)
$r$ = Distance from the center of the larger body (m)

Why is Gravitational Potential Energy Negative?

Firstly, at infinity, gravitational potential energy $U=0$
As an object moves closer to the mass, energy is lost (i.e., energy is released).
Since gravity is an attractive force, work must be done to move an object away from a mass, making GPE negative in bound systems.

Near-Earth Approximation

Near Earth’s surface, for small heights $h$ , we use the simpler formula:

$U = mgh$

where:

$g \approx 9.8 \text{ m/s}^2$ (acceleration due to gravity on Earth),
$h$ = height above the reference level.

This approximation is valid when $h \ll R_E$ (Earth’s radius).

Difference Between Gravitational Potential and Gravitational Potential Energy

Quantity	Gravitational Potential (V)	Gravitational Potential Energy (U)
Definition	Energy per unit mass	Energy stored in a mass due to gravity
Formula	V= -GM/r	U = -GMm/r
Units	Joules per kilogram (J/kg)	Joules (J)
Depends on	Mass creating the field	Mass of the object + mass of the field source

Example Problem

Problem:

If the mass of a satellite orbiting Earth at an altitude of 500 km is 1000 kg, find the gravitational potential energy.

Mass of Earth, $M = 5.972 \times 10^{24}$ kg
Radius of Earth, $R_E = 6.37 \times 10^6$ m
Distance from Earth’s center, $r = R_E + 500 \times 10^3 = 6.87 \times 10^6$ m

Solution:

Since:

$U = -\frac{(6.674 \times 10^{-11}) (5.972 \times 10^{24}) (1000)}{6.87 \times 10^6}$

$U \approx -5.81 \times 10^{10} \text{ J}$

Hence, this means the satellite has a gravitational potential energy of $-5.81 \times 10^{10}$ J, indicating it is bound to Earth’s gravity.

Gravitational Potential Energy of a Sphere

When dealing with a solid sphere of mass $M$ and radius $R$ , its gravitational potential energy ( $U$ ) is calculated differently depending on whether we consider:

A point mass outside the sphere
The self-gravitational potential energy of the sphere

1. Gravitational Potential Energy of a Mass Outside a Sphere

If a small mass $m$ is located at a distance $r$ from the center of a uniform sphere of mass $M$ , the gravitational potential energy is given by:

$U = -\frac{G M m}{r}$

First Case: Outside the Sphere ( $r \geq R$ )

The sphere behaves as if all its mass is concentrated at a point at its center.
The formula $U = -\frac{G M m}{r}$ applies as if it were a point mass.

Second Case: On the Surface ( $r = R$ )

$U = -\frac{G M m}{R}$

Third Case: Inside the Sphere ( $r < R$ )

When inside a uniform sphere, only the mass enclosed within radius $r$ contributes to the gravitational force. Therefore, the effective gravitational potential energy at distance $r$ inside the sphere is:

$U = -\frac{G M m}{2 R} \left(3 - \frac{r^2}{R^2} \right)$

2. Self-Gravitational Potential Energy of a Sphere

The self-gravitational potential energy is the energy required to assemble a sphere of mass $M$ and radius $R$ by bringing its mass together from infinity.

$U_{\text{self}} = -\frac{3 G M^2}{5 R}$

Consequently, this formula is derived by integrating the work required to bring infinitesimal mass shells together incrementally.

Key Takeaways:

A hollow sphere has constant gravitational potential inside.
A solid sphere exerts a gravitational pull inside it, and the potential energy follows a different equation as given above.

Example Calculation

Problem:

If mass $M = 5.972 \times 10^{24}$ kg and radius $R = 6.37 \times 10^6$ m, find the self-gravitational potential energy of Earth.

Solution:

$U_{\text{self}} = -\frac{3 G M^2}{5 R}$

Then, substituting values:

$U_{\text{self}} = -\frac{3 (6.674 \times 10^{-11}) (5.972 \times 10^{24})^2}{5 (6.37 \times 10^6)}$

$U_{\text{self}} \approx -2.24 \times 10^{32} \text{ J}$

Hence, this means that assembling Earth from dispersed particles would require an energy input of $2.24 \times 10^{32}$ Joules.

Escape Velocity

Escape velocity is the minimum velocity an object must have to escape a celestial body’s gravitational field without any further propulsion. That is, at this velocity, the object’s kinetic energy is equal to the gravitational potential energy, allowing it to reach infinity with zero remaining energy.

Derivation of Escape Velocity

For an object of mass $m$ to escape from a planet of mass $M$ and radius $R$ , the total energy at the surface should be zero at infinity.

Total Energy at the Surface:

$E = \text{Kinetic Energy} + \text{Gravitational Potential Energy}$

$E = \frac{1}{2} m v_e^2 - \frac{G M m}{R}$

That is, for the object to escape, its total energy must be zero at infinity:

$\frac{1}{2} m v_e^2 = \frac{G M m}{R}$

Then, canceling $m$ and solving for $v_e$ :

$v_e = \sqrt{\frac{2 G M}{R}}$

where:

$v_e$ = Escape velocity (m/s)
$G$ = Gravitational constant ( $6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2$ )
$M$ = Mass of the celestial body (kg)
$R$ = Radius of the celestial body (m)

Escape Velocity of Earth

For Earth:

$M = 5.972 \times 10^{24}$ kg
$R = 6.37 \times 10^6$ m

$v_e = \sqrt{\frac{2 (6.674 \times 10^{-11}) (5.972 \times 10^{24})}{6.37 \times 10^6}}$

$v_e \approx 11.2 \text{ km/s} \text{ or } 11,200 \text{ m/s}$

Thus, an object must travel at 11.2 km/s (about 40,320 km/h) to escape Earth’s gravity.

Key Observations

Escape velocity increases with mass ( $M$ ) and decreases with radius ( $R$ ).
A higher mass and smaller radius result in a higher escape velocity.
The Moon has a low escape velocity (2.4 km/s), which is why it cannot retain an atmosphere like Earth.
The Sun’s escape velocity is 617.5 km/s, meaning an object must move incredibly fast to leave the Sun’s gravity.

Relation to Orbital Velocity

Escape velocity is related to orbital velocity by:

$v_e = \sqrt{2} v_{\text{orbit}}$

where $v_{\text{orbit}}$ is the velocity required to stay in a circular orbit around the planet. Therefore, this means escape velocity is about 1.41 times the orbital velocity.

Example Problem

Problem:

Find the escape velocity for Mars.

$M = 6.42 \times 10^{23}$ kg
$R = 3.39 \times 10^6$ m

Solution:

$v_e = \sqrt{\frac{2 (6.674 \times 10^{-11}) (6.42 \times 10^{23})}{3.39 \times 10^6}}$

$v_e \approx 5.0 \text{ km/s}$

So, the escape velocity for Mars is 5.0 km/s.

Interesting Facts

Black holes have escape velocities greater than the speed of light ( $c$ ). This is why not even light can escape them.
Rockets do not need to reach escape velocity instantly. In short, they accelerate gradually using fuel over time.
The Voyager 1 spacecraft is moving at ~17 km/s, which is much faster than Earth’s escape velocity, allowing it to leave the solar system. Consequently, this is known as the third cosmic velocity or interstellar speed.

Black Hole — **Fig 3.Black holes have escape velocities greater than the speed of light**

Frequently Asked Questions

1. Why is gravitational potential always negative?

Gravitational potential is negative because work must be done against gravity to move a mass from a given point to infinity (where potential is defined as zero). Since gravity is always attractive, potential energy decreases as objects move closer, making it negative.

2. What is escape velocity, and why does it not depend on the escaping object’s mass?

Escape velocity is the minimum speed needed for an object to break free from a planet’s gravitational field without further propulsion. That is given by:

$v_e = \sqrt{\frac{2 G M}{R}}$

Since mass $m$ cancels out in the derivation, escape velocity depends only on the planet’s mass and radius, not on the mass of the escaping object.

3. Can an object escape Earth’s gravity at a velocity lower than escape velocity?

Yes, but only with continuous propulsion (e.g., a rocket). Additionally, escape velocity applies to objects launched without excess thrust (like a projectile). Rockets gradually build up speed over time, rather than needing 11.2 km/s instantly.

Sub escape velocity — **Fig 4. Rockets are gradually accelerated to escape the planet’s gravity**

4. Does escape velocity change with altitude?

Certainly, escape velocity is inversely proportional to the square root of distance from the center of the planet:

$v_e = \sqrt{\frac{2 G M}{r}}$

As altitude ( $h$ ) increases, $r = R + h$ also increases, so escape velocity decreases.

5. How does gravitational potential energy change in a multi-body system (e.g., the Earth-Moon system)?

Gravitational potential energy is a pairwise additive quantity. That is, for multiple bodies, the total energy is the sum of individual interactions:

$U_{\text{total}} = U_{\text{Earth-Moon}} + U_{\text{Earth-Sun}} + U_{\text{Moon-Sun}} + \dots$

Hence, each term follows $U = -\frac{G M m}{r}$ , accounting for the gravitational effect between each pair of bodies.

6. Why do astronauts in orbit feel weightless if gravity is still acting on them?

Astronauts feel weightless because they are in free fall around Earth. That is, the spaceship and astronauts accelerate toward Earth at the same rate, meaning there is no normal force acting on them, creating a sensation of weightlessness.

7. If a spacecraft reaches escape velocity, does it mean it will leave the solar system?

Not necessarily. Escape velocity is relative to a specific celestial body (e.g., Earth). Consequently, to leave the solar system, a spacecraft must achieve the Sun’s escape velocity at Earth’s distance, which is about 42.1 km/s (16.7 km/s counting speed of Earth’s revolution). This is much higher than Earth’s escape velocity (11.2 km/s).

Conclusion

In conclusion, understanding gravitational potential energy and its related concepts is essential for astrophysics, space exploration, and planetary science. Consequently, these principles help in predicting planetary motion, designing satellites, and understanding cosmic phenomena. Escape velocity is a crucial concept utilized in many practical situations such as the launching of satellites and rockets. Consequently, understanding the first, second and third cosmic velocities plays a crucial role in space exploration, now and in the future.

References

Gron, Ø. (2009). Newton’s law of universal gravitation. In Lecture notes in physics. Springer. https://doi.org/10.1007/978-0-387-88134-8_1
Borghi, R. (2014). On Newton’s shell theorem. European Journal of Physics, 35(2), 028003. https://doi.org/10.1088/0143-0807/35/2/028003
Longuski, J., Hoots, F., & IV, G. (2021). The gravitational potential. In Introduction to orbital perturbations (pp. 143–171). Springer. https://doi.org/10.1007/978-3-030-89758-1_7

Additionally, to stay updated with the latest developments in STEM research, visit ENTECH Online. This is our digital magazine for science, technology, engineering, and mathematics. Further, at ENTECH Online, you’ll find a wealth of information.

Author
Latest Posts

Uddhav Thakore

Gravitational Potential Energy: Fields, Potential and Escape Velocity

Newton’s Law of Universal Gravitation

What is a Gravitational Field?

Definition of Gravitational Field

Gravitational Field Due to a Point Mass

Gravitational Field Near Earth’s Surface

Direction of the Gravitational Field

Gravitational Potential Energy

Formula for Gravitational Potential Energy

Why is Gravitational Potential Energy Negative?

Near-Earth Approximation

Difference Between Gravitational Potential and Gravitational Potential Energy

Example Problem

Problem:

Solution:

Gravitational Potential Energy of a Sphere

1. Gravitational Potential Energy of a Mass Outside a Sphere

First Case: Outside the Sphere ()

Second Case: On the Surface ()

Third Case: Inside the Sphere ()

2. Self-Gravitational Potential Energy of a Sphere

Key Takeaways:

Example Calculation

Problem:

Solution:

Escape Velocity

Derivation of Escape Velocity

Total Energy at the Surface:

Escape Velocity of Earth

Key Observations

Relation to Orbital Velocity

Example Problem

Problem:

Solution:

Interesting Facts

Frequently Asked Questions

1. Why is gravitational potential always negative?

2. What is escape velocity, and why does it not depend on the escaping object’s mass?

3. Can an object escape Earth’s gravity at a velocity lower than escape velocity?

4. Does escape velocity change with altitude?

5. How does gravitational potential energy change in a multi-body system (e.g., the Earth-Moon system)?

6. Why do astronauts in orbit feel weightless if gravity is still acting on them?

7. If a spacecraft reaches escape velocity, does it mean it will leave the solar system?

Conclusion

References

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First Case: Outside the Sphere ( $r \geq R$ )

Second Case: On the Surface ( $r = R$ )

Third Case: Inside the Sphere ( $r < R$ )