# Derivation of Gravitational Escape Velocity Equation

by Ron Kurtus (revised 15 November 2015)

In order to overcome the gravitational pull of a planet, moon or sun, an object must be accelerated to the gravitational escape velocity for that celestial body at a given altitude above the body's surface.

You can derive the equation for the gravitational escape velocity by considering the initial gravitational potential energy at the given altitude and the initial kinetic energy of the object. By comparing this total initial energy with the sum of the potential and kinetic energies at an infinite separation, you can determine the escape velocity equation.

Questions you may have include:

• What are the initial potential and kinetic energies of the object?
• What are the potential and kinetic energies at an infinite distance?
• What determines the final equation?

This lesson will answer those questions. Useful tool: Units Conversion

## Initial considerations

Typically, the projected object has a much smaller mass than the body from which it is escaping, such that the center of mass between them is located close to the geometric center of the larger body. Thus, instead of both objects moving away from their center of mass, the smaller object is considered to be moving away from the larger body.

The equation for the escape velocity can be derived by applying the Law of Conservation of Energy. This Law states that the total of the object's potential and kinetic energy is a constant. The escape velocity equation is also a function of the separation between the centers of the object and the celestial body from which it is escaping.

The derivation starts with the initial gravitational potential energy at the given altitude and the initial kinetic energy of the object. This total initial energy is then compared with the sum of the potential and kinetic energies at an infinite separation, in order to determine the escape velocity equation.

Once an object—such as a rocket—has reached a sufficient velocity above the surface of a moon, planet or sun and is no longer being powered, it has an initial gravitational potential energy and an initial kinetic energy.

## Initial energy of the object

It is then assumed to be moving freely with only the gravitational force from the larger object being applied.

### Initial potential energy

The gravitational potential energy between two objects at some separation is defined as the work required to move them from a zero reference point to that given separation. The zero reference point from gravitation is at an infinite separation.

However, since there is a great difference in mass between the objects, the potential energy can also be considered as the work required to move the smaller object to the zero reference point. The separation will still be infinite.

Note: You can also consider the motion of the smaller object with respect to the larger object instead of with respect to the center of mass. In either case, since M >> m, the results are the same.

The initial gravitational potential energy between two objects at some separation is:

PEi = −GMm/Ri

where

• PEi is the initial gravitational potential energy in kg-km2/s2
• G is the Universal Gravitational Constant = 6.674*10−20 km3/kg-s2
• M is the mass of the attracting object in kilograms (kg)
• m is the mass of the escaping object in kg
• M is much greater than m (M >> m)
• Ri is the initial separation between the centers of the objects in kilometers (km)

Note: Since the work required to move an object from the zero reference point of PE to some point in space is negative, the PE at that point is also considered negative.

Also note: Since escape velocity is usually stated in km/s, the units of PE and R have been changed kilometers. Likewise, the value to G has been changed to reflect use of kilometers.

### Initial kinetic energy

The initial kinetic energy of an object projected at some velocity away from the Earth or other astronomical body is:

KEi = mve2/2

where

• KEi is the initial kinetic energy in kg-km2/s2
• m is the mass of the object in kg
• ve is the initial velocity—and thus the escape velocity—in km/s

### Illustration of factors

The following picture shows the relationship of factors involved. Factors involved in gravitational escape velocity

Note: Our direction convention states that a velocity vector that is "up" or away from the larger mass is in the negative direction.

### Total initial energy

The total initial energy is the sum of the potential and kinetic energies at the release point:

TEi = KEi + PEi

TEi = mve2/2 GMm/Ri

## Final energy

Gravitational fields hypothetically extend to infinity. Thus, if the initial velocity is great enough, the object will travel to an infinite separation and thus "escape" the gravitational force.

### Potential energy at infinity

The object's potential energy at an infinite separation or displacement is:

PE = −GMm/R

where

• PE is the gravitational potential energy at infinity
• R is the infinite separation between the centers of the objects

Since R = ∞ (infinity), then PE = 0.

### Kinetic energy at infinity

The object's kinetic energy at an infinite displacement is:

KE = mv2/2

where

• KE is the final kinetic energy
• v is the final velocity at an infinite distance

At infinity, the velocity of the object is zero: v = 0. Thus KE = 0.

### Total final energy

Since the kinetic energy is moving upward and the potential energy is acting downward, the total energy at the initial position is:

TE = KE + PE

TE = 0 + 0

## Escape velocity equation

The Law of Conservation of Energy states that the total energy of a closed system remains constant. In this case, the closed system consists of the two objects with the gravitational force between them and no outside energy or force affecting either object.

Thus the total final energy—potential energy plus kinetic energy—must equal the total initial energy:

TEi = TE

KEi + PEi = 0

Substitute values:

mve2/2 − GMm/Ri = 0

Add GMm/Ri to both sides of equation:

mve2/2 = GMm/Ri

Solve for ve2:

ve2 = 2GM/Ri

Take the square root of each expression to get:

ve = ± √(2GM/Ri)

Considering our gravitational convention for direction, ve is upward or away from the other object and is thus negative:

ve = − √(2GM/Ri)

Note: Although convention-wise, the negative version of the equation is correct, most textbooks will give the positive version of the equation. You should be aware of this fact.

### Altitude factor

The equation can also be written considering the initial altitude of the escaping object:

ve = − √[2GM/(r + h)]

where

• r is the radius of the celestial body in km
• h is the altitude or separation from the surface of the body in km

The altitude factor is necessary since the escaping object must accelerate over some displacement to reach the escape velocity.

## Summary

The derivation of the gravitational escape velocity of an object from a much larger mass is achieved by comparing the potential and kinetic energy values at some given point with the values at infinity, applying the Law of Conservation of Energy.

The equation for the gravitational escape velocity is:

ve = − √(2GM/Ri)

Taking altitude into account, the equation can be written as:

ve = − √[2GM/(r + h)]

Know that you can overcome obstacles

## Resources and references

The following resources can be used for further study on the subject.

### Web sites

What is escape velocity? - From PhysLink

Gravitational Potential Energy - HyperPhysics

Escape Velocity - From Wikipedia

Gravitation Resources

### Books

Top-rated books on Escape Velocity and Space Travel Do you have any questions, comments, or opinions on this subject? If so, send an email with your feedback. I will try to get back to you as soon as possible.

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www.school-for-champions.com/science/
gravitation_escape_velocity_derivation.htm

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