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Overview of Gravitational Escape Velocity

by Ron Kurtus (revised 15 February 2016)

Gravitational escape velocity is the release velocity of a freely moving object that has been accelerated away from a sun, planet or moon, such that this initial velocity is sufficient to prevent it from being overcome by gravitational force and falling back to the surface.

Although the escape velocity can be considered with respect to the center of mass (CM) between the objects, it is usually measured with respect to the larger of the two objects. Also, the mass of the escaping object is considered as much less than the mass of the attracting object.

A simple equation provides the escape velocity as a function of the initial separation of the objects and the mass of the larger body. There are several conditions for the equation to be valid.

The equation allows you to calculate the escape velocity from any celestial body, provided you know the body's mass and radius, as well as the altitude of the object. Applications of the equation include the calculations of the escape velocity from the Earth, Moon and Sun.

Questions you may have include:

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



Escape velocity equation

When an object is projected at a sufficient velocity in a direction away from a much larger object at some altitude, it can escape the gravitational attraction between the two and fly off into space. This initial velocity is called the escape velocity.

The direction of the velocity can be in the radial or vertical direction, in the tangential direction or directions in between, as long as it results in a direction away from the larger object. Considering our convention that velocity vectors moving in the opposite direction of gravitation are negative, the standard gravitational escape velocity equation is:

ve = − √(2GM/R)

where

Escape speed

Since the equation applies for a variety of directions resulting in motion away from the larger body, some sources refer to it as escape speed and use the magnitude of the velocity equation:

se = √(2GM/R)

where se is the escape speed.

However, this implies that direction is not a factor. It overlooks the fact that the projected object may crash into the larger body when sent in some directions.

Note: I feel the velocity equation is more correct, and the negative sign indicates the object is moving in a direction opposite the gravitational force.

Considering altitude

When considering the separation of the objects, it is often more convenient to use the radius of the larger object plus the altitude or height instead of separation. For example, let:

R = r + h

where

Thus, the equation becomes:

ve = − √[2GM/(r + h)]

This is shown in the illustration below:

Rocket reaches escape velocity

Rocket reaches escape velocity

Mass of object not a factor

Surprisingly, the mass of the object projected upward is not a factor in the escape velocity. But the escape velocity does depend on the mass of the body from which it is escaping.

Conditions and assumptions

There are several other conditions or assumptions for the gravitational escape velocity equation:

Freely moving

Although the object or rocket may be accelerated up to the escape velocity, any means of propulsion is turned off and the object is moving freely.

High altitude required to reach velocity

It is usually necessary for an object to reach a high altitude before it achieves the gravitational escape velocity of the celestial body.

Acceleration from surface

If an object is accelerated from the surface of the planet or sun, the object will often need to travel a great distance to some high altitude in order to reach a sufficient velocity to escape.

Note: Some textbooks refer to surface escape velocity. Unfortunately, that concept is incorrect, because objects of any size do not instantaneously accelerate to such a high velocity. Also, many calculations use the escape velocity from gravity, which also are measured from the surface.

(See Escape Velocity from Gravity for more information.)

Higher altitudes reduces required velocity

The further the object is from the center of the celestial body, the lower the required escape velocity.

In the situation of a rocket blasting off from the Earth, Moon or some planet, the point where the engines shut off and the rocket starts coasting is where the escape velocity is determined. The higher the rocket goes before the engines shut off, the lower the required initial velocity to escape.

Overcoming air resistance

Also, when escaping the Earth's gravitation, air resistance must be overcome. Often a rocket will first go into orbit at some high altitude—typically, around 190 km or 120 mi, where air resistance is no longer a factor. Then it will blast off into space to reach the escape velocity.

(See Gravitational Escape Velocity with Saturn V Rocket for more information.)

Ions and subatomic particles escaping from Sun's gravitation are usually sent upward in turbulent solar storms until they reach the escape velocity for the altitudes reached.

Rotation not included

The effect of planet rotation and orbital motion are not figured into the escape velocity equation we are using. Those factors can decrease or increase the escape velocity but also complicates calculations.

For example, the escape velocity of an object in the direction of the rotation of the Earth is less than when not calculating the rotation. It also varies with the latitude on Earth from which a rocket is fired. That is why it is preferred to launch rockets near the Earth's equator.

Effect of other objects not included

The effect of gravitation forces from other objects is not considered. Gravitation from the Sun on an object leaving the Earth influences the escape velocity but is not included in our simple equation.

(See Effect of Sun on Escape Velocity from Earth for more information.)

Common escape velocities

You can use the escape velocity equation to determine the necessary velocity for an object projected upward to escape the Earth, Moon or Sun.

Earth

The Saturn V rocket that was used to go to the Moon employed its first two stages to reach a speed that was close to the escape velocity at an altitude of about 191 km (119 mi).

What is the required escape velocity at that altitude, such that the rocket will continue rising without more propulsion and not fall back to Earth?

Solution

The radius of the Earth is about r = 6371 km or 6.371*103 km, and its mass is approximately M = 5.974*1024 kg.

Add the radius of the Earth to the altitude of the rocket:

R = r + h

R = (6371 + 191) = 6562 km

Substitute the values into the escape velocity equation:

ve = − √(2GM/R)

ve = − √[2*(6.674*10−20 km3/kg-s2)*(5.974*1024 kg)/
(6.562*103 km)]

ve= − √(121.519 km2/s2)

ve = −11.02 km/s

In other words, the calculated escape velocity from 191 km above the Earth's surface is about 39,685 km/hr or 24,684 mi/hr.

This velocity is less than the surface escape velocity from gravity of 11.2 km/s, which is also an impossible scenario.

Moon

Suppose a rocket landed on the Moon and then blasted off to return to Earth. The rocket's engines only operated for about 10 km to build up its speed. What velocity must the rocket attain to escape the Moon?

Solution

The approximate radius of the Moon is 1737 km and its mass is about
7.347*1022 kg. The separation between the center of the Moon and the center of the rocket is:

R = (1737 + 10) = 1.747*103 km

Substitute values into the escape velocity equation:

ve = − √(2GM/R)

ve = − √[2*(6.674*10−20)*(7.347*1022)/(1.747*103)] km/s

ve = − √(5.613) km/s

ve = − 2.37 km/s

Thus, the escape velocity from near the surface of the Moon is about 8529 km/hr or 5305 mi/hr.

Sun

Solar flares usually reach about 5000 km from the Sun's surface, as seen from telescopes on Earth. These flares project subatomic particles out into space as a coronal mass ejection or massive burst of solar wind.

If the storm pushed particles 100,000 km (105 km) until they reached their escape velocity from the Sun, what would that velocity be?

Solution

The radius of the Sun is about 6.955*105 km and its mass is approximately
1.989*1030 kg. The separation between the center of the Sun and a subatomic particle is:

R = (6.955*105 + 1*105) = 7.955*105 km

Substitute values into the escape velocity equation:

ve = − √(2GM/R)

ve = − √[2*(6.674*10−20)*(1.989*1030)/(7.955*105)] km/s

ve = − √(33.374*104) km/s

ve = − 5.777*102 = 577.7 km/s

Thus, the escape velocity at 100,000 km from the surface of the Sun is
2,079,720 km/hr or 1,293,586 mi/hr.

Summary

Gravitational escape velocity is the velocity of an object that is sufficient to escape the gravitation of a much larger body, so that it flies off into space.

The escape velocity equation is:

ve = − √(2GM/R)

or

ve = − √[2GM/(r + h)]

The escape velocity equation assumes that the object is not being propelled, it is released at a high altitude and rotation of the large body is not considered. Also, gravitation from other objects is not considered.

The equation allows you to calculate the escape velocity from any planet, moon or sun. Applications of the equation include the calculations of the escape velocity from the Earth, Moon and Sun.


Celebrate your successes


Resources and references

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

Web sites

What is escape velocity? - From PhysLink

Escape Velocity - From Wikipedia

Gravitation Resources

Books

Top-rated books on Escape Velocity and Space Travel


Questions and comments

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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