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Gravitational Escape Velocity
by Ron Kurtus (revised 15 January 2010)
Gravitational escape velocity is the speed an object projected upward must attain such that it will not be overcome by gravitational force and fall back to the ground. The gravitational escape velocity equation allows you to calculate the escape velocity from any planet, moon or sun, provided you know its mass and the distance to the object.
There are some restrictions or conditions applied to the equation to keep it simple. Application of the equation is the calculation of the escape velocity from the Earth, Moon and Sun.
Questions you may have include:
- What is the equation for calculating the escape velocity?
- What are some restrictions on the equation?
- What are the escape velocities for the Earth, Moon and Sun?
This lesson will answer those questions. There is a mini-quiz near the end of the lesson.
Useful tools: Metric-English Conversion | Scientific Calculator.
Escape velocity equation
When an object—such as a rocket—is projected directly upward from a planet, moon or sun until it reaches some velocity, it will start to slow due to the gravitational force acting on it, pulling it back toward the surface. However, if the velocity is great enough, the object will never slow to zero velocity or return to the ground. That velocity is called the gravitational escape velocity.
Note: In the case of a rocket that continually provides thrust, the velocity required to escape the gravitational pull can be much less than the escape velocity. Calculations for that velocity can be quite complex.
Equation
The gravitational escape velocity equation is:
ve = √(2GM/R)
where
- ve is the escape velocity in kilometers/second (km/s)
- G is the Universal Gravitational Constant = 6.67*10−20 km3/kg-s2
- M is the mass of the planet or sun in kilograms (kg)
- R is the distance from the center of mass of the planet or sun to the center of the object in kilometers (km)
- √(2GM/R) is the square root of the quantity (2GM/R)
Important note: In previous Universal Gravitation Equations, G was stated in N-m2/kg2 and R in meters (m). However, it is more convenient to define escape velocity in kilometers/second (km/s). Thus, G is defined in km3/kg-s2 and R in km.
Rocket reaches escape velocity
Distance can be substantial
You often see the escape velocity stated from the surface of the celestial mass, with R simply as the radius of the planet or sun. However, it is impossible for an object to start at the surface with a given upward velocity. The object must accelerate to reach the escape velocity, and that means it must travel some distance above the surface of the planet before it attains the escape velocity. Often, that distance is substantial.
For example, a rocket blasting off from the Earth may take several thousand kilometers until it reaches the 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.
Restrictions on equation
There are some restrictions or conditions that must be fulfilled to keep the escape velocity calculations simple.
Not continually powered
One condition on establishing an equation to calculate the escape velocity is that once the object achieves a given velocity, it no longer is powered and moves freely. For example, at that point the rocket will turn off its engines.
Moving straight up
It is assumed that the object is moving straight up. It is possible to achieve an escape velocity if the object is projected at an angle, however orbital mechanics then come into play.
Rotation not included
To keep the equation simple, the effect of rotation and orbital motion are not figured into the equation. Those factors can decrease or increase the escape velocity.
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.
No air resistance
Finally, air resistance on a rocket fired from Earth is not considered.
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 radius of the Earth is about 6370 km or 63.7*102 km, and its mass is approximately 5.97*1024 kg.
Consider a rocket blasting off from the Earth’s surface until it reaches thirty kilometers in altitude, when the engines turn off. Thus, R = (6370 + 30) km = 6400 km.
What velocity must the rocket attain to continue rising without falling back to Earth?
Solution
Substitute values into the escape velocity equation:
ve = √(2GM/R)
ve = √[2*(6.67*10−20 km3/kg-s2)*(5.97*1024 kg)/(6400 km)]
ve = √[(796400 km3/s2)/6400 km)]
ve= √(124.4 km2/s2)
ve = 11.16 km/s
Thus, the calculated escape velocity from 30 km above the Earth's surface is about 11.2 km/s or about 26,000 miles per hour.
Moon
The approximate radius of the Moon is 1737 km and its mass is about 7.3*1022 kg.
Suppose a rocket landed on the Moon and then blasted off to return to Earth. The engines only operated for about one kilometer to build up the speed. What velocity must the rocket attain to escape the Moon?
Solution
Substitute values into the escape velocity equation:
ve = √(2GM/R)
ve = √[2*(6.67*10−20 km3/kg-s2)*(7.3*1022 kg)/(1738 km)]
ve = √[5.6 km2/s2)]
ve = 2.37 km/s
Thus, the escape velocity from near the surface of the Moon is about 2.4 km/s.
Sun
The radius of the Sun is about 0.696*106 km and its mass is approximately
1.99*1030 kg.
A solar storm shoots atomic particles into space. What is their escape velocity near the surface of the Sun?
Also, if the storm pushed particles 100,000 km (105 km) from the Sun, what would their escape velocity be?
Solution near the surface
Substitute values near the Sun's surface into the escape velocity equation:
ve = √(2GM/R)
ve = √[2*(6.67*10−20)*(1.99*1030)/(0.696*106)] km/s
ve = √(38.1*104) km/s
ve = 6.176*102 = 617.6 km/s
Thus, the escape velocity from near the surface of the Sun is 617.6 km/s (about 1,500,000 miles per hour).
Away from surface
When the particles escape at 100,000 km from the surface of the Sun, the value of R in the equation is (6.96*105 + 1*105) km = 7.96*105 km.
Substitute values into the escape velocity equation:
ve = √(2GM/R)
ve = √[2*(6.67*10−20)*(1.99*1030)/(7.96*105)]
ve = √(3.335*105)
ve = √(33.35*104)
ve = 5.775*102 = 577.5 km/s
You can see how the escape velocity changes with distance from the Sun.
Summary
Gravitational escape velocity is the speed an object projected upward must attain, such that it will not be overcome by
gravitational force. The equation
ve = √(2GM/R) allows you to calculate the escape velocity from any planet, moon or sun.
There are restrictions or conditions on the equation. Also, you can readily calculate the escape velocity from the Earth, Moon and Sun.
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Resources
The following resources can be used for further study on the subject.
Web sites
What is escape velocity? - From PhysLink
Escape Velocity - From Wikipedia
Books
Top-rated
books on Escape Velocity and Space Travel
Mini-quiz to check your understanding
If you got all three correct, you are on your way to becoming a Champion in Physics. If you had problems, you had better look over the material again.
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