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Explanation of Gravitational Escape Velocity - Succeed in Understanding Physics. Also refer to physical science, Earth, planets, moon, sun, mass, gravity, rockets, surface, Ron Kurtus, School for Champions. Copyright © Restrictions
Gravitational Escape Velocity
by Ron Kurtus (revised 30 October 2009)
Escape velocity is the initial velocity that an object projected upward must attain in order to overcome the gravitational pull of the Earth, a planet, moon or sun. Since it is actually impossible for an object to escape the various gravitational forces, so what this means is that the velocity of the object is great enough that the object will not fall back to the ground.
A simple equation allows you to calculate the escape velocity of any object. This velocity is independent of the mass of the object and is a function of the mass of the planet or sun, as well the distance to the object. You can easily calculate the escape velocity from the Earth, our Moon and our Sun.
In the case of a sun or star that is a Black Hole, the gravitational force is so strong that escape velocity is greater or equal to the speed of light. You can calculate the radius of a Black Hole of a given mass.
Questions you may have include:
- What is the equation for calculating the escape velocity?
- What are the escape velocities for the Earth, Moon and Sun?
- What would be the radius of the Sun if it was a Black Hole?
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
Although the gravitational force decreases with distance from its source, it only becomes zero at an infinite distance. An object projected upward cannot escape this gravitational force, but its velocity can be sufficient that the object will not fall back down to the ground.
Surprisingly, the escape velocity is not a function of the mass of the object. Instead it is a function of the mass of the body from which it is escaping.
The escape velocity equation is:
ve = √(2GM/R)
where
- ve is the escape velocity in kilometers/second (km/s)
- G is the Universal Gravitational Constant; G = 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 object 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.
The escape velocity equation is really an approximation, based on Newton's Universal Gravitation Equation, where the masses of the objects are assumed to be concentrated at their center point.
Escape velocity related to density
From the equation, you can see that the greater the mass of the planet, the greater the velocity required to escape. But also, the equation shows that the greater the radius of the planet, the lower the escape velocity.
The ratio of M/R is important in comparing the escape velocity of two heavenly bodies. As M/R increases, the required escape velocity also increases.
Since density is mass divided by volume and volume is a function of the radius, the escape velocity is related to the density of the planet or moon.
Thus, it is possible that a planet consisting gases or liquid and having a large mass and very large radius would have a lower escape velocity of a smaller planet that had less mass but was made of densely-packed solids.
Escape from Earth
For objects relatively close to Earth, you can substitute g = GM/R2 in the gravitational escape velocity equation, where g is the acceleration due to gravity and M is the mass of the Earth. The equation for the escape velocity from Earth is then:
ve = √(2gR)
(See Gravity Equation Comes From Universal Gravitation for more information.)
Factors for escape velocity from Earth
Surface escape velocity
It is sometimes called surface escape velocity, since the initial velocity occurs at or very near the surface of the Earth or other planet.
Continuous thrust
In the case of a rocket that continually provides thrust, the velocity required to escape the gravitational pull can be much less than the surface escape velocity. Calculations for that velocity can be quite complex.
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 and its mass is approximately
6*1024 kg.
ve = √(2GM/R)
ve km/s = √[2*(6.67*10−20 km3/kg-s2)*(6*1024 kg)/(6370 km)]
ve km/s = √[80.04*104 km3/s2)/(63.7*10−2 km)]
ve= √(125.65 km2/s2)
ve = 11.2 km/s
Thus, the escape velocity from the Earth is 11.2 km/s or about 26,000 miles per hour. This is the initial velocity of a rocket shooting upward from the Earth's surface.
You can also solve for the escape velocity from Earth with the equation:
ve = √(2gR)
ve = √[2*(9.8*10−3 km/s2)*(6370 km)]
ve = √(124.852 km2/s2)
ve = 11.17 km/s
Note: Since g is typically in m/s2, we had to convert it to km/s2 for these calculations.
Moon
The radius of the Moon is 1737 km and its mass is about 7.3*1022 kg. Thus, the escape velocity from the surface of the Moon is 2.4 km/s.
Sun
The radius of the Sun is about 7*105 km and its mass is approximately 2*1030 kg.
Surface escape velocity
Thus, the escape velocity from the surface of the Sun is 617.5 km/s (about 1,500,000 miles per hour).
Obviously, a rocket couldn't land on the Sun, but atomic particles do escape the surface of the Sun during a solar storm.
Away from surface
If an object was 100,000 km (105 km) from the Sun, its escape velocity would be:
ve = √(2GM/R)
ve = √[2*(6.67*10−20)*(2*1030)/(7*105 + 1*105)]
ve = √[26.68*1010)/(8*105)]
ve = √(3.35*105)
Multiply 3.35 by 10 and divide 105 by 10 to get an even exponent for determining square root.
ve = √(33.5*104)
ve = 5.8*102 = 580 km/s
Application to Black Holes
A Black Hole is a star that has a gravitational field so great that even light rays cannot escape its pull. Such a star appears as a black hole in space.
The escape velocity of a Black Hole is greater than the speed of light. Substituting v = c, the escape velocity equation results in:
c < √(2GM/R)
where
- c is the speed of light (approximately 300,000 km/s or 186,000 mi/s)
- < is the "less than" symbol
Note: This equation is in non-relativistic terms. The equation from the General Theory of Relativity is more complex and considers other factors.
Finding the radius for a given mass
An interesting application of the escape velocity for a Black Hole is finding the radius of such an object, if you know its mass.
Squaring both sides of the equation and rearranging the items results in the equation:
R < 2GM/c2
Substituting values, you get:
R < 2*(6.67*10−20)M/(3*105)2
R < 1.5*10−30M km
Black Hole with mass of our Sun
Thus, if the mass of the black hole equaled the mass of our Sun (2*1030 kg), the radius would be:
R < (1.5*10−30)*(2*1030) km = 3 km
In other words, a star with the mass of our Sun with its matter compressed to a radius of less than 3 km would be a Black Hole, because the escape velocity would be greater than the speed of light.
Summary
Escape velocity is the initial velocity that an object must be accelerated in order to escape the gravitational pull of a planet, moon or sun. The surface escape velocity equation is ve = √(2GM/R).
You can easily calculate the escape velocity from the Earth, our Moon and our Sun. The escape velocity of a Black Hole is greater than the speed of light, but using the escape velocity equation, you can calculate the radius of a Black Hole of a given mass.
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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
You can purchase these books in your local bookstore or through Amazon.com.
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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