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Effect of Sun on Escape Velocity from Earth
by Ron Kurtus (7 January 2010)
The escape velocity equation allows you to calculate the velocity an object—such as a rocket—must attain in order to completely overcome the gravitational pull of the Earth. However, it is a simplistic view that the rocket will continually move outward into space, once it achieves the escape velocity from Earth. This is because there are other objects—such as the Sun—that can effect the rocket with their gravitational attraction.
When a rocket blasts off from the Earth in a direction away from the Sun, it must not only escape the Earth's gravitation but also the gravitational pull from the Sun. Surprisingly, the escape velocity from the Sun at the Earth's surface is greater than the escape velocity from the Earth. The two factors must be combined to give the true escape velocity.
Questions you may have include:
- What is the force of attraction between the Earth and the Moon?
- What is the force of attraction between a boy and a girl?
- What is the force of attraction between the Moon and a girl?
This lesson will answer those questions. There is a mini-quiz near the end of the lesson.
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Escape velocity from Earth
The escape velocity from Earth can be calculated from the equation:
vEarth = √(2GM/R)
where
- vEarth is the escape velocity from Earth in km/s
- G is the Universal Gravitational Constant = 6.67*10−20 km3/kg-s2
- M is the approximate mass of the Earth = 6*1024 kg
- R is the approximate radius of the Earth = 6370 km
Substituting values into the equation results in:
vEarth = √[2*(6.67*10−20)*(6*1024)/(6370)] km/s
vEarth = 11.2 km/s
A rocket would have to achieve this velocity before shutting off its engines, if it were to escape from the gravitational pull of the Earth.
Escape velocity from Sun at Earth
Suppose a rocket blasted off the Earth from its far side away from the Sun.
Rocket leaves Earth away from Sun
The escape velocity from the Sun would be:
vSun = √(2GMSun/D)
where
- MSun is the approximate mass of the Sun = 200*1028 kg
- D is the approximate distance from the Sun to the far side of the Earth = 150*106 km
Note: The distance from the Sun to the Earth is approximately 150*106 km. Addition of the 6370 km would make a negligible contribution.
Substituting values into the equation results in:
vSun = √(2*6.67*10−20*200*1028/150*106) km/s
vSun = √(17.79*102) km/s
vSun = 42.2 km/s
Note that the escape velocity from the Sun at the Earth's surface is larger than the escape velocity from the Earth itself.
Combined escape velocity
If the rocket did not exceeded the Escape velocity from the Earth but not the escape velocity from the Sun, it would move off into space and then soon reverse directions and fall into the Sun.
The combined escape velocity from the Earth, adding in the effect from the Sun, is:
ve = √(vEarth2 + vSun2) km/s
ve = √(11.22 + 42.22) km/s
ve = √(1906.28) km/s
ve = 43.7 km/s
This velocity does not take into account the rotation of the Earth and its orbital velocity, which will affect the escape velocity. This is beyond the scope of our studies.
Also, the contribution from the gravitation of the Moon is negligible.
Summary
The escape velocity from the Earth dies not take into account the escape velocity from the Sun that an object—such as a rocket—must attain in order to completely overcome the gravitational pull of both the Earth and the Sun.
When a rocket blasts off from the Earth in a direction away from the Sun, you must combine the escape velocity from Earth and the escape velocity from the Sun to get the true escape velocity.
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Resources
The following resources provide information on this subject:
Websites
Acceleration due to Gravity Calculations - from Western Washington University
Gravity and Gravitation Resources
Books
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books on Simple Gravity Science
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books on Advanced Gravity Physics
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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Effect of Sun on Escape Velocity from Earth