F=G M1 M2 /(r^2)


by korican
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korican
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#1
Dec1-06, 02:20 PM
P: 3
I have a question.
The equation for calculating the force due to gravity.
F=G M1 M2 /(r^2)

Are M1 and M2 the mass's at rest or relativistic masses? So basically if the objects are traveling reaaaaaaaaaally fast. And their relativistic masses' increase does the force due gravity between them increase?

Thanks in advance.
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dextercioby
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Dec3-06, 11:34 AM
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m_{1} and m_{2} are the gravitational masses of the objects under study. In Newtonian physics, where the law you wrote is valid, by a postulate, they are set equal to the inertial masses of the 2 bodies (i.e the masses appearing in Newton's second law of dynamics). They are the m_{0}'s (invariant, rest masses) appearing in special relativity.

Daniel.
pervect
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Dec3-06, 12:55 PM
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Quote Quote by korican View Post
I have a question.
The equation for calculating the force due to gravity.
F=G M1 M2 /(r^2)

Are M1 and M2 the mass's at rest or relativistic masses? So basically if the objects are traveling reaaaaaaaaaally fast. And their relativistic masses' increase does the force due gravity between them increase?

Thanks in advance.

That's an equation from Newtonian gravity. It does not give the correct result when the two masses are moving at relativistic velocities. Basically, for consistency, if you are using the Newtonian formula for gravitational "force", you should use the Newtonian mass. The formula will give the correct answer only if the two masses are moving "slowly".

The Newtonian mass will be equal to both the relativistic mass and the rest mass in the region in which the formula is valid. So in the region in which the formula is valid, it doesn't matter which mass you use. When the two masses differ, because one of the objects is moving "really fast", the formula given above simply won't work. (The formula also won't work under some other exotic conditions that make the Newtonian mass different from the relativisitic or rest mass, but we can hopefully avoid getting into that).

If one of the objects is moving really fast, you will even run into problems attempting to unambiguously define the "gravitational force". What you can easily define is the tidal gravitational force. You will find that the tidal gravitational field of a fast moving body is "squished" by its motion in much the same manner that the electrical field of a fast moving charge is squished.

See for instance the following two posts from the thread below

http://www.physicsforums.com/showpos...36&postcount=9
http://www.physicsforums.com/showpos...6&postcount=10

If you are not familiar with how relativity distorts (squishes) the electric field of a rapidly moving object, see for instance

http://www.phys.ufl.edu/~rfield/PHY2...ativity_15.pdf

Other slides in this series also contain useful information (replace the 15 in the above link by other numbers) including some mathematical formula, derivations, etc.

Also note that if you have two rapidly moving masses, you will get forces that are quite similar to the magnetic forces between two moving charges. In general, both masses must be moving (or one of them must be spinning) to observe these "gravitomagnetic forces".

korican
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#4
Dec4-06, 09:36 AM
P: 3

F=G M1 M2 /(r^2)


Thanks for the responses. I read all the links too and the lectures 1-19. I skimmed the beginning it was just basic math. Now I have more key words to use when I research this stuff. I'll come back with questions if they come up.


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