Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Gravitational Attraction

  1. Jan 28, 2009 #1
    2 particles are moving very fast in an "inertial reference frame." What is the force of gravitational attraction between them?

    Note that at low speeds it is G(m1)(m2)/r^2. Also note that the two particles are not aware of the inertial reference frame with respect to which they are moving very fast.
     
  2. jcsd
  3. Jan 28, 2009 #2
    How fast? It's kind of important? Oh and what are the rest masses of the particles?

    I presume you also mean they are moving at the same speed and in the same direction with reference to the inertial reference frame? We also need to know how close they are together still as well. Is this a minimum information problem. Because if so it was the cat who done it.
     
  4. Jan 28, 2009 #3
    Can't we say arbitrary speed and direction in IRF?

    Is'nt it simply G(m1')(m2')/r^2 where m1' and m2' are the relativistic masses and r is separation?
     
  5. Jan 28, 2009 #4
    Just say two electrons travelling at .0002c with a gap of .2 microns? Yeah we could do that, but since as speed increases so does mass, then if you want an answer that isn't arbitrary.

    It could be simply that I just thought you might want to use the more accurate equations of relativity?

    Newtons laws aren't usually applied to objects travelling at close to c, just asking how fast really. Up to you though?
     
  6. Jan 28, 2009 #5
    Let's say .75c. Or are you saying G(m1')(m2')/r^2 doesn't apply at high speeds. That's OK. I'll accept that. But then what is the attractive force between them? Let's stick with special relativity.
     
  7. Jan 28, 2009 #6

    jtbell

    User Avatar

    Staff: Mentor

    You can't. The relativistic theory of gravity is general relativity.
     
  8. Jan 28, 2009 #7
    To be more practical. Suppose we wanted to correct the orbit of a satellite for mass relative to coordinate system fixed in the earth. In principle, even if it's a matter of micrometers. That is, correct the orbit with respect to the classical Newtonian calculation. Is it as simple as using the relativistic mass wrt the earths coordinate system?
     
  9. Jan 28, 2009 #8
    I guess jtbell answered my question. Thanks.
     
  10. Jan 28, 2009 #9

    jtbell

    User Avatar

    Staff: Mentor

    Recall that classical mechanics, the orbit of a satellite doesn't depend on its mass, unless it's so massive that we need to consider the satellite and the earth as both revolving around their mutual center of mass.

    I think any general-relativistic effects would come in by way of the curvature of space-time produced by the satellite, but you'd need a very massive satellite for that to be significant, and a very massive Earth to go along with it.

    In general, I'm very skeptical of trying to predict relativistic gravitational effects by combining Newtonian gravitational formulas with special relativity.

    I do remember seeing someone who knows GR pretty well doing a calculation involving the gravitational effect of a fast-moving mass. The idea is to consider an initially stationary object being passed by a moving one, and calculating the "kick" or impuse that the gravitational field of the moving object delivers to the originally-stationary one. As I recall, the impulse turns out to be proportional to [itex]\gamma m_0[/itex] of the moving object, i.e. what is often called the "relativistic mass."

    Not being able to do any GR calculations myself, I don't know how well this result generalizes to other situations. Also, I don't remember what approximations were used, if any. It may very well be that this result applies only approximately for some range of mass and/or velocity.

    If you want to track this down, try searching for posts here made by former poster 'pervect' with keywords like "relativistic mass."
     
    Last edited: Jan 28, 2009
  11. Jan 28, 2009 #10
    From my physics book, mdV/dt=-GMm/r^2, and m cancels. My question could have been does the same apply with relativistic mass m' substituted for m? As you point out, no, it is a problem in general relativity. Whether the effect is significant or not doesn't matter, I was just interested in the principle.

    Thanks for response.
     
    Last edited: Jan 28, 2009
  12. Jan 28, 2009 #11
    I wouldn't like to take on the calculations myself, in four space it looks a bit above my weight. This does highlight a problem between general relativity and quantum mechanics though. I see no reason you couldn't make pretty good estimates at masses above say 1kg and speeds < c, but it looks messy. You have x,y,z and t in tensor form and you have to apply the relativistic masses of the two particles at time (t). I wonder if some maths guru knows?

    Interestingly with satellites that are travelling at much less than c it's actually relatively easy to adjust the clocks to synch in time, I wonder if anyone could do it if the satellites were approaching c? I wonder if anyone's worked out a way around the problems? Or at least a method to guestimate?
     
    Last edited: Jan 28, 2009
  13. Jan 28, 2009 #12

    Dale

    Staff: Mentor

    No, it doesn't, energy increases. And energy is only 1 of 16 components of the stress-energy tensor which is the source of gravity in GR.
     
  14. Jan 29, 2009 #13
    Apologies I was talking about relativistic mass, should of made that clear, of course a better way of looking at it is to assume that the effects come from space-time concerns in both special and general relativity, not from an increase in mass exactly.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Gravitational Attraction
  1. Gravitation ? (Replies: 2)

Loading...