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Acceleration of gravity

  1. Feb 11, 2010 #1
    there Is a force causing the andromeda galaxy and the milky way to move away from each other with a magnitude of 2x10^28N. How close must the galaxies be to each other for the force of gravity overtake this repulsive force?

    Mass of andromeda=
    6 x 10^41

    Mass of milky way= 3 x 10^41

    g= 6.7x 10^-11

    Would I used the acceleration of gravity formula for this question?

    Gx M1M2/d^2
     
  2. jcsd
  3. Feb 11, 2010 #2

    Matterwave

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    You could use that for a rough first order approximation. Galaxies aren't spherical (or solid), though, so the actual description is vastly more complicated.
     
  4. Feb 11, 2010 #3
    Oh I know that galaxies aren't perfect spheres, but this problem is for an intro astronomy course.

    I'm still unclear what gets plugged in where...
    M1=mass of andromeda
    m2= mass of milky way
    G= 6.7 x 10^-11

    so I solve for d^2?
     
  5. Feb 11, 2010 #4

    Matterwave

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    You want d^2 such that the force due to gravity is balancing the force of 2*10^28N
     
  6. Feb 11, 2010 #5

    BillSaltLake

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    Please note that Andromeda is moving toward us at about 120 km/sec, so that "force" is not yet causing the galaxies to move away from each other.
     
  7. Feb 11, 2010 #6

    marcus

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    Sounds like a plan.

    d^2 = GM1M2/ (2x10^28 Newtons)

    = G (18 x 10^82 kg^2)/(2x10^28 Newtons).

    So take the square root:
    sqrt(G (18 x 10^82 kg^2)/(2x10^28 Newtons))

    If you want to use the google calculator, replace the "x" times sign by "*" and put this in the window:

    sqrt(G*(18 * 10^82 kg^2)/(2*10^28 Newtons))

    If you want the answer in lightyears then put this in the window instead:

    sqrt(G*(18*10^82 kg^2)/(2*10^28 Newtons)) in lightyears

    That blue thing is what I typed in.

    I'm still not clear what your problem is. It doesn't make sense to me. Where does this imagined force come from, that is supposed to be driving the two apart? Why is it assumed to be 2x10^28 Newtons?

    I think in reality there is no such force, driving the two apart. Indeed they are falling together, as someone said, at something roughly on the order of 100 km/s (I don't remember exactly how fast but something like that.) The expansion of distances does not affect objects which are gravitationally bound together and expansion operates on a much larger scale. Andromeda is part of our local group of galaxies---not being carried away from us by expansion. So what is this hypothetical 2x10^28 newtons?
     
    Last edited: Feb 11, 2010
  8. Feb 11, 2010 #7

    BillSaltLake

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    The question might be this: If H0 is constant, then the second time derivative (due to expansion, IF it were comoving) of Andromeda's distance D equals DH02. This acts like a repulsive force. Find the distance that balances the gravitational attraction.
     
  9. Feb 12, 2010 #8

    Chalnoth

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    There isn't any repulsive force due to expansion, though. The repulsive force, if any, would be due to the cosmological constant.
     
  10. Feb 12, 2010 #9

    Ich

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    There is such a force, but it's rather 4*10^27 N acting on Andromeda (with m=6*10^41 kg) at the momentary distance. It's a fictitious force, though, but gravity is a fictitious force, too. And it's proprtional to distance.
     
  11. Feb 12, 2010 #10

    Chalnoth

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    There's no reason to believe that gravity is any more a fictitious force than electricity and magnetism or the strong and weak nuclear forces.
     
  12. Feb 12, 2010 #11

    Ich

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    There is a reason: it's proportional to (inertial) mass. "Fictitious force" is a well defined concept in physics, and "gravitational force" is fictitious, electric force is not.
     
  13. Feb 12, 2010 #12

    Chalnoth

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    Most consider a fictitious force to be one whose appearance depends upon the choice of reference frame. There are real gravitational effects that are entirely independent of reference frame.
     
  14. Feb 12, 2010 #13

    Ich

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    Which is consistent with what I say. That's just two versions of the equivalence principle, and both are true only locally.
    Did I say that there is no gravity?
    I said that "gravitational force" is a fictitious force. Gravity itself is not a force, it's spacetime curvature. Gravitation in itself does not produce proper acceleration, or force. These things are produced by real forces, like electricity - in some cases as a reaction to spacetime curvature.
     
  15. Feb 12, 2010 #14

    Chalnoth

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    Ich, if there weren't coordinate-independent quantities such as the Ricci curvature scalar that are definitively non-zero in gravitating systems, I might agree with you. The problem is that they are.

    Basically, gravity and acceleration are only equivalent at a single point. You can get appreciable differences between the two even in a small region around a point.

    Perhaps most crucially, pseudo forces don't follow Newton's third law, which gravity does.
     
  16. Feb 12, 2010 #15

    Ich

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    That's not a problem, as they do not enter the equation of motion. There are only Christoffel symbols, and these you can transform away at every point.
    Freely moving particles - those not affected by a force - move along a geodesic, I'm sure you agree. Geodesic deviation and such are a matter of geometry, not of force.
    If you're not talking about test particles, well, there there isn't even an equation of motion, so it's definitely even less appropriate to treat gravity as a real force.

    I'm quite sure that I'm in line with the standard philosophy of GR. If you have some sources that say otherwise, we should discuss the issue in a different thread. This one isn't really about the GR viewpoint of gravity.
    I just cautioned that the 4*10^27 N I was talking about are frame-dependent, and nobody at Andromeda could measure them there - they vanish in an inertial frame centered at Andromeda. And these 4*10^27 N are of gravitational origin (Dark Energy).
    They might be what the OP's problem is about (the numbers are close), and marcus was wondering what this force could mean, so I mentioned it.
     
  17. Feb 12, 2010 #16

    Chalnoth

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    I don't see why you wouldn't call geodesic deviation a matter of force. Yes, in GR, gravity is a matter of geometry. But this doesn't make it not a matter of force. Nor does it necessarily make it qualitatively different from other forces.

    The main problem here is that as of right now, we just don't know how to talk about the three quantum forces of which we are aware of and gravity in the same language. This doesn't mean they aren't fundamentally different aspects of the same thing. It just means that we haven't found out how to talk about them in the same language. Not entirely, at least: we do know how to talk about classical electricity and magnetism in the context of General Relativity, in which case E&M basically can be seen to take on the same curved-space properties as GR.

    Well, it is frame dependent, but in this case there's a natural frame choice (you can just pick an inertial frame).
     
  18. Feb 12, 2010 #17

    Ich

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    OK, and until the true nature of all forces is known, let's use the gold standard models along with their way to view things. I gather that you agree with "in GR, gravitational force is a fictitious force", while the other three interactions are not being described this way. (Kaluza-Klein and derivatives are not really gold standard today.)
    I did pick an inertial frame. That of the Milky Way, and I don't see how this is less natural that that of Andromeda. There is relative gravitational acceleration between both systems, that's why I get fictitious forces when I use the Newtonian gravity picture. Because Newtonian physics insists that if there is relative acceleration, I have to assign absolute, proper, acceleration (along whith a cause, namely gravitational force) to one or the other. It doesn't make a difference, though, that's why the GR concept (no absolute acceleration in this case) is in better shave. (Attention, pun!)
     
  19. Feb 12, 2010 #18

    Chalnoth

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    I don't agree, because to me a fictitious force has to be one which can be entirely removed by a gauge transformation, which cannot be done in general in GR.

    You'd get the same answer either way, though.
     
  20. Feb 12, 2010 #19

    Ich

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    By a global gauge transformation, you say. As I said, if you find sources that support this point of view, we can discuss it in another thread. A quick web search will provide you with a number of sources supporting my point of view.
    Remember, I'm not claiming that gravity is an illusion. I say that gravitational force is a fictitious force in GR. As can be seen in the next point:


    Nope.
    In Milky Way's frame, that's 0 N for Milky Way and 4*10^27 N for Andromeda.
    In Andromeda's frame, that's 0 N for Andromeda and 2*10^27 N for Milky Way.
    That's as fictitious as it can get.
    The relative acceleration is the same, of course, or else I'd have blundered. I didn't say that relative acceleration is fictitious. But the force is that Newton attributes as a cause.
     
  21. Feb 12, 2010 #20

    Chalnoth

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    I was actually thinking of a local gauge transformation, where you make a change in the coordinates at every point in space-time. The point here is that you can get rid of coordinate system dependent quantities like the Christoffel symbols just fine, but you can't get rid of coordinate-independent quantities like the curvature.

    By contrast, with the pseudo forces that arise from rotating reference frames, those forces disappear entirely if you just transform to an inertial frame.

    Well, the issue here is that this isn't the force of the Milky Way on Andromeda, or vice versa (so this isn't a third law issue). What we have instead is gravity interacting with vacuum energy producing the "force", which I suppose I have to admit acts very much like a pseudo force (you can get rid of it entirely by transforming to comoving coordinates). However, this just isn't the case for all gravitational systems. Particularly for, say, our solar system, gravity acts very much like a real force, as no amount of coordinate transformation does away with it.

    Of course, the exact forces on different objects do look quite a bit different depending upon which reference frame you talk about, but this is as true of E&M as it is with gravity.
     
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