0
$\begingroup$

If we were to calculate the force that one galaxy exerted onto another, would we consider the individual masses within the galaxies, or the masses of the galaxies as a whole?

Do the individual stars within one galaxy have control over the individual stars within another?

Or, do we consider the galaxies to be single bodies, each with it's own mass, and the distance between them as two uniform bodies?

If so, could we call galaxy clusters uniform bodies as well?

And if a galaxy cluster were to all of a sudden lose a star (hypothetically), would the net force exerted onto the other galaxy cluster be the same if you counted each star individually?

As another curiosity, I would like to pose this question: if a galaxy cluster were to lose one fourth of its mass, $2.0\times10^3$ light years away from the other galaxy cluster's centre of mass, how would the effect be different than if the mass was lost only $0.50\times10^3$ light years away?

What I mean is, the distance would be different, so the force would be different (**(if you can address this as well please)and information would have to travel faster than light for this to occur, yet with a mass loss this great, the other cluster would be sure to notice it almost immediately).

| cite | improve this question | |
$\endgroup$
  • 1
    $\begingroup$ I think you need to be much more specific about the applications you are interested in studying. The assumptions that go into the model are dictated by the resources and desired outcomes from the model. $\endgroup$ – tpg2114♦ May 28 '14 at 21:09
  • $\begingroup$ You can't have a cluster "lose" mass. It can eject some of the stars, but the mass doesn't disappear. It slowly recedes from the cluster. The gravitation field is slowly altered to reflect the new distribution. Also 2000 ly or 500 ly are tiny distances on a galactic cluster scale-they are right at the center. Much less than a quarter of the mass of the cluster is anywhere near that close to the center. $\endgroup$ – Ross Millikan May 28 '14 at 22:49
2
$\begingroup$

A useful way to represent a mass distribution is a multipole expansion. For the earth, the monopole term represents the overall mass of the earth. The dipole term represents (roughly) the flattening at the poles. Higher order multipoles represent more complicated parts of the mass distribution. The gravitational effects of the monopole fall off as $\frac 1{r^2}$. The effects of the dipole fall off as $\frac 1{r^3}$ and higher order terms fall off faster yet. Far enough away, the monopole term dominates, so you can consider the earth (or a galaxy) to be a mass point. .

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I see, that helps a lot. So, what if you were in such a large scale reference frame that photons appeared to be stationary (and then zoom out even further)? If gravity can only act at light speed, how would the various portions of the reference frame interact? $\endgroup$ – Gödel May 28 '14 at 21:50
  • $\begingroup$ There is no reference frame where photons appear stationary. The point here is that if you are far from (basically compared to its dimensions) a source of gravity you can consider it to be a point. Where that point is can change with time and its motion will cause the gravitational field to change. These changes propagate with light speed. Under GR all interactions are local as they are described by differential equations. It is not portions of a reference frame that interact, it is masses and space. $\endgroup$ – Ross Millikan May 28 '14 at 21:58
  • $\begingroup$ Yes I'm sorry I have been camping for quite some time, so I haven't gotten a chance to respond(forgive me). I must not have been clear with what I was asking. What I meant was something like this: Imagine a T shaped mass in hyperspace on such a large scale that 3.08x108 ms-1 is negligible (so a photon would appear to be still, because the scale is so massive that it would get no where). HYPOTHETICALLY, if one side of that T were to disappear(I am not as daft as to think such a thing could actually occur), how would we see the rest of the T react? Would it just take a long time? $\endgroup$ – Gödel Jun 19 '14 at 23:09
0
$\begingroup$

Galaxies can't just spontaneously lose mass. Gravitational effects can only travel at the speed of light. As for the rest of it, as tpg2114 says, you have to consider the resources available to do calculations. Ultimately, every particle in the universe exerts a gravitational force on every other particle. We can usually simplify this out by considering them to be extended bodies with some average mass distribution, or even by amalgamating many objects into point masses. But this stuff is done for mathematical simplicity, not for any on high a priori reason.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.