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If Hubble's constant is $2.33 \times 10^{-18} \text{ s}^{-1}$ and the earth orbits the sun with average distance of 150 million kilometers; Does that mean the earth's orbital radius increases approximately $11\text{ m}/\text{year}$? Does the earth's angular momentum change? If so, where does the torque come from? If the angular momentum doesn't change, does the earth's orbital velocity (length of a year) change? If so, where does the lost kinetic energy go?

Aside: the 11 meters per year figure comes from Hubble expansion of space the distance of the earth's orbital radius integrated over an entire year.

$$(2.33 \times 10^{-18}\text{ s}^{-1}) (1.5 \times 10^{11} \text{ m}) (3.15 \times 10^7 \text{ s}/\text{year}) = 11 \text{ m}/\text{year}$$

rae
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No. Hubble's constant roughly says how the distance between two objects at rest with the universe grows. It does not say that the distant between everything is growing - the size of the hydrogen atom is not increasing. (My size is increasing, but from dietary rather than cosmological sources.) The size of objects and orbits are maintained by a balance of forces (classically). To whatever extent one can think of the expansion of the universe as pushing the Earth and Sun apart, it is already taken into account in setting the Earth's orbit.

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The change in the Hubble constant can effect the orbit, see the paper linked by Ben Crowell. But just taking the Hubble constant and multiplying it by the Earth's radius, as I believe you have done, does not give you anything sensible.

BebopButUnsteady
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    looks like you don't believe in the Big Rip. Regardless, there does appear to be some evidence for hubble expansion of the moon's orbit. Though it's a bit more difficult to laser measure the distance from the earth to the sun. – rae Jul 22 '11 at 21:37
  • Actually a system like the solar system is predicted to expand due to cosmological expansion, but the effect is calculated to be incredibly small, much too small to detect: http://arxiv.org/abs/astro-ph/9803097v1 The size of a hydrogen atom doesn't change at all, because it's fixed by fundamental constants. Since the solar system does expand by some tiny amount, there is predicted to be a small violation of conservation of energy. That's OK, because energy isn't conserved in general relativity. –  Jul 22 '11 at 21:44
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    @rae: The paper by Dumin was never published, and looks just plain wrong to me. It contradicts the Cooperstock paper that I referenced above, which was published in a peer-reviewed journal. There is not a scrap of GR anywhere in the Dumin paper; to my knowledge, no competent relativist has ever suggested that GR leads to an effect of the order of magnitude of the discrepancy that Dumin attributes to cosmological effects. The Big Rip is not really relevant. We don't know if the laws of physics are such as would cause a Big Rip, and the OP is not asking about the remote future. –  Jul 22 '11 at 21:54
  • @Ben Crowell: I agree that a changing Hubble constant effects the orbit which is what that paper derives (see Eqn. 4.2 depends only on the second derivative of the scale factor). I'm dealing with only the effects of a constant Hubble 'constant', since I believe that what the original poster was asking about. His number 11m/year comes I believe from multiplying the earth's orbital radius by the Hubble constant which is certainly not correct. – BebopButUnsteady Jul 22 '11 at 22:30
  • The Cooperstock paper is right. I haven't read the Dumin paper. A key point to observe from the Cooperstock paper: not only is the effect very small, it's proportional to $\ddot a$, not $\dot a$ (eq. 4.2, for instance). That means that there's no effect do to expansion; the effect is due to acceleration. We could've guessed this from the get-go: one example of a uniformly expanding Universe ($a(t)\propto t$) is just good old Minkowski spacetime, written in funny coordinates. But there's certainly no funny effect on the solar system in that cosmology! So there can't be an $\dot a$ term. – Ted Bunn Jul 22 '11 at 22:32
  • @BebopButUnsteady: Sounds like we're in agreement. I wouldn't attach any significance to the OP's use of the word "constant," since everyone calls it the Hubble "constant," knowing full well that it isn't constant. Realistic models have $\ddot{a}\ne 0$, so it wouldn't be physically meaningful to interpret the OP's question as assuming $\ddot{a}=0$. –  Jul 22 '11 at 22:50
  • @Ted Bunn: Any opinion on Ostvang's quasi-metric relativity, http://arxiv.org/abs/gr-qc/0112025v6 ? It predicts significant solar-system effects, was intended to explain the Pioneer anomaly. Iorio's analysis of the outer solar system data says that the Pioneer anomaly can't be gravitational if it obeys the equivalence principle, but Ostvang says QMR is OK with that because it doesn't obey the equivalence principle. (Less motivation from experiment now that thermal models have made the Pioneer anomaly appear to have been bogus all along.) –  Jul 22 '11 at 23:23
  • Ok, so while the space between the earth and the sun is indeed expanding and if the earth would return to the same point in its orbit the following year it would indeed be futher out than it was the previous year, but not nearly the 11m the simple calculation would lead one to believe. This means that the distance between the earth and the sun is not keeping up with the expansion of space, but rather the earth is captured in its orbit and mostly determined by its total energy. Even though its orbital radius is expanding, but much less than the simply calculated 11m/year? – rae Jul 23 '11 at 00:17
  • @Ben Crowell -- I haven't studied Ostvang's theory. Just to be clear, everything I've said is based on standard general relativity. (I do think that studying alternative theories is a salutary thing to do, although I've never found the Pioneer anomaly to be a particularly strong motivation.) – Ted Bunn Jul 23 '11 at 17:39
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    @rae -- The problem is that there's no clearly-defined meaning to be attached to the phrase "the same point in its orbit the following year." If you use comoving coordinates (i.e., coordinates that expand with the Universe), then "the same point" will be further out than before. If you use local Minkowski coordinates, it won't. And of course there are infinitely many other choices. The common mistake people make is to think that comoving coordinates are what space is "really" doing, but the central idea of relativity is that coordinate systems are just conveniences, not "Truth." – Ted Bunn Jul 23 '11 at 17:43
  • @Ted -- Sorry, poor choice of wording. I meant same angular position around the sun the following year. The earth's orbital radius would not be 11m further away from the sun as my simple-minded calculation, but it wouldn't be the same as it was a year earlier, either. Only a slight increase due to the 2nd derivative effects of Hubble expansion? At least that is what I gleaned from the above discussion. – rae Jul 25 '11 at 19:33
  • @rae -- I see. Sorry for misunderstanding. Yes, I think what you say is right. – Ted Bunn Jul 25 '11 at 19:36
  • A point that I hadn't understood until today is that the predicted secular trend is for the earth's orbit to contract. Cooperstock was writing in 1998, and assumed $\ddot{a}<0$, which leads to an expanding trend in the orbital radius. We now know that $\ddot{a}$ is positive, so the prediction is contraction over time! I think this makes it clear that the effect is not just a brute-force competition between cosmological expansion and binding. It's more of an intricate dynamical effect. –  Sep 03 '12 at 18:23
  • Oops, it's even more complicatewd than I thought. According to the Cooperstock paper, the sign of the anomalous radial force acting on an object in a circular orbit depends on the sign of $\ddot{a}$, but the sign of the secular trend in the orbital radius depends on the sign of $(d/dt)(\ddot{a}/a)$. I don't even know the sign of this quantity in the present cosmological epoch. (It's zero in a vacuum-dominated cosmology.) –  Sep 05 '12 at 05:49
  • While indeed the effect of a constant H is already "baked in" to the orbital parameters, this answer doesn't address the question of to what extent they produce a deviation wrt the Newtonian prediction. While the Cooperstock paper shows that the effect of changing H is totally negligible, it does not show that the "baked in" discrepancy from the Newtonian prediction is negligible, which may be of interest to readers coming across this question. – user1247 Jun 27 '21 at 15:31
  • There seems to be much discussion from a pedagogic angle (e.g. see here) that the "balloon" analogy can be confusing because the balloon isn't sticky, so there isn't any friction-like "pull" from hubble expansion on bound states. But they neglect to address the fact that even on a frictionless balloon, the expansion of the balloon between test masses should produce an effective acceleration. My (limited) understanding is that the more correct resolution is that the balloon is simply not in fact expanding at all in local bound state regions. – user1247 Jun 27 '21 at 17:55
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For objects smaller than cosmic scale, such as atoms, planets and solar systems, the electromagnetic and gravitational forces that hold them together are not changing (as far as we know) and so those objects do not change size.

Between galaxies, so widely separated, there's just gravity, and that tends to average out due to every galaxy being surrounded by other galaxies in all directions. On a cosmic scale, galaxies are like a gas, with galaxies being the "molecules" and described by the idea gas equation. To account for gravity and finite size of the galaxies, we might use the Van der Waals equation or some other variation, but that's beside the point, useful only for increasing accuracy.

Hubble's constant describes the rate at which the "container" of the galactic gas is expanding, the way the density of galaxies decreases over time. In an ordinary gas such as air, when in an expanding chamber, certainly the molecules are not expanding. Likewise, neither are the galaxies changing their sizes, at least not for Hubble-related reasons.

DarenW
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  • You're oversimplifying by treating atoms and solar systems as being the same. GR does predict that solar systems will expand, just not by very much: http://arxiv.org/abs/astro-ph/9803097v1 The size of a hydrogen atom is set by fundamental constants. The size of a solar system is not. –  Jul 25 '11 at 00:39
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The reason the universe expands is gravitation, as described by Einstein's field equation. The evolution of the universe is governed by gravitation, as described by Einstein's field equation. Over cosmological scale, the universe can be seen as homogeneous and isotropic, with very small density of matter and radiation. The density of matter and radiation is too small to counteract the expansion, an effect of initial condition. In local areas, however, the density is many magnitudes higher, and the effect of expansion is all but counteracted by the binding gravitational attraction.

Siyuan Ren
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  • Everything you've said is true, but it fails to answer the OP's question. One way to see that it doesn't answer it is that although you mention the Einstein field equation, everything you say is equally valid in a Newtonian expanding universe. The fractional rate of change in size due to cosmological expansion is 0 for a hydrogen atom, $\sim 10^{-41}\ \text{s}^{-1}$ for the earth-sun system (http://arxiv.org/abs/astro-ph/9803097v1 ), and $\sim H_o$ for a photon. I don't see how you would get that from "tends to confine." –  Mar 11 '12 at 01:44
  • I also wouldn't agree that "the reason the universe expands is gravitation." The reason it has been expanding, ever since the Big Bang, is inertia, and this would be just as true in a Newtonian model as in one based on GR. –  Mar 11 '12 at 01:49
  • @BenCrowell: To your second comment: By inertial the expansion will slow down, whereas in fact the expansion is speeding up, currently modeled by a non-zero cosmological constant, an effect of gravitation. – Siyuan Ren Mar 11 '12 at 01:58
  • @BenCrowell: To your first comment: I read your paper, and all I see is that according to the authors themselves, this topic, the effect of cosmological expansion on local systems, is highly contentious. I doubt your paper has settled the problem and become the consensus. – Siyuan Ren Mar 11 '12 at 02:04
  • @BenCrowell: Regardless, it is well known that gravitation bounds the solar system. Even cosmological expansion has an effect, it is infinitesimal small. I don't see how that invalidates my phrase "tends to confine" at local scale. – Siyuan Ren Mar 11 '12 at 02:08
  • @BenCrowell: OK, I stand corrected. Gravitation is not the reason for expansion. It merely accelerates it. – Siyuan Ren Mar 11 '12 at 02:54