# Threshold of gravitational pull over dark energy

• B

## Main Question or Discussion Point

I’m only an interested layman and my math is pretty basic so please excuse my lack expertise in this subject but I was wondering how to calculate the threshold distance an object of mass m would have to be to a system of mass M to overcome the effects of dark energy and not be receding?

kimbyd
Gold Member
In the Newtonian limit, the modified form of Newton's gravitation is (the mass $m$ is the object accelerating, in response to M's gravity):

$$\vec{F} = -{G m M \over r^2}\hat{r} + {m \Lambda \vec{r} \over 3}$$

From this, the threshold distance you suggest is found by setting the force to zero, so we can set the classical Newtonian gravity (proportional to $1/r^2$) equal to the repulsion from the cosmological constant:

$${G M \over r^2} = {\Lambda r \over 3}$$

Solving for $r$ we get:

$$r = \left({3 G M \over \Lambda}\right)^{1\over 3}$$

So, it depends upon the mass we're accelerating away from. Dark energy creates a constant acceleration between any two objects in the universe regardless of their masses. Plugging in some numbers, the distance is given by:

$$r = (260,000 km) \left({M \over 1kg}\right)^{1 \over 3}$$

Last edited:
• Dr. Courtney and ohwilleke
PeterDonis
Mentor
2019 Award
Dark energy creates a constant acceleration between any two objects in the universe regardless of their masses.
More precisely, a constant acceleration at a given distance.

• ohwilleke
JMz
So, it depends upon the mass we're accelerating away from. Dark energy creates a constant acceleration between any two objects in the universe regardless of their masses. Plugging in some numbers, the distance is given by:

$$r = (260,000 km) \left({M \over 1kg}\right)^{1 \over 3}$$
Nice. I submit, however, that kg & km may not be the most convenient units of mass & distance for this problem. :-) Perhaps solar masses & LY, or Milky Way masses & Mpc?

kimbyd
Gold Member
Nice. I submit, however, that kg & km may not be the most convenient units of mass & distance for this problem. :-) Perhaps solar masses & LY, or Milky Way masses & Mpc?
Haha, sure :)

If I'm doing my unit conversions correctly:
$$r = (350ly) \left({M \over M_\odot}\right)^{1 \over 3}$$

• ohwilleke and PeroK
Haha, sure :)

If I'm doing my unit conversions correctly:
$$r = (350ly) \left({M \over M_\odot}\right)^{1 \over 3}$$
Thank you for that. But as my math is pretty crap coukd I ask you do an example so I can see how it works. Say how far would an object of 2 solar masses have to be from an object of say 10 solar masses before they were no longer gravitationally bound.

JMz
Thank you for that. But as my math is pretty crap coukd I ask you do an example so I can see how it works. Say how far would an object of 2 solar masses have to be from an object of say 10 solar masses before they were no longer gravitationally bound.
For practical purposes, it's the lighter one that is, or isn't, bound. So the relevant mass is 10 solar. The cube root of 10 is 2 (close enough!), so that means 700 LY.

JMz
Haha, sure :)

If I'm doing my unit conversions correctly:
$$r = (350ly) \left({M \over M_\odot}\right)^{1 \over 3}$$
Or just about (1 Mpc) * (mass/Milky Way's mass)1/3 -- convenient! [Yes, if you did the conversions correctly, which I didn't check -- but doesn't this seem a little too small?]

For practical purposes, it's the lighter one that is, or isn't, bound. So the relevant mass is 10 solar. The cube root of 10 is 2 (close enough!), so that means 700 LY.
Ok great thanks.

So taking that the Milky Way is about 700 billion solar masses, then that works out that anything within 3.1 million light years would be gravitational bound to it.

Considering that the observable universe is about 46 billion LY’s in any direction (96 across) and there are over 100 billion galaxies it seems odd that we do see galaxies receding. Especially as the universe would have been a lot smaller some 10 million years ago. Even now it would only take around 15,000 Milky Way size galaxies equally spaced between us and the edge of the observable universe for them all to be gravitationally bound.

By the way that doesn’t mean I dispute expansion at all. It’s just seems surprising there is expansion.

Anyway, thanks for the help. It’s much appreciated:)

Orodruin
Staff Emeritus
Homework Helper
Gold Member
Wolfram Alpha is your friend. The region bound to the local galactic group would be of a size of order 1/100000 of the diameter of the observable Universe. That works out to a volume that is about $1/10^{15}$ of the volume of the observable Universe. $10^{15} \gg 10\cdot 10^9$.

Wolfram Alpha is your friend. The region bound to the local galactic group would be of a size of order 1/100000 of the diameter of the observable Universe. That works out to a volume that is about $1/10^{15}$ of the volume of the observable Universe. $10^{15} \gg 10\cdot 10^9$.
Do you have the actual numbers as I get something different so would like to see where I’m going wrong.

Thanks.

kimbyd
Gold Member
Ok great thanks.

So taking that the Milky Way is about 700 billion solar masses, then that works out that anything within 3.1 million light years would be gravitational bound to it.
This calculation isn't enough to state that something is gravitationally-bound. What the calculation suggests is that nothing which is much lighter than the Milky Way can possibly be gravitationally-bound to it if it is further than 3.1 million light years away. This number is, on its face, not absurd, given that all of the Milky Way's satellite dwarf galaxies are substantially closer than this:
https://en.wikipedia.org/wiki/Satellite_galaxies_of_the_Milky_Way

Things that are closer can still be unbound if they are moving quickly enough. Things that are further might become bound in the future if they're on a trajectory towards the Milky Way and will slow down due to interactions with the galaxy (e.g. when the Andromeda galaxy collides with us in a few billion years).

Anyway, here's the math that I used in the calculation.

First, the constants (grabbed from Ned Wright's cosmology calculator):
$$\Omega_\Lambda = 0.714$$
$$H_0 = 69.6 km/s/Mpc$$

From the Friedmann equations, we can determine the value of $\Lambda/3$:
$${\Lambda \over 3} = \Omega_\Lambda H_0^2$$

Then:

$${3GM \over \Lambda} = {GM \over \Omega_\Lambda H_0^2}$$

Then, I could use Google's calculator to plug in the above given an appropriate test mass by searching for:
(G * (mass of sun) / (0.714 * (69.6 km/s/(1e6 parsecs))^2))^(1/3) in light years

This calculation isn't enough to state that something is gravitationally-bound. What the calculation suggests is that nothing which is much lighter than the Milky Way can possibly be gravitationally-bound to it if it is further than 3.1 million light years away. This number is, on its face, not absurd, given that all of the Milky Way's satellite dwarf galaxies are substantially closer than this:
https://en.wikipedia.org/wiki/Satellite_galaxies_of_the_Milky_Way

Things that are closer can still be unbound if they are moving quickly enough. Things that are further might become bound in the future if they're on a trajectory towards the Milky Way and will slow down due to interactions with the galaxy (e.g. when the Andromeda galaxy collides with us in a few billion years).
Yes of course. My thinking was along the lines.... if we say for the sake of simplicity that the universe is 90 billion ly across and there are around 90 billion galaxies. And assuming the cosmological principle then that would give a grid of galaxies around 300,000 ly apart in any direction. So if they were all of similar mass to the Milky Way then they would, without any prior momentum, all be gravitationally bound.

Of course the universe is a bit more lumpy than that but the outcome should still be the same. So there must have been something that gave early mass momentum away from each other.

I always thought of inflation as just the grid of co-moving objects getting bigger. But obviously it must be more than that.

kimbyd
Gold Member
Yes of course. My thinking was along the lines.... if we say for the sake of simplicity that the universe is 90 billion ly across and there are around 90 billion galaxies. And assuming the cosmological principle then that would give a grid of galaxies around 300,000 ly apart in any direction. So if they were all of similar mass to the Milky Way then they would, without any prior momentum, all be gravitationally bound.

Of course the universe is a bit more lumpy than that but the outcome should still be the same. So there must have been something that gave early mass momentum away from each other.

I always thought of inflation as just the grid of co-moving objects getting bigger. But obviously it must be more than that.
Oh, yeah. Dark energy is only a late-time effect. The early expansion was vastly faster than it is today. Dark energy puts a floor on that rate of expansion, preventing it from decreasing further. In the far future, the expectation is that the rate of a expansion will be a constant.

Orodruin
Staff Emeritus
Homework Helper
Gold Member
Yes of course. My thinking was along the lines.... if we say for the sake of simplicity that the universe is 90 billion ly across and there are around 90 billion galaxies. And assuming the cosmological principle then that would give a grid of galaxies around 300,000 ly apart in any direction. So if they were all of similar mass to the Milky Way then they would, without any prior momentum, all be gravitationally bound.
90 billion ly means a volume of order $10^{33}$ ly3. With 2 trillion galaxies (again, Wolfram Alpha, which cites NASA - but it is larger than your number so it is conservative) this means your grid constant is of the order 10 million lightyears, not 300000 ly.

Of course the universe is a bit more lumpy than that but the outcome should still be the same. So there must have been something that gave early mass momentum away from each other.
Whether galaxies have momentum relative to one another is a question of choice of coordinates. In standard cosmological coordinates, comoving distances are growing due to metric expansion, not due to motion of the comoving galaxies.

I always thought of inflation as just the grid of co-moving objects getting bigger. But obviously it must be more than that.
Accelerated expansion is a late time effect. What you are computing is the distance over which the cosmological constant is relevant. It was irrelevant in the early Universe. Also, you should probably not use the word "inflation" for the accelerated expansion due to our current cosmological constant. The connotation of the word is to the hypothetical period of inflation in the early Universe before the onset of the standard Big Bang.

Dark energy puts a floor on that rate of expansion, preventing it from decreasing further. In the far future, the expectation is that the rate of a expansion will be a constant.
I think it should be stressed that this is only true in a universe that grows to be large enough for the cosmological constant to become relevant (i.e., where it becomes a significant portion of the contribution to the energy density). It is perfectly possible to imagine a universe where the initial matter density is large enough for the universe to collapse and contract before the cosmological constant has had any relevance whatsoever.

• PeterDonis
JMz
What the calculation suggests is that nothing which is much lighter than the Milky Way can possibly be gravitationally-bound to it if it is further than 3.1 million light years away. This number is, on its face, not absurd, given that all of the Milky Way's satellite dwarf galaxies are substantially closer than this:
https://en.wikipedia.org/wiki/Satellite_galaxies_of_the_Milky_Way
Yes, almost exactly one Mpc for one "galactic mass". :-)

But this distance does seem a bit too small to my eyes. The implication is that the Local Group is just barely bound against DE, a conclusion that I would have expected to read long before now if true, and have not. I have seen the conclusion that the Norma Cluster at the core of Laniakea is beyond the critical distance, but that Virgo is not. (More mass in both cases, of course, but tens of thousands of times more for Virgo?)

PeterDonis
Mentor
2019 Award
I have seen the conclusion that the Norma Cluster at the core of Laniakea is beyond the critical distance, but that Virgo is not.

PeterDonis
Mentor
2019 Award
The implication is that the Local Group is just barely bound against DE
No, it isn't. The relevant mass $M$ in the formula is not the mass of one galaxy, but the mass of the entire Local Group, which includes two Milky-Way sized galaxies plus many smaller ones, totalling 54 galaxies in all. The total mass of the group seems to be about 5 times the mass of the Milky Way, which gives an $r$ of about 15 Mpc.

JMz
No, it isn't. The relevant mass $M$ in the formula is not the mass of one galaxy, but the mass of the entire Local Group, which includes two Milky-Way sized galaxies plus many smaller ones, totalling 54 galaxies in all. The total mass of the group seems to be about 5 times the mass of the Milky Way, which gives an $r$ of about 15 Mpc.
I'll take your word for it that it's ~ 5x. (I'd have guessed 3x.) But if @kimbyd's equation is right, then r ~ 1 Mpc x ("galactic masses")1/3, right? So that's a distance r ~ 51/3 Mpc, somewhat < 2 Mpc.

JMz
"Specific": can't. Almost certainly either Tyson's Cosmos or History Channel's Universe, in one of their episodes on the long(er)-term fate of the universe. A practicing astrophysicist, yes, but not a scholarly presentation.

FWIW, I did find one ApJ paper that reproduces @kimbyd's result wrt the scale factor and Virgo, and therefore reaches the opposite conclusion about Virgo (as it should, since it's just what her result predicts): http://iopscience.iop.org/article/10.1086/378043/fulltext/. The problem, in the context of the pop-culture series, is that the paper is considerably older, coming just a few years after the detection of DE. I didn't find anything newer that either supports or contradicts that. (I was somewhat surprised not to find more. OTOH, I didn't search journals generally, only ApJ.)

PeterDonis
Mentor
2019 Award
I'll take your word for it that it's ~ 5x.
That's the best estimate I can come up with from what I can find online; the most detailed paper I can find is here:

https://arxiv.org/abs/1312.2587

This seems to be giving a total mass of stars of about 2.4 trillion Suns, which is roughly 5 times the Milky Way. However, it also gives a total mass of dark matter (which should probably be included) of about 4.2 trillion Suns; the sum of the two, which would be the total mass overall, is 6.6 trillion Suns, or about 13 times the Milky Way.

if @kimbyd's equation is right, then r ~ 1 Mpc x ("galactic masses")1/3, right? So that's a distance r ~ 51/3 Mpc
Ah, yes, you're right, r goes like the cube root of the mass.

kimbyd
Gold Member
Yes, almost exactly one Mpc for one "galactic mass". :-)

But this distance does seem a bit too small to my eyes. The implication is that the Local Group is just barely bound against DE, a conclusion that I would have expected to read long before now if true, and have not. I have seen the conclusion that the Norma Cluster at the core of Laniakea is beyond the critical distance, but that Virgo is not. (More mass in both cases, of course, but tens of thousands of times more for Virgo?)
I'd be very curious to see if I've made an error in the above calculations.

One confounding factor is that as you go further away, the total mass of the system grows because more mass is included.

JMz
I'd be very curious to see if I've made an error in the above calculations.

One confounding factor is that as you go further away, the total mass of the system grows because more mass is included.
Yes, not only a confound, but a potential observation bias, as the population of small but numerous dwarfs, low- vs. high-gas spirals, and stellar streams --not to mention DM -- become increasingly hard to weigh.

Note that the paper I mentioned above does agree with your numbers. (But it's 15 YO, whereas my memory about other conclusions dates from incidental reading or viewing materials from several years later.)

JMz
That's the best estimate I can come up with from what I can find online; the most detailed paper I can find is here:

https://arxiv.org/abs/1312.2587

This seems to be giving a total mass of stars of about 2.4 trillion Suns, which is roughly 5 times the Milky Way. However, it also gives a total mass of dark matter (which should probably be included) of about 4.2 trillion Suns; the sum of the two, which would be the total mass overall, is 6.6 trillion Suns, or about 13 times the Milky Way.

Ah, yes, you're right, r goes like the cube root of the mass.
[It's nice that the equation just comes out as a variant of Kepler's 3rd Law. :-)]

On the mass: Nice find. I would have guessed a similar total mass, but more for the MW. Probably 3 T Suns total vs. 1 T for MW, if I had thought about it. I see that WP lists ~ 1 T, albeit with substantial uncertainty. Of course, that's only WP, but it does give a few journal references over the last several years.

FWIW, the last news article about the MW mass that I can recall seeing (probably on Phys.org, several months ago) had 1 T, remarking that it reflected a significantly increased DM estimate that did not hold for M31. The result was that the two galaxies were estimated to have rather similar masses, but with different LM/DM ratios. (Speculation: Less DM around M31 because it's apparently had a much more dynamic/violent assembly history than the MW. That's not a finding, of course, but I find the narrative helpful for linking the DM quantity and the assembly history, which would otherwise be independent items.)

JMz
On rereading, I see that you included DM in your total estimate (which is obviously appropriate for the binding distance). I cannot resolve the paradox between M31 ~ 1 T Suns (DM+LM), MW ~ 1 T, and your numbers ~ 6 T for the whole LG.

My personal bias is to favor larger estimates, because most of the observation biases seem to go in the opposite direction. On that basis, I can easily buy the 6 T value.