<p>Yes and no.</p>

<p>First, the <a href="https://astronomy.stackexchange.com/q/7939/2188">answers here</a> should give you some understanding of dark energy and its effects. </p>

<p>The net effect of dark energy is tiny over small distances. If it is given by a cosmological constant, then its effects on gravitationally bound systems is independent of time (hence 'constant'). Observational evidence puts us at least close to the constant case. Which means that even if we are not actually in the constant case, then it will be many billions of years until there would be a measurable difference.</p>

<p>We can, however, estimate the mass equivalent of dark energy in a given area by measuring the apparent amount in the universe itself.</p>

<p>As estimated <a href="http://hyperphysics.phy-astr.gsu.edu/hbase/astro/dareng.html" rel="nofollow noreferrer">here</a>:</p>

<blockquote>
  <p>If we take the WMAP value for critical density at $$\rho_{c,0} = 9.47 \times 10^{-27}\, kg/m^3$$
  and presume that dark energy makes up about 73% of that, then the effective density of the dark energy would amount to just over 4 hydrogen atoms ($m = 1.67 \times 10^{-27}$ kg) in a cubic meter of space. If we take $5.9 \times 10^9$ m as a mean radius of Pluto and calculate the volume of a sphere of that radius, then the dark energy in that sphere would be equivalent to just under 6000 kg distributed throughout a space representing the solar system. The density of the asteroid Ida has been measured to be about 2.7 g/cm$^3$. So all the dark energy in the solar system would amount to only about 2.2 m$^3$ of the material of Ida, or a sphere of about 0.8 m radius. This is a hundred times smaller than Ida's tiny moon Dactyl! </p>
</blockquote>

<p>Of course, that assumes dark energy is essentially the same on small scales as it is large scales, which is unknown. Nevertheless I think it is pretty clear that it would be extremely difficult to observe dark energy directly within the Earth-Moon system.</p>

<p>See also Stan's comment.</p>

<p>Our understanding of dark energy is minimal at best, to boot, giving us no solid ideas on how to detect it in the first place. We infer its existence from the large scale expansion of the universe, but we're otherwise short on solid ideas on its nature.</p>

CC BY-SA 3.0
2014-11-14T22:54:31.180Z
+1, though it would be better to say that it would have tiny effects that for cosmological constant are time-independent. Amusingly, orbits in the Kottler spacetime (isolated spherically symmetric body with Λ) have an effective potential that's just the Schwarzschild potential plus $-\Lambda r^2/6$, which is **exactly** what one would get by putting in $-M_\Lambda/r$, where $M_\Lambda$ is the mass-equivalent of dark energy enclosed at that radius by the Euclidean formula. So by a fortuitous coincidence, the calculation is even more correct than one might expect, at least for spherical bodies.
Stan Liou
CC BY-SA 3.0
2014-11-14T23:04:11.780Z
@StanLiou The time independence is what I meant by "what's already there", but I suppose that was needlessly imprecise. I will adjust it.
zibadawa timmy
CC BY-SA 3.0
2014-11-14T19:42:05.833Z
2014-11-14T23:12:23.943Z
2017-04-13T12:59:06.910Z
zibadawa timmy
<p>What are the experiments, or measurements, that can detect, or account for, dark energy involved in making calculations concerning Earth and its only natural satellite? </p>

<dark-energy>
Is the dark energy between the moon and Earth measurable in any capacity?
Billy
# Is the dark energy between the moon and Earth measurable in any capacity?

(Asked by: username_0 on 2014-11-14)

What are the experiments, or measurements, that can detect, or account for, dark energy involved in making calculations concerning Earth and its only natural satellite?

**Tags:** dark-energy


---

## 1 answers

### Answer by username_1 on 2014-11-14. Score: 4

Yes and no.

First, the [answers here](https://astronomy.stackexchange.com/q/7939/2188) should give you some understanding of dark energy and its effects.

The net effect of dark energy is tiny over small distances. If it is given by a cosmological constant, then its effects on gravitationally bound systems is independent of time (hence 'constant'). Observational evidence puts us at least close to the constant case. Which means that even if we are not actually in the constant case, then it will be many billions of years until there would be a measurable difference.

We can, however, estimate the mass equivalent of dark energy in a given area by measuring the apparent amount in the universe itself.

As estimated [here](http://hyperphysics.phy-astr.gsu.edu/hbase/astro/dareng.html):

> If we take the WMAP value for critical density at $$\rho\_{c,0} = 9.47 \times 10^{-27}\, kg/m^3$$
> and presume that dark energy makes up about 73% of that, then the effective density of the dark energy would amount to just over 4 hydrogen atoms ($m = 1.67 \times 10^{-27}$ kg) in a cubic meter of space. If we take $5.9 \times 10^9$ m as a mean radius of Pluto and calculate the volume of a sphere of that radius, then the dark energy in that sphere would be equivalent to just under 6000 kg distributed throughout a space representing the solar system. The density of the asteroid Ida has been measured to be about 2.7 g/cm$^3$. So all the dark energy in the solar system would amount to only about 2.2 m$^3$ of the material of Ida, or a sphere of about 0.8 m radius. This is a hundred times smaller than Ida's tiny moon Dactyl!

Of course, that assumes dark energy is essentially the same on small scales as it is large scales, which is unknown. Nevertheless I think it is pretty clear that it would be extremely difficult to observe dark energy directly within the Earth-Moon system.

See also Stan's comment.

Our understanding of dark energy is minimal at best, to boot, giving us no solid ideas on how to detect it in the first place. We infer its existence from the large scale expansion of the universe, but we're otherwise short on solid ideas on its nature.

> **Comments:**
> **username_2**: +1, though it would be better to say that it would have tiny effects that for cosmological constant are time-independent. Amusingly, orbits in the Kottler spacetime (isolated spherically symmetric body with Λ) have an effective potential that's just the Schwarzschild potential plus $-\Lambda r^2/6$, which is **exactly** what one would get by putting in $-M_\Lambda/r$, where $M_\Lambda$ is the mass-equivalent of dark energy enclosed at that radius by the Euclidean formula. So by a fortuitous coincidence, the calculation is even more correct than one might expect, at least for spherical bodies.  
> **username_1**: @StanLiou The time independence is what I meant by "what's already there", but I suppose that was needlessly imprecise. I will adjust it.  

---

