<p>Not very strong at all. I get a rough figure of $3.725\times 10^{-9} m/s^2$. </p>

<p>To perform that calculation I made a few simplifying assumptions.</p>

<ul>
<li>Assume that we can ignore everything outside the <a href="https://en.wikipedia.org/wiki/Observable_universe" rel="noreferrer">observable
universe</a>.   </li>
<li>Assume that the observable universe is a static homogeneous sphere
with a radius of 46.6 billion light-years, and a mean density of
$\rho = 0.85\times 10^{-26} kg/m^3$, that figure is the <a href="https://en.wikipedia.org/wiki/Observable_universe#Estimates_based_on_critical_density" rel="noreferrer">critical
density</a>
needed for the global curvature of the universe to be zero (i.e., on
the very large scale spacetime is flat).   </li>
<li>Assume that we can use Newtonian physics to calculate the
acceleration due to gravity.</li>
</ul>

<p>I didn't bother removing all the stuff closer than 10 billion light years.</p>

<p>We're standing at the centre of the flat surface of a hemisphere of radius $r$ and we want to know how strong the gravity is. We can pretend that all the mass of the hemisphere is concentrated at its centre of mass, which is $\frac{3r}{8}$ below us.</p>

<p>The Newtonian formula for gravitational acceleration is </p>

<p>$$a = \frac{GM}{r^2}$$</p>

<p>Mass is volume times density, i.e., $M=V \rho$ and the volume of a hemisphere is</p>

<p>$$V = \frac{2}{3}\pi r^3$$</p>

<p>Combining everything, not forgetting to use $\frac{3r}{8}$ as the distance term in the denominator of the acceleration formula, we get</p>

<p>$$a = G \rho \frac{128}{27} \pi r$$</p>

<p>Plugging in the values I mentioned earlier for $\rho$ and $r$, the <a href="https://www.google.com/#q=G*0.85E-26(kg/m%5E3)*128/27*pi*(46.6E9+light-years)" rel="noreferrer">Google Calculator</a> says $3.72484086\times 10^{-9} m/s^2$. That's about $3.8\times 10^{-10}$ standard gravities.</p>

CC BY-SA 3.0
2017-04-25T13:53:28.803Z
2017-04-25T13:58:53.327Z
2017-04-25T13:58:53.327Z
PM 2Ring
<p>I would like to know, how strong would the gravity be from the distant universe if we were somehow able to remove half of it.  For example, if everything behind us did not exist, how strongly would the stuff in front of us be pulling?  By distant, let us consider everything beyond 10 billion light years (e.g. remove all the closer stuff).  How many G's acceleration would I experience (discounting the fact that I would be in free fall and wouldn't feel it)?</p>

CC BY-SA 3.0
2017-04-25T12:15:43.390Z
@mikey Excellent starting point, yes, that factor will certainly come into play.  Note that the universe used to be extremely dense, and the gravity from that should be reaching us from the region where the cosmic microwave background was emitted and beyond.  We are effectively seeing and probably experiencing gravity from the past universe.
Jonathan
CC BY-SA 3.0
2017-04-25T13:40:08.190Z
Is this question sufficiently different from [your other question](https://astronomy.stackexchange.com/questions/20825/how-strong-is-the-gravitational-stretch-we-experience-from-the-edge-of-the-unive)?
zephyr
CC BY-SA 3.0
2017-04-25T23:00:35.220Z
How abstract are we talking about this?  Do we assume a uniformly dense universe (at sufficiently large distance scales)?
Brian Tung
CC BY-SA 3.0
2017-04-25T23:49:02.427Z
@mikey Gravity falls off by the square but the mass of everything within 2 light years vs 1 light year increases by the cube if we assuming uniform density.    It wouldn't be negligible at all.  It would start out negligible for near-by it would grow.   Dark energy however is the wild card.  It's impossible to work out an answer with Dark Energy.  Ignoring the dark energy and expanding universe, it could be done I've not wanted to do the math cause I do math badly and I don't speak Jax very well, but I think, off the top of my head, the gravitation would be surprisingly high.
userLTK
CC BY-SA 3.0
2017-04-26T01:29:57.937Z
@zephyr Yes, I believe so.  This question arose from an interesting variation on the original question where I consider the whole universe vs half of it.  In the original question, the full shell has the force cancelling out, except maybe for stretching space.
Jonathan
CC BY-SA 3.0
2017-04-26T02:12:32.213Z
@userLTK Although I assumed that the observable universe isn't expanding, I kind of included dark energy in my calculation because I used the full critical density, so my mass estimate is that of the total energy content of the observable universe - normal matter and energy as well as dark matter and energy.
PM 2Ring
CC BY-SA 3.0
2017-04-26T03:22:33.157Z
@PM2Ring Well, you did the calculations and I didn't, so props for that.  Dark energy changes the entire thing.   Straight gravity, I think the pull in this theoretical is enormous, which is why I replied to mikey's post.   With dark energy, the math is over my pay-grade.   Props for working it out though.
userLTK
CC BY-SA 3.0
2017-04-26T21:37:47.537Z
@Brian Tung I would especially like to see the effect of the "dense edge" of the universe, where the cosmic microwave background comes from, and beyond.  Not 100% sure if the approximation using the average density of the universe gets us there or not, but it is a reasonable approach.
Jonathan
<gravity><dark-energy><cosmological-horizon>
How strong is the gravity from half of the distant universe?
Jonathan
# How strong is the gravity from half of the distant universe?

(Asked by: username_0 on 2017-04-24)

I would like to know, how strong would the gravity be from the distant universe if we were somehow able to remove half of it. For example, if everything behind us did not exist, how strongly would the stuff in front of us be pulling? By distant, let us consider everything beyond 10 billion light years (e.g. remove all the closer stuff). How many G's acceleration would I experience (discounting the fact that I would be in free fall and wouldn't feel it)?

**Tags:** gravity, dark-energy, cosmological-horizon

> **Comments:**
> **username_0**: @mikey Excellent starting point, yes, that factor will certainly come into play.  Note that the universe used to be extremely dense, and the gravity from that should be reaching us from the region where the cosmic microwave background was emitted and beyond.  We are effectively seeing and probably experiencing gravity from the past universe.  
> **username_1**: Is this question sufficiently different from [your other question](https://astronomy.stackexchange.com/questions/20825/how-strong-is-the-gravitational-stretch-we-experience-from-the-edge-of-the-unive)?  
> **username_2**: How abstract are we talking about this?  Do we assume a uniformly dense universe (at sufficiently large distance scales)?  
> **username_3**: @mikey Gravity falls off by the square but the mass of everything within 2 light years vs 1 light year increases by the cube if we assuming uniform density.    It wouldn't be negligible at all.  It would start out negligible for near-by it would grow.   Dark energy however is the wild card.  It's impossible to work out an answer with Dark Energy.  Ignoring the dark energy and expanding universe, it could be done I've not wanted to do the math cause I do math badly and I don't speak Jax very well, but I think, off the top of my head, the gravitation would be surprisingly high.  
> **username_0**: @username_1 Yes, I believe so.  This question arose from an interesting variation on the original question where I consider the whole universe vs half of it.  In the original question, the full shell has the force cancelling out, except maybe for stretching space.  
> **username_4**: @username_3 Although I assumed that the observable universe isn't expanding, I kind of included dark energy in my calculation because I used the full critical density, so my mass estimate is that of the total energy content of the observable universe - normal matter and energy as well as dark matter and energy.  
> **username_3**: @PM2Ring Well, you did the calculations and I didn't, so props for that.  Dark energy changes the entire thing.   Straight gravity, I think the pull in this theoretical is enormous, which is why I replied to mikey's post.   With dark energy, the math is over my pay-grade.   Props for working it out though.  
> **username_0**: @username_2 I would especially like to see the effect of the "dense edge" of the universe, where the cosmic microwave background comes from, and beyond.  Not 100% sure if the approximation using the average density of the universe gets us there or not, but it is a reasonable approach.  

---

## 1 answers

### Answer by username_4 on 2017-04-25. Score: 5 | Accepted answer

Not very strong at all. I get a rough figure of $3.725\times 10^{-9} m/s^2$.

To perform that calculation I made a few simplifying assumptions.

* Assume that we can ignore everything outside the [observable
  universe](https://en.wikipedia.org/wiki/Observable_universe).
* Assume that the observable universe is a static homogeneous sphere
  with a radius of 46.6 billion light-years, and a mean density of
  $\rho = 0.85\times 10^{-26} kg/m^3$, that figure is the [critical
  density](https://en.wikipedia.org/wiki/Observable_universe#Estimates_based_on_critical_density)
  needed for the global curvature of the universe to be zero (i.e., on
  the very large scale spacetime is flat).
* Assume that we can use Newtonian physics to calculate the
  acceleration due to gravity.

I didn't bother removing all the stuff closer than 10 billion light years.

We're standing at the centre of the flat surface of a hemisphere of radius $r$ and we want to know how strong the gravity is. We can pretend that all the mass of the hemisphere is concentrated at its centre of mass, which is $\frac{3r}{8}$ below us.

The Newtonian formula for gravitational acceleration is

$$a = \frac{GM}{r^2}$$

Mass is volume times density, i.e., $M=V \rho$ and the volume of a hemisphere is

$$V = \frac{2}{3}\pi r^3$$

Combining everything, not forgetting to use $\frac{3r}{8}$ as the distance term in the denominator of the acceleration formula, we get

$$a = G \rho \frac{128}{27} \pi r$$

Plugging in the values I mentioned earlier for $\rho$ and $r$, the [Google Calculator](https://www.google.com/#q=G*0.85E-26(kg/m%5E3)*128/27*pi*(46.6E9+light-years)) says $3.72484086\times 10^{-9} m/s^2$. That's about $3.8\times 10^{-10}$ standard gravities.


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