# Conditions at the 10AU border of an Active Galactic Nucleus

Imagine that a ordinary human being was suddenly teleported to a distance 10AU from the center of the Milky Way. What would happen to them (in terms of things like pressure, temperature, radiation, stellar activity, etc.)?

I'm asking because I want to design a space station for my universe at that distance and need help understanding what the universe is like that close to the nucleus before I do.

• Hi @Saucy, welcome.. could you focus on the proximity-question please ? there are two different questions now, a "level of technical knowledge" is relative.. and difficult to find an answer for. Nov 1, 2021 at 18:47
• I reworded the main question in the second paragraph. Is this along the lines of what you were looking for? Edit: Upon a second read, I realize it is still too vague Nov 1, 2021 at 19:14
• Ok, I'll split this question up into two. Thanks! Nov 1, 2021 at 19:28
• Keep in mind that Milky Way's Sagittarius A* (as presently observed) is NOT an active galactic nucleus. So is this question about Sagittarius A* or some other, more active AGN? Nov 1, 2021 at 20:30
• @Alexander The question is asking about the Milky Way's nucleus (think, "center of the galaxy"). Saucy, it's simple and to the point. I've retracted my vote and upvoted. I've edited the question to add the reason you're asking as that's a requirement when asking Real World questions (which this now is), but you'll notice that it's not changing the question at all. With a little luck one of our experts in celestial mechanics and astrophysics will be online soon and give you a lot of details.
– JBH
Nov 1, 2021 at 23:49

While Sgr A* is not currently an active galactic nucleus, it is believed to have an accretion disk accompanied by a larger reservoir of gas surrounding it, with the latter extending well beyond a radius of 10 AU. We know from Chandra x-ray data that there is $$\sim10^7$$ Kelvin gas extending to $$\sim10^5R_S$$ (or $$\sim8000$$ AU, with $$R_S$$ the Schwarzschild radius$$^{\dagger}$$). More recent observations with the Atacama Large Millimeter Array (Murchikova et al. 2017 have found a cooler disk of $$\sim10^4$$ Kelvin gas extending to $$2\times10^5R_S$$ - still easily encompassing your poor human.

The ALMA measurements yield a number density of $$n\approx10^{6}$$ cm$$^{-3}$$, or a mass density, assuming a primarily hydrogen disk, of $$\rho\sim10^{-18}$$ kg cm$$^{-3}$$. If we were to naively apply the ideal gas law, this results in a pressure of $$P\approx10^{-7}$$ Pascals. In short, you've got gas that is - by human standards - extremely hot and extremely diffuse. Given that 10 AU is close to the very center of this cool disk, I would expect the temperature to actually be significantly hotter - so perhaps closer to that $$\sim10^7$$ K gas from the x-ray observations.

This is maybe 5 orders of magnitude denser than the interplanetary medium and 3-4 orders of magnitude hotter (depending on how hot you think this part of the disk will be). The x-ray emission, of course, would likely be catastrophic - that's what going to kill you and whatever spacecraft you're on, rather than the ambient gas.

Some general relativistic effects would likely be measurable but not easily noticeable. Things like time dilation and redshift scale with distance $$r$$ by a multiplicative factor of $$\sqrt{1-\frac{R_S}{r}}\approx1-\frac{1}{2}\frac{R_S}{r}$$ and the second term comes out to about 0.4% at 10 AU - certainly not catastrophic, but not invisible, so to speak.

$$^{\dagger}$$The Schwarzschild radius is $$R_S=\frac{2GM}{c^2}$$ and comes out to $$1.18\times10^{10}$$ meters assuming a black hole mass of $$M\sim4\times10^6M_{\odot}$$.

Gravity is proportional to the square of distance.

At 10 AU, gravity will be 100 times smaller than normal gravity from the sun.

Studies have estimated the super-heavy black hole in the milky way center has a mass of 3.7 million or 4.1 million solar masses. As a result, you'll get 4e6/100 = 40,000 times Solar gravity at the point where you teleport your person.

https://en.wikipedia.org/wiki/Galactic_Center

On Earth, at the distance we orbit the sun, the gravitational pull of the sun is only 0.0006 of the strength of the earth's gravity on the surface of the earth. So when teleported at 10 AU from the galaxy center we'd experience 0.0006 x 40,000 is a force of 24g. In free fall, you won't notice..

https://van.physics.illinois.edu/qa/listing.php?id=184

Temperature and radiation may be a much bigger issue.. near a black hole, you'll have 10 million degrees.

However, at 10Au you'll be far away from the Schwartzshield radius of the black hole:

"Located 26,000 light-years from the Sun, our galaxy’s central black hole, Sagittarius A(star), has a radius about 17 times that of the Sun, meaning that it would sit well within Mercury’s orbit."

https://astronomy.com/news/2020/04/how-much-space-does-a-black-hole-take-up

.. a distance of 10AU, far away from the Schwartzshield radius, will reduce the heat considerably.

Other potential issues, e.g. the Hawking radiation emitted,

This Hawking radiation is very low energy

An evaporating black hole would be detectable from Earth only if it went off within the solar system, or at best no further away than the nearest star.