<p>As far as I know, the issue is not that Sgr A* itself moves around so much, but rather it is the gas around Sgr A* (that the event horizon telescope tries to image) that moves around. The typically timescale for variations in the gas is set by the period of the innermost stable circular orbit (ISCO) around  Sgr A*. This period is roughly proportion with the mass of the black hole. For Sgr A* this period is of order minutes. For M87*, it is order days.</p>

CC BY-SA 4.0
2022-04-18T15:40:03.440Z
2022-04-18T15:40:03.440Z
TimRias
<p>If a star of mass <span class="math-container">$m$</span> hits a black hole of mass <span class="math-container">$M$</span> at velocity <span class="math-container">$v$</span>, conservation of momentum means that the black hole now moves at velocity <span class="math-container">$$\Delta v=\frac{mv}{m+M}=\frac{1}{1+\frac{M}{m}}v.$$</span> The bigger <span class="math-container">$M$</span> is compared to <span class="math-container">$m$</span>, the less the change in velocity.</p>
<p>Sag A* is a fairly small black hole, so the random mergers have given it a bigger velocity than a heavier black hole like M87.</p>
<p>(This is not 100% as straightforward as the above argument: to make a mass <span class="math-container">$M$</span> black hole out of <span class="math-container">$N=M/m$</span> stars you have to add their random velocities, and a heavy black hole will have more added momentum than a small one. To properly calculate it one needs to add a sum of random velocity vectors rescaled by the mass accumulated to that date. However, <span class="math-container">$N$</span> random Gauss-distributed momentum vectors will produce a Gaussian vector with mean 0 and a variance <span class="math-container">$\sigma^2/N$</span>: even if they all contributed equally the spread of black hole momentum around 0 would tend towards zero. Dividing by the growing mass gives an even slower velocity distribution.)</p>

CC BY-SA 4.0
2022-04-19T13:58:38.307Z
As mmeent’s answer points out, the context of the original quote is clearly about the variability of emission from rapidly moving gas orbiting around the SMBH, not motion of the SMBH itself.
Peter Erwin
CC BY-SA 4.0
2022-04-19T14:03:47.280Z
Also, SMBH growth is understood to be primarily from gas accretion via accretion disks, not from collisions with individual stars.
Peter Erwin
CC BY-SA 4.0
2022-04-18T20:14:51.363Z
2022-04-18T20:14:51.363Z
Anders Sandberg
<p><a href="https://physics.stackexchange.com/a/472128/72564">This answer</a> mentions that the supermassive black hole at the center of our galaxy is constantly moving.</p>
<blockquote>
<p>Sagittarius A Star is 1000 times faster and smaller. Its like a toddler who is moving constantly. In comparison, M87 is much slower, like a big bear</p>
</blockquote>
<p>What causes Sgr A* to move?</p>

CC BY-SA 4.0
2022-04-18T15:10:04.470Z
Could be related to the fact that, since it’s smaller, the mass of the whole galaxy around it makes it move, like in a similar way the sun moves just a little bit from the orbits of all the bodies around it, but I’m not 100% sure
Justin T
<sagittarius-a>
Why is Sgr A* moving so much?
usernumber
# Why is Sgr A* moving so much?

(Asked by: username_0 on 2022-04-18)

[This answer](https://physics.stackexchange.com/a/472128/72564) mentions that the supermassive black hole at the center of our galaxy is constantly moving.

> Sagittarius A Star is 1000 times faster and smaller. Its like a toddler who is moving constantly. In comparison, M87 is much slower, like a big bear

What causes Sgr A\* to move?

**Tags:** sagittarius-a
> **Comments:**
> **username_1**: Could be related to the fact that, since it’s smaller, the mass of the whole galaxy around it makes it move, like in a similar way the sun moves just a little bit from the orbits of all the bodies around it, but I’m not 100% sure  

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## 2 answers

### Answer by username_2 on 2022-04-18. Score: 7

As far as I know, the issue is not that Sgr A\* itself moves around so much, but rather it is the gas around Sgr A\* (that the event horizon telescope tries to image) that moves around. The typically timescale for variations in the gas is set by the period of the innermost stable circular orbit (ISCO) around Sgr A\*. This period is roughly proportion with the mass of the black hole. For Sgr A\* this period is of order minutes. For M87\*, it is order days.


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### Answer by username_3 on 2022-04-18. Score: 4

If a star of mass $m$ hits a black hole of mass $M$ at velocity $v$, conservation of momentum means that the black hole now moves at velocity $$\Delta v=\frac{mv}{m+M}=\frac{1}{1+\frac{M}{m}}v.$$ The bigger $M$ is compared to $m$, the less the change in velocity.

Sag A\* is a fairly small black hole, so the random mergers have given it a bigger velocity than a heavier black hole like M87.

(This is not 100% as straightforward as the above argument: to make a mass $M$ black hole out of $N=M/m$ stars you have to add their random velocities, and a heavy black hole will have more added momentum than a small one. To properly calculate it one needs to add a sum of random velocity vectors rescaled by the mass accumulated to that date. However, $N$ random Gauss-distributed momentum vectors will produce a Gaussian vector with mean 0 and a variance $\sigma^2/N$: even if they all contributed equally the spread of black hole momentum around 0 would tend towards zero. Dividing by the growing mass gives an even slower velocity distribution.)

> **Comments:**
> **username_4**: As mmeent’s answer points out, the context of the original quote is clearly about the variability of emission from rapidly moving gas orbiting around the SMBH, not motion of the SMBH itself.  
> **username_4**: Also, SMBH growth is understood to be primarily from gas accretion via accretion disks, not from collisions with individual stars.  

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