<p>It's very similar to <a href="https://astronomy.stackexchange.com/questions/33936/hills-mechanism-for-making-hypervelocity-stars">the question noted in the comments</a> which has a nice mathematical answer.</p>

<p>But some comments in your question are incorrect and worth addressing.</p>

<blockquote>
  <p>As far as I can see, the escaping star can at most end up with the
  kinetic energy of the other star on top of its own, accelerating it by
  a factor of 2–√ if the stars have the same mass.</p>
</blockquote>

<p>The combined kinetic energy remains unchanged, unless there's a collision or some kind of relativistic interaction, but that's not really important.    It's easier to think of a gravitational slingshot interaction between a larger object (like Jupiter) and a smaller object like a Voyager spacecraft.   The kinetic energy and momentum exchanged is equal between the two, but the acceleration to the larger object is negligible but the acceleration to the smaller object can be up to 2 times the velocity of the larger one plus the initial velocity, or 2U + V, which is more than enough to escape both the central black hole and the entire galaxy when the initial velocities are high enough. </p>

<p><a href="https://space.stackexchange.com/questions/23295/how-do-we-come-up-with-the-gravity-assist-or-slingshot-formula">See this question for more math</a>.</p>

<p>And you wrote this:</p>

<blockquote>
  <p>A "slingshot effect" is impossible because the central black hole is
  stationary relative to the galaxy.</p>
</blockquote>

<p>This isn't relevant, because it's not the central object's velocity that matters but the velocity of the stars orbiting the central black hole, which can be very high.   When two stars orbiting the central black hole pass close to one another, then a gravity assist happens and the two stars exchange velocity and momentum.</p>

<p>This can be a hard thing to picture.   It's often described as bouncing a ball off a moving train, the train's velocity (times 2) can be added to the ball's initial velocity.   The problem with that image is that there's a point of impact where velocity is reflected and that moment never happens in an orbit.    The way I like to think about it, is, imagine you're standing on a station and a train is approaching and the train is carrying a very heavy object - say a neutron star, and you throw a ball towards, but just behind the neutron star.</p>

<p>You throw the ball at 100 kph and the train is moving towards the station at 100 kph.   The neutron star on the train sees the ball approaching it at 200 kph, it then swings around the Neutron star's gravity and flys away from the Neutron star at 200 kph, because orbital speeds are conserved in the hyperbolic orbit.</p>

<p>But if the ball is now flying away from the train at 200 kph and the train is moving towards you at 100 kph, the ball, inevitably is traveling in your direction at 300 kph, or 2U + V.</p>

<blockquote>
  <p>Does that mean that this effect necessarily involves binary systems
  with very different masses so that more energy is available for
  transfer to the lighter star? Or are there other effects at play?</p>
</blockquote>

<p>This is an interesting question and the math is a bit messy so maybe somebody else could do it.   Two stars of equal mass and opposite directions could accelerate each other and one would fly away from the black hole and the other would fly towards the black hole.    I think it's easier with a larger and smaller object where the smaller objects is accelerated and the larger one only moved a tiny bit, but I see no reason why two equal mass objects couldn't exchange significant velocity to each other, changing both trajectories.   That's a good question whether that would be sufficient to achieve escape velocity for one of them.</p>

CC BY-SA 4.0
2019-11-19T07:31:54.177Z
(1) My "factor of sqrt(2)" argument was for binary stars with equal sizes; your Jupiter/Voyager reply seems to indicate that unequal masses result in higher speeds. I'm actually unsure whether the sqrt(2) argument holds; I'm still mulling over the Oberth effect. Or maybe tidal forces inject gravitational energy, like swinging one end of a fast-orbiting bola (= accelerating it in the inhomogeneous gravity) to accelerate the other end and then cutting the cord. (2) I meant no slingshot effect *from the stationary black hole* (which is why you need a binary star, or a binary black hole).
Peter - Reinstate Monica
CC BY-SA 4.0
2019-11-19T13:46:49.397Z
@Peter-ReinstateMonica my mistake.   I think it's a good question and I'm curious about the answer too.  You might want to say binary system of roughly equal mass for clarity.   I can delete my answer if you think it'll take away from an answer you're looking for.
userLTK
CC BY-SA 4.0
2019-11-19T14:12:08.430Z
Please let it stand, it is valuable. I just wanted to clarify my thoughts. Yes, I should clarify the post as well ;-). (Revisiting my post: I did say " a factor of √2 if the stars have the same mass", so I think that's ok.)
Peter - Reinstate Monica
CC BY-SA 4.0
2019-11-18T23:30:29.793Z
2019-11-18T23:30:29.793Z
userLTK
<p>This is basically just the <a href="https://en.wikipedia.org/wiki/Oberth_effect" rel="nofollow noreferrer">Oberth effect</a>. The momentum transfer between the stars is limited by their mutual orbital velocity, more or less, but that transfer is taking place deep in the gravity well of the black hole.</p>
<p>If we consider it from the perspective of the black hole (or an observer at rest with respect to it) the binary is moving at a high speed <span class="math-container">$V$</span> due to its orbit around the black hole. That velocity came from the potential energy lost as it fell in, so <span class="math-container">$PE = 1/2 mV^2$</span>. After they are separated, one may be moving at <span class="math-container">$V+v$</span> (where <span class="math-container">$v$</span> is related to the mutual orbital velocity), so KE <span class="math-container">$1/2 m (V+v)^2$</span>.</p>
<p>As it moves away from the black hole it will lose convert <span class="math-container">$1/2 mV^2$</span> of this back to PE, leaving a net energy of roughly <span class="math-container">$mVv$</span>, which remains as KE, so the star is moving at about <span class="math-container">$\sqrt{2Vv}$</span> which can obviously be a lot bigger than <span class="math-container">$V$</span>.</p>

CC BY-SA 4.0
2021-02-17T12:49:15.107Z
2021-02-17T12:49:15.107Z
Steve Linton
<p>A <a href="https://arxiv.org/pdf/1907.11725.pdf" rel="nofollow noreferrer">recent publication</a> reports that a hypervelocity star on course to leave the galaxy was found, likely accelerated by the black hole at the center of the Milky Way in a process <a href="https://www.nature.com/articles/331687a0" rel="nofollow noreferrer">predicted by Jack G. Hills in 1988</a>. A sentence from that abstract insightfully points out:</p>

<blockquote>
  <p>The discovery of even one such hyper-velocity star coming from the Galactic centre would be nearly definitive evidence for a massive black hole. </p>
</blockquote>

<p>The process involves a binary system coming close enough to the black hole that one of the stars is "captured" and the other one is flung out. </p>

<p>As far as I can see, the escaping star can at most end up with the kinetic energy of the other star on top of its own, accelerating it by a factor of <span class="math-container">$\sqrt{2}$</span> if the stars have the same mass. A "slingshot effect"  is impossible because the central black hole is stationary relative to the galaxy.</p>

<p>Does that mean that this effect necessarily involves binary systems with very different masses so that more energy is available for transfer to the lighter star? Or are there other effects at play?</p>

<sagittarius-a><runaway-star>
How can a star be "thrown out of the Milky Way" by Sagittarius A*?
Peter - Reinstate Monica
# How can a star be "thrown out of the Milky Way" by Sagittarius A*?

(Asked by: username_0 on 2019-11-17)

A [recent publication](https://arxiv.org/pdf/1907.11725.pdf) reports that a hypervelocity star on course to leave the galaxy was found, likely accelerated by the black hole at the center of the Milky Way in a process [predicted by Jack G. Hills in 1988](https://www.nature.com/articles/331687a0). A sentence from that abstract insightfully points out:

> The discovery of even one such hyper-velocity star coming from the Galactic centre would be nearly definitive evidence for a massive black hole.

The process involves a binary system coming close enough to the black hole that one of the stars is "captured" and the other one is flung out.

As far as I can see, the escaping star can at most end up with the kinetic energy of the other star on top of its own, accelerating it by a factor of $\sqrt{2}$ if the stars have the same mass. A "slingshot effect" is impossible because the central black hole is stationary relative to the galaxy.

Does that mean that this effect necessarily involves binary systems with very different masses so that more energy is available for transfer to the lighter star? Or are there other effects at play?

**Tags:** sagittarius-a, runaway-star


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## 2 answers

### Answer by username_1 on 2019-11-18. Score: 2

It's very similar to [the question noted in the comments](https://astronomy.stackexchange.com/questions/33936/hills-mechanism-for-making-hypervelocity-stars) which has a nice mathematical answer.

But some comments in your question are incorrect and worth addressing.

> As far as I can see, the escaping star can at most end up with the
> kinetic energy of the other star on top of its own, accelerating it by
> a factor of 2–√ if the stars have the same mass.

The combined kinetic energy remains unchanged, unless there's a collision or some kind of relativistic interaction, but that's not really important. It's easier to think of a gravitational slingshot interaction between a larger object (like Jupiter) and a smaller object like a Voyager spacecraft. The kinetic energy and momentum exchanged is equal between the two, but the acceleration to the larger object is negligible but the acceleration to the smaller object can be up to 2 times the velocity of the larger one plus the initial velocity, or 2U + V, which is more than enough to escape both the central black hole and the entire galaxy when the initial velocities are high enough.

[See this question for more math](https://space.stackexchange.com/questions/23295/how-do-we-come-up-with-the-gravity-assist-or-slingshot-formula).

And you wrote this:

> A "slingshot effect" is impossible because the central black hole is
> stationary relative to the galaxy.

This isn't relevant, because it's not the central object's velocity that matters but the velocity of the stars orbiting the central black hole, which can be very high. When two stars orbiting the central black hole pass close to one another, then a gravity assist happens and the two stars exchange velocity and momentum.

This can be a hard thing to picture. It's often described as bouncing a ball off a moving train, the train's velocity (times 2) can be added to the ball's initial velocity. The problem with that image is that there's a point of impact where velocity is reflected and that moment never happens in an orbit. The way I like to think about it, is, imagine you're standing on a station and a train is approaching and the train is carrying a very heavy object - say a neutron star, and you throw a ball towards, but just behind the neutron star.

You throw the ball at 100 kph and the train is moving towards the station at 100 kph. The neutron star on the train sees the ball approaching it at 200 kph, it then swings around the Neutron star's gravity and flys away from the Neutron star at 200 kph, because orbital speeds are conserved in the hyperbolic orbit.

But if the ball is now flying away from the train at 200 kph and the train is moving towards you at 100 kph, the ball, inevitably is traveling in your direction at 300 kph, or 2U + V.

> Does that mean that this effect necessarily involves binary systems
> with very different masses so that more energy is available for
> transfer to the lighter star? Or are there other effects at play?

This is an interesting question and the math is a bit messy so maybe somebody else could do it. Two stars of equal mass and opposite directions could accelerate each other and one would fly away from the black hole and the other would fly towards the black hole. I think it's easier with a larger and smaller object where the smaller objects is accelerated and the larger one only moved a tiny bit, but I see no reason why two equal mass objects couldn't exchange significant velocity to each other, changing both trajectories. That's a good question whether that would be sufficient to achieve escape velocity for one of them.

> **Comments:**
> **username_0**: (1) My "factor of sqrt(2)" argument was for binary stars with equal sizes; your Jupiter/Voyager reply seems to indicate that unequal masses result in higher speeds. I'm actually unsure whether the sqrt(2) argument holds; I'm still mulling over the Oberth effect. Or maybe tidal forces inject gravitational energy, like swinging one end of a fast-orbiting bola (= accelerating it in the inhomogeneous gravity) to accelerate the other end and then cutting the cord. (2) I meant no slingshot effect *from the stationary black hole* (which is why you need a binary star, or a binary black hole).  
> **username_1**: @username_0-ReinstateMonica my mistake.   I think it's a good question and I'm curious about the answer too.  You might want to say binary system of roughly equal mass for clarity.   I can delete my answer if you think it'll take away from an answer you're looking for.  
> **username_0**: Please let it stand, it is valuable. I just wanted to clarify my thoughts. Yes, I should clarify the post as well ;-). (Revisiting my post: I did say " a factor of √2 if the stars have the same mass", so I think that's ok.)  

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### Answer by username_2 on 2021-02-17. Score: 0

This is basically just the [Oberth effect](https://en.wikipedia.org/wiki/Oberth_effect). The momentum transfer between the stars is limited by their mutual orbital velocity, more or less, but that transfer is taking place deep in the gravity well of the black hole.

If we consider it from the perspective of the black hole (or an observer at rest with respect to it) the binary is moving at a high speed $V$ due to its orbit around the black hole. That velocity came from the potential energy lost as it fell in, so $PE = 1/2 mV^2$. After they are separated, one may be moving at $V+v$ (where $v$ is related to the mutual orbital velocity), so KE $1/2 m (V+v)^2$.

As it moves away from the black hole it will lose convert $1/2 mV^2$ of this back to PE, leaving a net energy of roughly $mVv$, which remains as KE, so the star is moving at about $\sqrt{2Vv}$ which can obviously be a lot bigger than $V$.


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