<p>The escape speed is found by summing together the kinetic energy and the gravitational potential energy and equating that to zero. i.e. Making the target total energy be zero when it escapes to infinity.
<span class="math-container">$$ \frac{1}{2} m v^2 + m\Phi(x,y,z) = 0\ ,$$</span>
where <span class="math-container">$\Phi$</span> is the gravitational potential, which is <em>negative</em> at any point within the Galaxy and is zero at infinity.</p>
<p>Since <span class="math-container">$\Phi$</span> is simply a function of position, then the escape velocity in any direction (starting from the same position) is the same.
<span class="math-container">$$ v = \sqrt{2|\Phi|}$$</span></p>
<p>NB: Assuming the object travels inertially (unpowered and unimpeded).</p>

CC BY-SA 4.0
2022-06-21T23:04:31.927Z
`+1` And for those like me who feel that this correct answer is counterintuitive, it is because the higher order gravitational potential multipoles decay faster than the monopole term at large distances. The $v(t)$ curves would be different and the arrival times at some very distant but finite radius would always be different, but they would "escape to infinity" similarly because it's just a question of $\phi_{\text{final}} - \phi_{\text{initial}}$.
uhoh
CC BY-SA 4.0
2022-06-21T22:05:08.460Z
2022-06-22T07:19:55.453Z
2022-06-22T07:19:55.453Z
ProfRob
<p>There is no difference. The escape velocity is the velocity is the velocity to arrive at infinity with velocity zero. The equal potential lines will tend to spherical symmetrical surfaces as you approach infinity because there the Milky-Way looks like a point. So in whatever direction you go, the initial kinetic energy needs to be the same. Which means that the escape velocity is the same in every direction, thus including the horizontal plane and the vertical polar direction.</p>

CC BY-SA 4.0
2022-06-23T01:41:26.047Z
I don't think this is correct because the shape of the mass plays a considerable factor.   For example, you have to travel much less far vertically to have the vast majority of the mass below you and pulling you back than if you travel horizontally.   Once outside the Milkyway, the escape velocity becomes much more uniform, but when one distance leads you outside faster than the other, that's a measurable difference.   That said, I really don't want to run the math.
userLTK
CC BY-SA 4.0
2022-06-23T07:38:01.793Z
@userLTK At infinity the equipotential surface is spherical. The Milkyway looks like a point there. So it doesn't make a difference inn which direction you leave. From each direction the potential energy will be zero at the spherical surface. So the initial velocity is the same in each direction, from whatever point you leave (same initial potential) . Counterintuitive as it might seem.
Felicia
CC BY-SA 4.0
2022-06-22T17:22:08.947Z
2022-06-22T17:22:08.947Z
Felicia
<p>At the risk of being downvoted into oblivion, I am going to &quot;ever-so-slightly&quot; disagree with other esteemed posters.</p>
<p>The gravitational potential <span class="math-container">$\Phi$</span> at a particular point in the galaxy is not constant over time, because the stars in the galaxy are moving. This allows for the potential of a natural <a href="https://en.wikipedia.org/wiki/Gravity_assist" rel="nofollow noreferrer">gravity assist</a> in which an object could be sped up or slowed down by interactions with nearby relatively massive bodies. This can lead to <a href="https://en.wikipedia.org/wiki/Asteroid_capture" rel="nofollow noreferrer">gravitational capture</a>.</p>
<p>One could make a simulation in which only altering the initial direction (and not the speed) could cause an object to become gravitationally bound when it otherwise would have escaped.  An object is more likely to  have close passes with other objects if its course is near the galactic plane rather than perpendicular to it (since there is a higher density of objects close to the galactic plane).</p>
<p>Suppose we performed a monte carlo experiment in which we</p>
<ol>
<li>selected random points near the Milky Way's galactic plane.</li>
<li>calculated an escape velocity magnitude <span class="math-container">$|v_e|$</span> for each point based on the gravitational potential at an initial time.</li>
<li>simulated a massless trajectory based on an initial velocity magnitude <span class="math-container">$|v_e|$</span>, and direction (I) nearly parallel to the galactic plane, and (II) nearly perpendicular to the galactic plane.</li>
<li>deleted all the points with trajectories that grazed or collided with other objects.</li>
</ol>
<p>For a sufficiently large sample size, we would see that fewer trajectories parallel to the galactic plane would actually escape the Milky Way's gravitational influence.</p>

CC BY-SA 4.0
2022-06-23T07:36:30.933Z
If you start applying time dependent, non-conservative forces to an object then the concept of an escape velocity is meaningless. However, besides that, I believe the chances of any kind of close encounter with a suitable object is negligible unless it were aimed in some way or perhaps fired at the Galactic centre.
ProfRob
CC BY-SA 4.0
2022-06-23T10:59:58.843Z
Of course. The escape velocity of the Earth varies in this case too.
Felicia
CC BY-SA 4.0
2022-06-22T21:30:22.417Z
2022-06-22T21:30:22.417Z
Connor Garcia
<p>Is there a difference in the escape velocity when leaving the Milky Way galaxy</p>
<ol>
<li>vertically (out of galactic plane)</li>
<li>horizontally (in the galactic plane)?</li>
</ol>

CC BY-SA 4.0
2022-06-21T14:34:38.830Z
Yes. Exactly. I have seen the terminology that I have used on a website online.
Jane N.
CC BY-SA 4.0
2022-06-22T04:19:45.827Z
@user438383 What's the "ecliptic plane of the Milky Way"? Do you mean the [galactic plane](https://en.wikipedia.org/wiki/Galactic_plane) of the Milky Way?
PM 2Ring
CC BY-SA 4.0
2022-06-23T01:36:29.137Z
Ignoring orbital speed or with orbital speed?       Most of the objects in the Milky Way or orbiting along the plane at pretty high speed.   I thought I'd throw that out there.     Escape Velocity from the Sun at 1 AU is 44.5 km, but if you launch a ship from Earth at 14.9 km, after you've escaped Earths' gravity, that's enough to escape the solar system.     Interesting question.
userLTK
CC BY-SA 4.0
2022-06-23T07:43:46.717Z
@userLTK that's like saying what is the escape velocity if I start with some velocity. If you want to change your frame of reference (I presume you chose a circular orbiting frame of reference) the escape velocity can be any value you choose.
ProfRob
CC BY-SA 4.0
2022-06-23T07:57:47.013Z
I know, but if you were to launch from our solar system, that velocity would already be there.   You could look at it as adding to the frame of reference or using our frame of reference.   I thought it was a distinction worth pointing out.
userLTK
<escape-velocity>
escape velocity of leaving the Milky Way galaxy vertically and horizontally
Jane N.
# escape velocity of leaving the Milky Way galaxy vertically and horizontally

(Asked by: username_0 on 2022-06-21)

Is there a difference in the escape velocity when leaving the Milky Way galaxy

1. vertically (out of galactic plane)
2. horizontally (in the galactic plane)?

**Tags:** escape-velocity
> **Comments:**
> **username_0**: Yes. Exactly. I have seen the terminology that I have used on a website online.  
> **username_1**: @user438383 What's the "ecliptic plane of the Milky Way"? Do you mean the [galactic plane](https://en.wikipedia.org/wiki/Galactic_plane) of the Milky Way?  
> **username_2**: Ignoring orbital speed or with orbital speed?       Most of the objects in the Milky Way or orbiting along the plane at pretty high speed.   I thought I'd throw that out there.     Escape Velocity from the Sun at 1 AU is 44.5 km, but if you launch a ship from Earth at 14.9 km, after you've escaped Earths' gravity, that's enough to escape the solar system.     Interesting question.  
> **username_3**: @userLTK that's like saying what is the escape velocity if I start with some velocity. If you want to change your frame of reference (I presume you chose a circular orbiting frame of reference) the escape velocity can be any value you choose.  
> **username_2**: I know, but if you were to launch from our solar system, that velocity would already be there.   You could look at it as adding to the frame of reference or using our frame of reference.   I thought it was a distinction worth pointing out.  

---

## 3 answers

### Answer by username_3 on 2022-06-21. Score: 11

The escape speed is found by summing together the kinetic energy and the gravitational potential energy and equating that to zero. i.e. Making the target total energy be zero when it escapes to infinity.
$$ \frac{1}{2} m v^2 + m\Phi(x,y,z) = 0\ ,$$
where $\Phi$ is the gravitational potential, which is *negative* at any point within the Galaxy and is zero at infinity.

Since $\Phi$ is simply a function of position, then the escape velocity in any direction (starting from the same position) is the same.
$$ v = \sqrt{2|\Phi|}$$

NB: Assuming the object travels inertially (unpowered and unimpeded).

> **Comments:**
> **username_4**: `+1` And for those like me who feel that this correct answer is counterintuitive, it is because the higher order gravitational potential multipoles decay faster than the monopole term at large distances. The $v(t)$ curves would be different and the arrival times at some very distant but finite radius would always be different, but they would "escape to infinity" similarly because it's just a question of $\phi_{\text{final}} - \phi_{\text{initial}}$.  

---

### Answer by username_5 on 2022-06-22. Score: 3

There is no difference. The escape velocity is the velocity is the velocity to arrive at infinity with velocity zero. The equal potential lines will tend to spherical symmetrical surfaces as you approach infinity because there the Milky-Way looks like a point. So in whatever direction you go, the initial kinetic energy needs to be the same. Which means that the escape velocity is the same in every direction, thus including the horizontal plane and the vertical polar direction.

> **Comments:**
> **username_2**: I don't think this is correct because the shape of the mass plays a considerable factor.   For example, you have to travel much less far vertically to have the vast majority of the mass below you and pulling you back than if you travel horizontally.   Once outside the Milkyway, the escape velocity becomes much more uniform, but when one distance leads you outside faster than the other, that's a measurable difference.   That said, I really don't want to run the math.  
> **username_5**: @userLTK At infinity the equipotential surface is spherical. The Milkyway looks like a point there. So it doesn't make a difference inn which direction you leave. From each direction the potential energy will be zero at the spherical surface. So the initial velocity is the same in each direction, from whatever point you leave (same initial potential) . Counterintuitive as it might seem.  

---

### Answer by username_6 on 2022-06-22. Score: 1

At the risk of being downvoted into oblivion, I am going to "ever-so-slightly" disagree with other esteemed posters.

The gravitational potential $\Phi$ at a particular point in the galaxy is not constant over time, because the stars in the galaxy are moving. This allows for the potential of a natural [gravity assist](https://en.wikipedia.org/wiki/Gravity_assist) in which an object could be sped up or slowed down by interactions with nearby relatively massive bodies. This can lead to [gravitational capture](https://en.wikipedia.org/wiki/Asteroid_capture).

One could make a simulation in which only altering the initial direction (and not the speed) could cause an object to become gravitationally bound when it otherwise would have escaped. An object is more likely to have close passes with other objects if its course is near the galactic plane rather than perpendicular to it (since there is a higher density of objects close to the galactic plane).

Suppose we performed a monte carlo experiment in which we

1. selected random points near the Milky Way's galactic plane.
2. calculated an escape velocity magnitude $|v\_e|$ for each point based on the gravitational potential at an initial time.
3. simulated a massless trajectory based on an initial velocity magnitude $|v\_e|$, and direction (I) nearly parallel to the galactic plane, and (II) nearly perpendicular to the galactic plane.
4. deleted all the points with trajectories that grazed or collided with other objects.

For a sufficiently large sample size, we would see that fewer trajectories parallel to the galactic plane would actually escape the Milky Way's gravitational influence.

> **Comments:**
> **username_3**: If you start applying time dependent, non-conservative forces to an object then the concept of an escape velocity is meaningless. However, besides that, I believe the chances of any kind of close encounter with a suitable object is negligible unless it were aimed in some way or perhaps fired at the Galactic centre.  
> **username_5**: Of course. The escape velocity of the Earth varies in this case too.  

---

