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# escape velocity of leaving the Milky Way galaxy vertically and horizontally

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)?
• Yes. Exactly. I have seen the terminology that I have used on a website online. Jun 21 at 14:34
• @user438383 What's the "ecliptic plane of the Milky Way"? Do you mean the galactic plane of the Milky Way? Jun 22 at 4:19
• 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. Jun 23 at 1:36
• @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. Jun 23 at 7:43
• 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. Jun 23 at 7:57

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).

• +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
Jun 21 at 23:04

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.

• 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. Jun 23 at 1:41
• @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. Jun 23 at 7:38

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 in which an object could be sped up or slowed down by interactions with nearby relatively massive bodies. This can lead to gravitational 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.

• 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. Jun 23 at 7:36
• Of course. The escape velocity of the Earth varies in this case too. Jun 23 at 10:59


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Geometry: Common Core (15th Edition)

$\overline{BC}$
Let's take a look first at the angles in this triangle. We have an exterior angle, which will be supplementary to the interior angle $\angle A$; therefore, $\angle A$ of the triangle can be calculated by the following equation: $m \angle A = 180 - 85$ Subtract to find the measure of $\angle A$: $m \angle A = 95$ We know that $\angle A$ is the largest angle because the sum of the measures of the other two angles has to be $180^{\circ} - 95^{\circ}$ or $85^{\circ}$. $\overline{BC}$ would be the longest because it is opposite the largest angle in a triangle where two of the angles are not congruent.

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## anonymous one year ago Consider the following problem. A bicyclist is riding on a path modeled by the function f(x) = 0.06x, where x and f(x) are measured in miles (see figure). Find the rate of change of elevation at x = 2.

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1. anonymous

we know the derivative is the rate of change for the function over a certain time, and because this is a linear function, the derivative is constant throughout. Therefore $\frac{d(f)}{d(x)} = = \frac{d}{dx}(0.06x) = 0.06$

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A differentiable version of the Michael selection theorem

Assume that $X$ and $Y$ are Banach spaces and $T:X\to Y$ is a bounded surjective linear map. Is there a Gateaux differentiable function $g:Y\to X$ such that $T\circ g=Id_{Y}$?

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No. Consider the case of a surjective bounded linear operator $T:X\to Y$ which is not a (top-linear) left inverse (that is, $\operatorname{ker}(T)$ does not split in $X$). However by classical selection theorems a surjective bounded linear operator $T$ has a continuous right inverse $g$, even $1$-homogeneous; but it can be G-differentiable at no point, otherwise differentiating you would get a bounded linear right inverse to $T$.
Rmk. I refer to the standard definition of the Gâteaux differential of $f:X \to Y$ at $x$, that is, a bounded linear operator $L$ such that for all $v\in X$ there holds $\frac{d}{dt}f(x+tv)\Big|_{t=0}=Lv$. In particular the chain rule holds. (Warning: some adopt a weaker definition, where $L$ is not even linear.)
I can understand from your answer if $T$ is not a left inverse(there is no bounded operator $S:Y\to X$ with TS=id) then there is no a Frechete differentiable map $g$ with $T\circ g=Id$. (using chain rulle) But I do not understand why your argument work for Gateau differentiability? –  Ali Taghavi Mar 27 at 0:31
In the literuture is there a property "P", stronger than continuity, such that every right inverse "g" to a bounded surjective operator "T" on a banach space $X$, satisfies "P"? Is it true to say that a banach space $X$ is Hilbertable if every bounded surjective map on $X$ has a Frechete diferentiable righth inverse? –  Ali Taghavi Mar 27 at 20:59