<p>The motion of the Sun in the direction perpendicular to the Galactic plane is deduced from a mixture of measurement and theoretical calculation.</p>
<p>The measurements are (i) deducing the Solar System's current position and speed with respect to the galactic plane (see <a href="https://astronomy.stackexchange.com/questions/822/how-far-is-the-earth-sun-above-below-the-galactic-plane-and-is-it-heading-towar">How far is the Earth/Sun above/below the galactic plane, and is it heading toward/away from it?</a> ) and (ii) deducing how much mass is present in the galactic plane which acts to accelerate/decelerate the motion of the Solar System.</p>
<p>From there it is a relatively simple application of Newtonian mechanics to work out what the motion of the Sun will be in the future (and where it came from in the past). The solution is an oscillatory motion above and below the plane with an amplitude of about 100 pc and a period of 70 million years.</p>
<p>A flavour of the calculation can be done analytically using some simple approximations.</p>
<p>Very roughly, you can characterise the density in and around the Galactic plane as <span class="math-container">$\rho_0 \exp(-|z|/h)$</span>, where <span class="math-container">$z$</span> is the height above or below the Galactic plane and <span class="math-container">$h$</span> is a scale-height of perhaps 200 pc or so and <span class="math-container">$\rho_0$</span> is the density at the plane mid-line.</p>
<p>Applying Gauss' law for gravity, assuming the Galactic disc is uniform on radial scales of thousands of parsecs (so looks effectively like an infinite plane when you are close to it), then
<span class="math-container">$$g(z) \simeq -2\pi G\int^{z}_{0} \rho\ dz = -2\pi G \rho_0 h[1-\exp(-|z|/h)]\ .$$</span></p>
<p>Thus, when a star (e.g. the Sun) has <span class="math-container">$|z|&gt;0$</span>, there is a restoring acceleration that acts towards the Galactic plane at all times.</p>
<p>Even more roughly, we can approximate <span class="math-container">$\exp(-|z|/h) \sim 1-|z|/h$</span>, which simplifies the acceleration to
<span class="math-container">$$ g \simeq -2\pi G\rho_0 |z|\ .$$</span></p>
<p>This is just simple harmonic motion of the form acceleration <span class="math-container">$= -\omega^2 |z|$</span> and thus stars oscillate up and down through the Galactic plane, with a period of <span class="math-container">$2\pi/\omega = (2\pi/G\rho_0)^{1/2}$</span>.</p>
<p>The stellar mass density near the Galactic plane is something like <span class="math-container">$0.1M_\odot$</span> per cubic parsec. This gives an oscillation period of <span class="math-container">$\sim 100$</span> million years. A more accurate calculation gives more like 70 million years.</p>
<p>To estimate the amplitude of the motion requires a measurement of the current motion of the Solar System with respect to the galactic plane. This is currently estimated as about 7 km/s directed out of the plane and the Sun is currently about 20 pc above the plane (see <a href="https://astronomy.stackexchange.com/a/8336/2531">here</a>). Again, assuming simple harmonic motion, then the sum of potential and kinetic energy is constant
<span class="math-container">$$\pi G \rho_o z^2 +  \frac{v^2}{2} = {\rm constant}\ .$$</span></p>
<p>Substituting <span class="math-container">$z=20$</span> pc and <span class="math-container">$v= 7$</span> km/s we find a constant of <span class="math-container">$2.5\times 10^7$</span> J/kg. Then setting the speed to zero this is the potential energy per unit mass at the highest/lowest point of the oscillation, which yields an amplitude of 136 pc.</p>

CC BY-SA 4.0
2021-06-20T21:51:40.640Z
So the oscillation does not require any non-uniform distribution of masses in the Milky Way(?). Would it also happen if the stars of Milky Way were completely uniformly distributed (and of the same mass and not moving) within a 600 x 60,000 parsec flat cylinder? Or what is the simplest star/mass distribution that will produce the oscillation? A rotational symmetric flat disk of uniform mass density, ***thinner*** than the oscillation amplitude?
Peter Mortensen
CC BY-SA 4.0
2021-06-20T23:00:47.790Z
@PeterMortenson the mass distribution isn't uniform, it approaches the simple exponential model in my answer. Other mass distributions might produce something different, but I think your proposed mass distribution would also lead to an oscillatory motion.
ProfRob
CC BY-SA 4.0
2021-06-17T15:05:07.190Z
2021-06-17T15:21:36.930Z
2021-06-17T15:21:36.930Z
ProfRob
<p>According to research, excluding the motion of the entire galaxy, astronomers found the motion of the solar system in the Milky Way.</p>
<ol>
<li>The solar system revolve around the galactic bulge of the Milky Way.</li>
<li>Due to the axis of the sun, the mass of the galactic plane or something else, the path of the solar system passing thorugh the galactic plane looks like a wave as shown in the pictures below.</li>
</ol>
<p><a href="https://i.sstatic.net/hafcp.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/hafcp.jpg" alt="enter image description here" /></a></p>
<p><sub><a href="https://astronomy.com/magazine/ask-astro/2020/07/in-which-direction-does-the-sun-move-through-the-milky-way" rel="nofollow noreferrer">Image source.</a></sub></p>
<p>From the research, a single wave takes about 60 million years. The solar system passed through the galactic plane about 30,000 years ago, currently it is above the galactic plane.</p>
<p>How did the astronomers calculate the period? How did they find the highest and lowest point of the path?</p>
<p>I can understand the wave's path of the solar system due to the mass of the galaxy, I'm just curious how to calculate it.</p>

CC BY-SA 4.0
2021-06-20T21:21:58.820Z
The obliquity (axial tilt) of [the sun is 7° to the ecliptic and 67° to the galactic plane](https://en.wikipedia.org/wiki/Sun). The 60° is Earth's orbit (the ecliptic) to the galactic plane. The text should read *"The Sun orbits the center of our galaxy tipped at an angle of 67°"* (not 60°)
Peter Mortensen
CC BY-SA 4.0
2021-06-20T21:25:28.047Z
But isn't the tilt of Earth's orbit or the axial tilt of Sun ***completely irrelevant*** in this context?
Peter Mortensen
<solar-system><milky-way>
How do astronomers calculate the period about the wave motion of the solar system in the Milky Way?
JohnConLee
# How do astronomers calculate the period about the wave motion of the solar system in the Milky Way?

(Asked by: username_0 on 2021-06-17)

According to research, excluding the motion of the entire galaxy, astronomers found the motion of the solar system in the Milky Way.

1. The solar system revolve around the galactic bulge of the Milky Way.
2. Due to the axis of the sun, the mass of the galactic plane or something else, the path of the solar system passing thorugh the galactic plane looks like a wave as shown in the pictures below.

[![enter image description here](https://i.sstatic.net/hafcp.jpg)](https://i.sstatic.net/hafcp.jpg)

[Image source.](https://astronomy.com/magazine/ask-astro/2020/07/in-which-direction-does-the-sun-move-through-the-milky-way)

From the research, a single wave takes about 60 million years. The solar system passed through the galactic plane about 30,000 years ago, currently it is above the galactic plane.

How did the astronomers calculate the period? How did they find the highest and lowest point of the path?

I can understand the wave's path of the solar system due to the mass of the galaxy, I'm just curious how to calculate it.

**Tags:** solar-system, milky-way
> **Comments:**
> **username_1**: The obliquity (axial tilt) of [the sun is 7° to the ecliptic and 67° to the galactic plane](https://en.wikipedia.org/wiki/Sun). The 60° is Earth's orbit (the ecliptic) to the galactic plane. The text should read *"The Sun orbits the center of our galaxy tipped at an angle of 67°"* (not 60°)  
> **username_1**: But isn't the tilt of Earth's orbit or the axial tilt of Sun ***completely irrelevant*** in this context?  

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## 1 answers

### Answer by username_2 on 2021-06-17. Score: 10 | Accepted answer

The motion of the Sun in the direction perpendicular to the Galactic plane is deduced from a mixture of measurement and theoretical calculation.

The measurements are (i) deducing the Solar System's current position and speed with respect to the galactic plane (see [How far is the Earth/Sun above/below the galactic plane, and is it heading toward/away from it?](https://astronomy.stackexchange.com/questions/822/how-far-is-the-earth-sun-above-below-the-galactic-plane-and-is-it-heading-towar) ) and (ii) deducing how much mass is present in the galactic plane which acts to accelerate/decelerate the motion of the Solar System.

From there it is a relatively simple application of Newtonian mechanics to work out what the motion of the Sun will be in the future (and where it came from in the past). The solution is an oscillatory motion above and below the plane with an amplitude of about 100 pc and a period of 70 million years.

A flavour of the calculation can be done analytically using some simple approximations.

Very roughly, you can characterise the density in and around the Galactic plane as $\rho\_0 \exp(-|z|/h)$, where $z$ is the height above or below the Galactic plane and $h$ is a scale-height of perhaps 200 pc or so and $\rho\_0$ is the density at the plane mid-line.

Applying Gauss' law for gravity, assuming the Galactic disc is uniform on radial scales of thousands of parsecs (so looks effectively like an infinite plane when you are close to it), then
$$g(z) \simeq -2\pi G\int^{z}\_{0} \rho\ dz = -2\pi G \rho\_0 h[1-\exp(-|z|/h)]\ .$$

Thus, when a star (e.g. the Sun) has $|z|>0$, there is a restoring acceleration that acts towards the Galactic plane at all times.

Even more roughly, we can approximate $\exp(-|z|/h) \sim 1-|z|/h$, which simplifies the acceleration to
$$ g \simeq -2\pi G\rho\_0 |z|\ .$$

This is just simple harmonic motion of the form acceleration $= -\omega^2 |z|$ and thus stars oscillate up and down through the Galactic plane, with a period of $2\pi/\omega = (2\pi/G\rho\_0)^{1/2}$.

The stellar mass density near the Galactic plane is something like $0.1M\_\odot$ per cubic parsec. This gives an oscillation period of $\sim 100$ million years. A more accurate calculation gives more like 70 million years.

To estimate the amplitude of the motion requires a measurement of the current motion of the Solar System with respect to the galactic plane. This is currently estimated as about 7 km/s directed out of the plane and the Sun is currently about 20 pc above the plane (see [here](https://astronomy.stackexchange.com/a/8336/2531)). Again, assuming simple harmonic motion, then the sum of potential and kinetic energy is constant
$$\pi G \rho\_o z^2 + \frac{v^2}{2} = {\rm constant}\ .$$

Substituting $z=20$ pc and $v= 7$ km/s we find a constant of $2.5\times 10^7$ J/kg. Then setting the speed to zero this is the potential energy per unit mass at the highest/lowest point of the oscillation, which yields an amplitude of 136 pc.

> **Comments:**
> **username_1**: So the oscillation does not require any non-uniform distribution of masses in the Milky Way(?). Would it also happen if the stars of Milky Way were completely uniformly distributed (and of the same mass and not moving) within a 600 x 60,000 parsec flat cylinder? Or what is the simplest star/mass distribution that will produce the oscillation? A rotational symmetric flat disk of uniform mass density, ***thinner*** than the oscillation amplitude?  
> **username_2**: @PeterMortenson the mass distribution isn't uniform, it approaches the simple exponential model in my answer. Other mass distributions might produce something different, but I think your proposed mass distribution would also lead to an oscillatory motion.  

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