<blockquote>
  <p>What is the inclination of Mercury's orbit w.r.t. the Sun's equatorial plane?</p>
</blockquote>

<p>From <a href="https://en.wikipedia.org/wiki/Orbital_inclination#Orbits" rel="nofollow noreferrer">this page</a>, we can see that it is 3.38°.</p>

<blockquote>
  <p>Additionally, what is the orientation of Mercury's orbit (long axis) w.r.t. the Milky Way?</p>
</blockquote>

<p><strike>This is a bit harder to estimate precisely, but we can refer to <a href="http://coolcosmos.ipac.caltech.edu/image_galleries/IRAS/allsky.html" rel="nofollow noreferrer">this view from the IRAS satellite</a> to get a rough figure of 63° between the ecliptic and the plane of the Milky Way (for more information on how to calculate that, check out <a href="https://en.wikipedia.org/wiki/Aitoff_projection" rel="nofollow noreferrer">Aitoff projection</a>). Now just add/subtract Mercury's 7.01° of inclination to ecliptic and you have your result.</strike></p>

<p>EDIT: Because the angles of 63 and 7 degrees may not be along the same axis, there are ways to be a bit more precise: you can take Mercury's coordinates in the ecliptic coordinates system, and transform them into <a href="https://en.wikipedia.org/wiki/Galactic_coordinate_system" rel="nofollow noreferrer">galactic coordinates</a>. If you do that with only 3 different points on Mercury's orbit, you can find the angle made by the plane defined by those 3 points and the horizontal galactic plane.</p>

<p>You can use <a href="https://omniweb.gsfc.nasa.gov/cgi/models/planet.cgi" rel="nofollow noreferrer">this website</a> to get your 3 sets of ecliptic coordinates for Mercury, and <a href="https://ned.ipac.caltech.edu/forms/calculator.html" rel="nofollow noreferrer">that website</a> to transform them into galactic coordinates. You can transform those into regular Cartesian coordinates since the website that gave you the ecliptic coordinates also gives you the distance from Mercury to the Sun! Once you get there, it's easy to get the angle made by that plane and the horizontal: good luck!</p>

CC BY-SA 3.0
2018-02-02T10:07:29.660Z
Dear Romain, thank you. I was also thinking of the orientation of Mercury's long axis w.r.t. the center of galaxy, indeed, by taking into account the 63° incilination. I expect that the long side of the ellipse, away from the sun (aphelion), is the closest near the galaxy's center, pointing to it, and the perihelion is at the other side of the sun, away from the galaxy's center.
Ted
CC BY-SA 3.0
2018-02-02T13:48:32.727Z
Is this so, or is the orientation different?
Ted
CC BY-SA 3.0
2018-02-02T15:55:04.130Z
You're right, the angles could very well be along different axes! But finding out the inclination of Mercury's orbit directly wrt. the galactic plane might require some calculus (I edited my original reply to explain a bit more)
Romain P.
CC BY-SA 3.0
2018-02-02T06:14:33.690Z
2018-02-02T16:37:10.027Z
2018-02-02T16:37:10.027Z
Romain P.
<p>What is the inclination of Mercury's orbit w.r.t. the Sun's equatorial plane? 
Additionally, what is the orientation of Mercury's orbit (long axis) w.r.t. the Milky Way?</p>

<the-sun><mercury>
What is the inclination of Mercury's orbit w.r.t. the Sun's equatorial plane?
Ted
# What is the inclination of Mercury's orbit w.r.t. the Sun's equatorial plane?

(Asked by: username_0 on 2018-02-01)

What is the inclination of Mercury's orbit w.r.t. the Sun's equatorial plane?
Additionally, what is the orientation of Mercury's orbit (long axis) w.r.t. the Milky Way?

**Tags:** the-sun, mercury

---

## 1 answers

### Answer by username_1 on 2018-02-02. Score: 2

> What is the inclination of Mercury's orbit w.r.t. the Sun's equatorial plane?

From [this page](https://en.wikipedia.org/wiki/Orbital_inclination#Orbits), we can see that it is 3.38°.

> Additionally, what is the orientation of Mercury's orbit (long axis) w.r.t. the Milky Way?

This is a bit harder to estimate precisely, but we can refer to [this view from the IRAS satellite](http://coolcosmos.ipac.caltech.edu/image_galleries/IRAS/allsky.html) to get a rough figure of 63° between the ecliptic and the plane of the Milky Way (for more information on how to calculate that, check out [Aitoff projection](https://en.wikipedia.org/wiki/Aitoff_projection)). Now just add/subtract Mercury's 7.01° of inclination to ecliptic and you have your result.

EDIT: Because the angles of 63 and 7 degrees may not be along the same axis, there are ways to be a bit more precise: you can take Mercury's coordinates in the ecliptic coordinates system, and transform them into [galactic coordinates](https://en.wikipedia.org/wiki/Galactic_coordinate_system). If you do that with only 3 different points on Mercury's orbit, you can find the angle made by the plane defined by those 3 points and the horizontal galactic plane.

You can use [this website](https://omniweb.gsfc.nasa.gov/cgi/models/planet.cgi) to get your 3 sets of ecliptic coordinates for Mercury, and [that website](https://ned.ipac.caltech.edu/forms/calculator.html) to transform them into galactic coordinates. You can transform those into regular Cartesian coordinates since the website that gave you the ecliptic coordinates also gives you the distance from Mercury to the Sun! Once you get there, it's easy to get the angle made by that plane and the horizontal: good luck!

> **Comments:**
> **username_0**: Dear Romain, thank you. I was also thinking of the orientation of Mercury's long axis w.r.t. the center of galaxy, indeed, by taking into account the 63° incilination. I expect that the long side of the ellipse, away from the sun (aphelion), is the closest near the galaxy's center, pointing to it, and the perihelion is at the other side of the sun, away from the galaxy's center.  
> **username_0**: Is this so, or is the orientation different?  
> **username_1**: You're right, the angles could very well be along different axes! But finding out the inclination of Mercury's orbit directly wrt. the galactic plane might require some calculus (I edited my original reply to explain a bit more)  

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