<p>The asymmetry in the terrestrial <a href="https://en.wikipedia.org/wiki/Bow_shock" rel="nofollow noreferrer">bow shock</a> is due to aberration from Earth's orbit about the <a href="https://en.wikipedia.org/wiki/Sun" rel="nofollow noreferrer">Sun</a> (it's on the order of ~29-30 km/s).  Thus, the &quot;ram side&quot; of the bow shock transverse to the Earth-Sun line will be more compressed than the opposite side.</p>
<p>Note, there is a north-south asymmetry as well, but that is caused by the dipole tilt of Earth's <a href="https://en.wikipedia.org/wiki/Magnetosphere" rel="nofollow noreferrer">magnetic field</a>.</p>

CC BY-SA 4.0
2024-01-17T17:05:44.390Z
The OP is primarily asking about the asymmetry north/south of the ecliptic plane.
PM 2Ring
CC BY-SA 4.0
2024-01-17T17:56:43.317Z
Ah okay, I did note why there is a north-south asymmetry though...
honeste_vivere
CC BY-SA 4.0
2024-01-17T14:08:51.643Z
2024-01-17T14:08:51.643Z
honeste_vivere
<p><a href="https://jwst-docs.stsci.edu/jwst-observatory-characteristics/jwst-orbit" rel="nofollow noreferrer">These diagrams</a> (both confusingly called &quot;Figure 1&quot;) of the James Webb Space Telescope's orbit around the L2 Lagrange point show the Earth's bow shock. It makes sense that the bow shock is asymmetric in the second diagram, which depicts the ecliptic plane, because of Earth's orbit.</p>
<p>My question is: why is the bow shock asymmetric in the first diagram, which I've copied below, that shows the ecliptic edge-on? At the distance of L2, the bow shock is less 450 Mm away &quot;above&quot; the plane, while it's more than 550 Mm away &quot;below&quot; the plane. Why are these two distances different?</p>
<p><a href="https://i.sstatic.net/XqHEY.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/XqHEY.png" alt="Diagram of JWST orbit around L2, shown with the ecliptic edge-on" /></a></p>

CC BY-SA 4.0
2023-12-12T13:56:28.760Z
If the bow shock is shown for a given time of year, like for example whenever "Launch + 200 days" was supposed to be, then I'd suggest it might be related to the tilt of Earth's magnetic dipole (with respect to the ecliptic) *at that time.* Of course parts of the diagram span many years, since each of those orbits of JWST with respect to L2 takes almost exactly 1/2 year. [1](https://space.stackexchange.com/q/58020/12102), [2](https://space.stackexchange.com/q/57890/12102), [3](https://space.stackexchange.com/q/56043/12102)
uhoh
CC BY-SA 4.0
2023-12-14T03:21:34.210Z
The vertical axis is labeled as normal to the ecliptic.
Doradus
CC BY-SA 4.0
2023-12-14T04:12:36.673Z
I can't find a detailed diagram of Earth's magnetosphere which clearly shows the impact of Earth's magnetic dipole on the bow shock, but this artist's rendering kind of shows the (rotational) polar axis: https://en.wikipedia.org/wiki/Earth%27s_magnetic_field#Magnetosphere Also see https://en.wikipedia.org/wiki/Van_Allen_radiation_belt
PM 2Ring
<ecliptic><solar-wind><bow-shock>
Why is Earth's bow shock wider on one side of the ecliptic than the other?
Doradus
# Why is Earth's bow shock wider on one side of the ecliptic than the other?

(Asked by: username_0 on 2023-12-12)

[These diagrams](https://jwst-docs.stsci.edu/jwst-observatory-characteristics/jwst-orbit) (both confusingly called "Figure 1") of the James Webb Space Telescope's orbit around the L2 Lagrange point show the Earth's bow shock. It makes sense that the bow shock is asymmetric in the second diagram, which depicts the ecliptic plane, because of Earth's orbit.

My question is: why is the bow shock asymmetric in the first diagram, which I've copied below, that shows the ecliptic edge-on? At the distance of L2, the bow shock is less 450 Mm away "above" the plane, while it's more than 550 Mm away "below" the plane. Why are these two distances different?

[![Diagram of JWST orbit around L2, shown with the ecliptic edge-on](https://i.sstatic.net/XqHEY.png)](https://i.sstatic.net/XqHEY.png)

**Tags:** ecliptic, solar-wind, bow-shock

> **Comments:**
> **username_1**: If the bow shock is shown for a given time of year, like for example whenever "Launch + 200 days" was supposed to be, then I'd suggest it might be related to the tilt of Earth's magnetic dipole (with respect to the ecliptic) *at that time.* Of course parts of the diagram span many years, since each of those orbits of JWST with respect to L2 takes almost exactly 1/2 year. [1](https://space.stackexchange.com/q/58020/12102), [2](https://space.stackexchange.com/q/57890/12102), [3](https://space.stackexchange.com/q/56043/12102)  
> **username_0**: The vertical axis is labeled as normal to the ecliptic.  
> **username_2**: I can't find a detailed diagram of Earth's magnetosphere which clearly shows the impact of Earth's magnetic dipole on the bow shock, but this artist's rendering kind of shows the (rotational) polar axis: https://en.wikipedia.org/wiki/Earth%27s_magnetic_field#Magnetosphere Also see https://en.wikipedia.org/wiki/Van_Allen_radiation_belt  

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## 1 answers

### Answer by username_3 on 2024-01-17. Score: 2

The asymmetry in the terrestrial [bow shock](https://en.wikipedia.org/wiki/Bow_shock) is due to aberration from Earth's orbit about the [Sun](https://en.wikipedia.org/wiki/Sun) (it's on the order of ~29-30 km/s). Thus, the "ram side" of the bow shock transverse to the Earth-Sun line will be more compressed than the opposite side.

Note, there is a north-south asymmetry as well, but that is caused by the dipole tilt of Earth's [magnetic field](https://en.wikipedia.org/wiki/Magnetosphere).

> **Comments:**
> **username_2**: The OP is primarily asking about the asymmetry north/south of the ecliptic plane.  
> **username_3**: Ah okay, I did note why there is a north-south asymmetry though...  

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