<p>I presume what you mean is how does the plane of the orbit compare to the equatorial rotation plane of the star?</p>

<p>The answer is, you can sort of estimate this, by using something called the <a href="https://exoplanetmusings.wordpress.com/2013/03/25/764/" rel="nofollow noreferrer">Rossiter-McLaughlin</a> effect (see also <a href="http://adsabs.harvard.edu/abs/1924ApJ....60...15R" rel="nofollow noreferrer">Rossiter 1924</a>; <a href="http://adsabs.harvard.edu/abs/1924ApJ....60...22M" rel="nofollow noreferrer">McLaughlin 1924</a>). </p>

<p>You can find plenty of information on the web - I'll add a couple of links when I have a moment - but to summarise:</p>

<p>The rotation of a star broadens its spectral absorption lines. The hemisphere coming towards us emits light that is is blue shifted, the hemisphere receding is redshifted. If we now take a <em>transiting</em> planet, during the eclipse it crosses the disc of the star and obscures regions that are blue or red shifted by various amounts.</p>

<p>Now what you do is measure the line-of-sight velocity of the star. During the transit you would not expect this to vary due to the "Doppler wobble" caused by the exoplanet, except that if the planet obscures a blue shifted portion of the stellar disc, the net spectral absorption line that remains shifts to the red, and vice-versa. The pattern of red, then blue shift (or vice versa) as the transit progresses is known as the Rossiter-Mclaughlin effect.</p>

<p><em>A schematic showing how the Rossiter-McLaughlin effect works and how a different transit geometry with respect to the rotation axis of the star leads to a different line-of-sight-velocity signature in the spectral lines of the star. (Image credit: Subaru Telescope, National Observatories of Japan.)</em></p>

<p><img src="https://i.sstatic.net/X5eOF.jpg" alt="Schematic of Rossiter McLaughlin effect."></p>

<p>If the planet orbits in the same plane and in the same direction as the stellar rotation (as the planets in our solar system nearly do), then the blue shifted limb of the parent star is obscured first, followed by an equal amount of redshift as the planets moves to obscure the receding stellar limb (see image above, left). Thus the stellar absorption lines show a redshift followed by a symmetric blueshift. If the planet was retrograde it would be symmetric but occur in the opposite order. A Polar orbit would show no RM effect. Inclined orbits would have an asymmetric RM effect - ie perhaps more blueshift than redshift (see image above, right).</p>

<p>The RM effect cannot give an exact orientation, it gives the projected angle between orbital and rotation axes on the plane of the sky. Nevertheless, that is sufficient for us to know that a lot of the transiting exoplanets have highly misaligned orbital and stellar rotation axes.</p>

CC BY-SA 3.0
2015-04-15T15:14:47.877Z
Nice explanation. This effect is cool.
Yashbhatt
CC BY-SA 3.0
2015-04-12T20:47:47.850Z
2015-04-13T08:43:03.573Z
2015-04-13T08:43:03.573Z
ProfRob
<p><em>Is there any way to find the angle of the orbital plane of the exoplanet orbiting its parent star?</em> It's unclear what precisely you're asking. The planet will be orbiting (to very good approximation) on a Keplerian orbit (closed ellipse co-focal with the parent star). Such an orbit has <em>two relevant angles</em>. One is the angle between the orbital plane and the line of sight to Earth (often called the inclination $i$); the other is the direction of the orbital periapse (usually measured as the angle in the orbital plane between the line of nodes and the periapse). </p>

<p>In case of transit detections, $i\sim 90^\circ$ (for otherwise, there will be no ecclipse/transit), but it cannot (usually) inferred for radial-velocity detections. Since the amplitude of the observed radial velocity is proportional to $M\sin i$, this implies that the planet mass is only limited but could be larger than for $i=90^\circ$.</p>

CC BY-SA 3.0
2015-04-12T16:59:13.323Z
2015-04-12T16:59:13.323Z
Walter
<p>Astronomers can tell if there are any exoplanets orbiting its parent star using but not limited to the following methods:</p>

<p><strong>Transit</strong></p>

<p>As an massive object usually a planet passing in front of a star causing the amount of light emitted from the star to dim, by measuring the dimness the data can reveal how big the planet is.</p>

<p><strong>Radial velocity</strong></p>

<p>The star will seems to wobble as if there is a massive object usually a planet in orbit playing tug of war in a game of whose gravity is mightier. However since this wobbling effect is difficult to observe nerds prefer to measure the blueshift of the light (as star accelerates in our direction) and redshift (as star accelerates away from our direction) which seems to also tells how massive the planet is.</p>

<p><strong>Note</strong>
Astronomers can derive the planet's density from its mass and radius(see transit).</p>

<p><strong>Question</strong></p>

<p>So far astronomers are able to determine the size, mass as well as density thus even its composition, is there any way to find the angle of the orbital plane of the exoplanet orbiting its parent star?</p>

<star><orbit><planet>
Can we know the orbital planes of extraterrestrial planetary bodies?
user6760
# Can we know the orbital planes of extraterrestrial planetary bodies?

(Asked by: username_0 on 2015-04-12)

Astronomers can tell if there are any exoplanets orbiting its parent star using but not limited to the following methods:

**Transit**

As an massive object usually a planet passing in front of a star causing the amount of light emitted from the star to dim, by measuring the dimness the data can reveal how big the planet is.

**Radial velocity**

The star will seems to wobble as if there is a massive object usually a planet in orbit playing tug of war in a game of whose gravity is mightier. However since this wobbling effect is difficult to observe nerds prefer to measure the blueshift of the light (as star accelerates in our direction) and redshift (as star accelerates away from our direction) which seems to also tells how massive the planet is.

**Note**
Astronomers can derive the planet's density from its mass and radius(see transit).

**Question**

So far astronomers are able to determine the size, mass as well as density thus even its composition, is there any way to find the angle of the orbital plane of the exoplanet orbiting its parent star?

**Tags:** star, orbit, planet


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## 2 answers

### Answer by username_1 on 2015-04-12. Score: 2 | Accepted answer

I presume what you mean is how does the plane of the orbit compare to the equatorial rotation plane of the star?

The answer is, you can sort of estimate this, by using something called the [Rossiter-McLaughlin](https://exoplanetmusings.wordpress.com/2013/03/25/764/) effect (see also [Rossiter 1924](http://adsabs.harvard.edu/abs/1924ApJ....60...15R); [McLaughlin 1924](http://adsabs.harvard.edu/abs/1924ApJ....60...22M)).

You can find plenty of information on the web - I'll add a couple of links when I have a moment - but to summarise:

The rotation of a star broadens its spectral absorption lines. The hemisphere coming towards us emits light that is is blue shifted, the hemisphere receding is redshifted. If we now take a *transiting* planet, during the eclipse it crosses the disc of the star and obscures regions that are blue or red shifted by various amounts.

Now what you do is measure the line-of-sight velocity of the star. During the transit you would not expect this to vary due to the "Doppler wobble" caused by the exoplanet, except that if the planet obscures a blue shifted portion of the stellar disc, the net spectral absorption line that remains shifts to the red, and vice-versa. The pattern of red, then blue shift (or vice versa) as the transit progresses is known as the Rossiter-Mclaughlin effect.

*A schematic showing how the Rossiter-McLaughlin effect works and how a different transit geometry with respect to the rotation axis of the star leads to a different line-of-sight-velocity signature in the spectral lines of the star. (Image credit: Subaru Telescope, National Observatories of Japan.)*

![Schematic of Rossiter McLaughlin effect.](https://i.sstatic.net/X5eOF.jpg)

If the planet orbits in the same plane and in the same direction as the stellar rotation (as the planets in our solar system nearly do), then the blue shifted limb of the parent star is obscured first, followed by an equal amount of redshift as the planets moves to obscure the receding stellar limb (see image above, left). Thus the stellar absorption lines show a redshift followed by a symmetric blueshift. If the planet was retrograde it would be symmetric but occur in the opposite order. A Polar orbit would show no RM effect. Inclined orbits would have an asymmetric RM effect - ie perhaps more blueshift than redshift (see image above, right).

The RM effect cannot give an exact orientation, it gives the projected angle between orbital and rotation axes on the plane of the sky. Nevertheless, that is sufficient for us to know that a lot of the transiting exoplanets have highly misaligned orbital and stellar rotation axes.

> **Comments:**
> **username_2**: Nice explanation. This effect is cool.  

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### Answer by username_3 on 2015-04-12. Score: 0

*Is there any way to find the angle of the orbital plane of the exoplanet orbiting its parent star?* It's unclear what precisely you're asking. The planet will be orbiting (to very good approximation) on a Keplerian orbit (closed ellipse co-focal with the parent star). Such an orbit has *two relevant angles*. One is the angle between the orbital plane and the line of sight to Earth (often called the inclination $i$); the other is the direction of the orbital periapse (usually measured as the angle in the orbital plane between the line of nodes and the periapse).

In case of transit detections, $i\sim 90^\circ$ (for otherwise, there will be no ecclipse/transit), but it cannot (usually) inferred for radial-velocity detections. Since the amplitude of the observed radial velocity is proportional to $M\sin i$, this implies that the planet mass is only limited but could be larger than for $i=90^\circ$.


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