<p>If you go to the <a href="https://www.nature.com/articles/s41550-021-01495-w" rel="noreferrer">original article</a>, you can see that it's something like this</p>
<ol>
<li>They have a plausible size for the X-ray-emitting source (that is, the X-ray-bright part of the accretion disk): <span class="math-container">$R = 2.5 \times 10^{9}$</span> cm, which is about 4 times the size of the Earth. (I think this comes from fitting the X-ray broadband spectrum before and after the putative transit, plus the known distance to the galaxy.)</li>
<li>They have a fairly good (X-ray) light curve, allowing them to work out the duration of the transit, along with the fact that the X-ray source is completely eclipsed during the transit.</li>
<li>Putting these together, they run a bunch of simulations where they vary things like the size of the eclipsing object and its orbital speed. The best-fitting values give a planet size roughly the same as Saturn or Jupiter and a velocity of roughly 15 or 20 km/s.</li>
<li>If you make reasonable assumptions for the mass of the X-ray binary system (= companion star plus neutron star or black hole), then you can work out the size and period of the planet's (circumbinary) orbit, given its transiting velocity. For an assumed X-ray binary mass of <span class="math-container">$20 M_{\odot}$</span>, this works out to a period of about 70 years.</li>
</ol>
<p>(I believe the X-ray binary mass comes mostly from the previous identification, from <em>HST</em> imaging of the X-ray source's location, of a B supergiant as the likely companion star; its mass would dominate over the compact object's mass, which would be only <span class="math-container">$\sim 1.5$</span>-<span class="math-container">$2 M_{\odot}$</span> for a neutron star and a few <span class="math-container">$M_{\odot}$</span> for a black hole.)</p>

CC BY-SA 4.0
2021-10-27T10:54:02.803Z
2021-10-27T10:54:02.803Z
Peter Erwin
<p>The answer can be found in the &quot;Methods&quot; section of the original Nature paper (<a href="https://www.nature.com/articles/s41550-021-01495-w" rel="nofollow noreferrer">Stefano et. al</a>). I will try to give a brief sketch of what they did:</p>
<p>First thing is, that the eclipsing object is not much larger that that of the emitting source. Only about 0.74 vs 0.43 Jupiter radii.
This means you have not just an eclipsed point source, but two circles overlapping each other. This allows to estimate both the velocity <span class="math-container">$vpl$</span>, size <span class="math-container">$f_\mathrm{ec} $</span> and offset <span class="math-container">$b$</span> of the eclipse from the shape of the dip. In their model the overlap / coverage is give by:
<span class="math-container">$$
A(t)= (\alpha_X-\cos \alpha_X \sin \alpha_X) + f^2_\mathrm{ec}(\alpha_\mathrm{ec}-\cos \alpha_\mathrm{ec} \sin \alpha_\mathrm{ec}) \\
$$</span>
where <span class="math-container">$\alpha_X$</span> and <span class="math-container">$a_\mathrm{ec}$</span> are the angles subtended by the intersecting arcs of the star and
the foreground object and <span class="math-container">$d$</span> is the distance between the two spheres:<span class="math-container">$$
a_X = \arccos \frac{d(t)^2+1-f^2_\mathrm{ec}}{2 d(t)}\\
a_\mathrm{ec} = \arccos \frac{d(t)^2+f^2_\mathrm{ec}-1}{2 d(t) f^2_\mathrm{ec}}\\
d(T) = \sqrt{b² + (|t-T_\mathrm{mid}| v))}
$$</span>
Everything, however, relative to the size of the X-ray emitting region.</p>
<p>Luckily the X-ray source is a ultraluminous supersoft source (ULS) and its spectrum can be described by a simple black-body.
And from the <a href="https://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law" rel="nofollow noreferrer">Stefan-Boltzman Law</a> we can simply get the size of the Black-body, if we know the temperature &amp; luminosity of our source (see <a href="https://iopscience.iop.org/article/10.1088/0004-637X/782/2/98/pdf" rel="nofollow noreferrer">here</a> for the distance estimate to M51).
<span class="math-container">$$P \propto A T^{4}
$$</span>
Both we get from the spectrum.</p>
<p>And now, that we have a ruler for absolute sizes in the system, we can simply fit the shape of the dip with the above model that depends on the absolute size and velocity of your eclipser.</p>
<p>After we know the velocity of the source, we use Kepler's law to get the semi-major axis and period.</p>
<p><span class="math-container">$$
\alpha = 45 \mathrm{au} \,\,(\dfrac{M_\mathrm{tot}}{20 M_\circ}) (\dfrac{20 \mathrm{km s}^{-1}}{v_\mathrm{pl}})^2 \\
P_\mathrm{orb} = 68 \mathrm{yr} \,\, (\frac{\alpha}{45 \mathrm{au}})^{3/2}(\dfrac{20 M_\circ }{M_\mathrm{tot}})^{1/2}$$</span>
Probably assuming a circular orbit, as orbits with high eccentricity would not be stable in a system with a central binary star.</p>

CC BY-SA 4.0
2021-10-27T11:03:24.243Z
2021-10-27T11:03:24.243Z
SpaceCore
<p>Phys.org's <a href="https://phys.org/news/2021-10-astronomers-planet-galaxy.html" rel="nofollow noreferrer">Astronomers may have discovered the first planet outside of our galaxy</a> links to  Di Stefano et al. 2021 <a href="https://www.nature.com/articles/s41550-021-01495-w" rel="nofollow noreferrer">A possible planet candidate in an external galaxy detected through X-ray transit</a> (earlier <a href="https://arxiv.org/abs/2009.08987" rel="nofollow noreferrer">arXiv preprint</a>) and says:</p>
<blockquote>
<p>While this is a tantalizing study, more data would be needed to verify the interpretation as an extragalactic exoplanet. <strong>One challenge is that the planet candidate's large orbit means it would not cross in front of its binary partner again for about 70 years, thwarting any attempts for a confirming observation for decades.</strong></p>
<p>&quot;Unfortunately to confirm that we're seeing a planet we would likely have to wait decades to see another transit,&quot; said co-author Nia Imara of the University of California at Santa Cruz. &quot;And because of the uncertainties about how long it takes to orbit, we wouldn't know exactly when to look.&quot;</p>
<p>Can the dimming have been caused by a cloud of gas and dust passing in front of the X-ray source? The researchers consider this to be an unlikely explanation, as the characteristics of the event observed in M51-ULS-1 are not consistent with the passage of such a cloud. The model of a planet candidate is, however, consistent with the data.</p>
</blockquote>
<p>The result is very exciting! For optical transits of a planet across a usually much larger stellar disk, the duration is a function of orbital velocity, stellar diameter and inclination. But for the present techniques I think the X-ray source is supposed to be much smaller than the planet, so the radius determining the duration is now the planet's; the time is always less than <span class="math-container">$R_{planet}/v_{planet}$</span> since inclination factors in.</p>
<p>Have I got that right?</p>
<p><strong>Question:</strong> After only one eclipse of its X-ray bright primary, how can astronomers estimate the first extragalactic exoplanet's period to be about 70 years?</p>
<p>Related:</p>
<ul>
<li><a href="https://astronomy.stackexchange.com/q/47212/7982">Was the First Extragalactic Planet discovered in Andromeda Galaxy</a></li>
</ul>

<observational-astronomy><exoplanet><x-ray><eclipse-non-earth-related>
After only one eclipse of its X-ray bright primary, how can astronomers estimate the first extragalactic exoplanet's period to be about 70 years?
uhoh
# After only one eclipse of its X-ray bright primary, how can astronomers estimate the first extragalactic exoplanet's period to be about 70 years?

(Asked by: username_0 on 2021-10-26)

Phys.org's [Astronomers may have discovered the first planet outside of our galaxy](https://phys.org/news/2021-10-astronomers-planet-galaxy.html) links to Di Stefano et al. 2021 [A possible planet candidate in an external galaxy detected through X-ray transit](https://www.nature.com/articles/s41550-021-01495-w) (earlier [arXiv preprint](https://arxiv.org/abs/2009.08987)) and says:

> While this is a tantalizing study, more data would be needed to verify the interpretation as an extragalactic exoplanet. **One challenge is that the planet candidate's large orbit means it would not cross in front of its binary partner again for about 70 years, thwarting any attempts for a confirming observation for decades.**
>
> "Unfortunately to confirm that we're seeing a planet we would likely have to wait decades to see another transit," said co-author Nia Imara of the University of California at Santa Cruz. "And because of the uncertainties about how long it takes to orbit, we wouldn't know exactly when to look."
>
> Can the dimming have been caused by a cloud of gas and dust passing in front of the X-ray source? The researchers consider this to be an unlikely explanation, as the characteristics of the event observed in M51-ULS-1 are not consistent with the passage of such a cloud. The model of a planet candidate is, however, consistent with the data.

The result is very exciting! For optical transits of a planet across a usually much larger stellar disk, the duration is a function of orbital velocity, stellar diameter and inclination. But for the present techniques I think the X-ray source is supposed to be much smaller than the planet, so the radius determining the duration is now the planet's; the time is always less than $R\_{planet}/v\_{planet}$ since inclination factors in.

Have I got that right?

**Question:** After only one eclipse of its X-ray bright primary, how can astronomers estimate the first extragalactic exoplanet's period to be about 70 years?

Related:

* [Was the First Extragalactic Planet discovered in Andromeda Galaxy](https://astronomy.stackexchange.com/q/47212/7982)

**Tags:** observational-astronomy, exoplanet, x-ray, eclipse-non-earth-related

---

## 2 answers

### Answer by username_1 on 2021-10-27. Score: 5

If you go to the [original article](https://www.nature.com/articles/s41550-021-01495-w), you can see that it's something like this

1. They have a plausible size for the X-ray-emitting source (that is, the X-ray-bright part of the accretion disk): $R = 2.5 \times 10^{9}$ cm, which is about 4 times the size of the Earth. (I think this comes from fitting the X-ray broadband spectrum before and after the putative transit, plus the known distance to the galaxy.)
2. They have a fairly good (X-ray) light curve, allowing them to work out the duration of the transit, along with the fact that the X-ray source is completely eclipsed during the transit.
3. Putting these together, they run a bunch of simulations where they vary things like the size of the eclipsing object and its orbital speed. The best-fitting values give a planet size roughly the same as Saturn or Jupiter and a velocity of roughly 15 or 20 km/s.
4. If you make reasonable assumptions for the mass of the X-ray binary system (= companion star plus neutron star or black hole), then you can work out the size and period of the planet's (circumbinary) orbit, given its transiting velocity. For an assumed X-ray binary mass of $20 M\_{\odot}$, this works out to a period of about 70 years.

(I believe the X-ray binary mass comes mostly from the previous identification, from *HST* imaging of the X-ray source's location, of a B supergiant as the likely companion star; its mass would dominate over the compact object's mass, which would be only $\sim 1.5$-$2 M\_{\odot}$ for a neutron star and a few $M\_{\odot}$ for a black hole.)


---

### Answer by username_2 on 2021-10-27. Score: 4

The answer can be found in the "Methods" section of the original Nature paper ([Stefano et. al](https://www.nature.com/articles/s41550-021-01495-w)). I will try to give a brief sketch of what they did:

First thing is, that the eclipsing object is not much larger that that of the emitting source. Only about 0.74 vs 0.43 Jupiter radii.
This means you have not just an eclipsed point source, but two circles overlapping each other. This allows to estimate both the velocity $vpl$, size $f\_\mathrm{ec} $ and offset $b$ of the eclipse from the shape of the dip. In their model the overlap / coverage is give by:
$$
A(t)= (\alpha\_X-\cos \alpha\_X \sin \alpha\_X) + f^2\_\mathrm{ec}(\alpha\_\mathrm{ec}-\cos \alpha\_\mathrm{ec} \sin \alpha\_\mathrm{ec}) \\
$$
where $\alpha\_X$ and $a\_\mathrm{ec}$ are the angles subtended by the intersecting arcs of the star and
the foreground object and $d$ is the distance between the two spheres:$$
a\_X = \arccos \frac{d(t)^2+1-f^2\_\mathrm{ec}}{2 d(t)}\\
a\_\mathrm{ec} = \arccos \frac{d(t)^2+f^2\_\mathrm{ec}-1}{2 d(t) f^2\_\mathrm{ec}}\\
d(T) = \sqrt{b² + (|t-T\_\mathrm{mid}| v))}
$$
Everything, however, relative to the size of the X-ray emitting region.

Luckily the X-ray source is a ultraluminous supersoft source (ULS) and its spectrum can be described by a simple black-body.
And from the [Stefan-Boltzman Law](https://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law) we can simply get the size of the Black-body, if we know the temperature & luminosity of our source (see [here](https://iopscience.iop.org/article/10.1088/0004-637X/782/2/98/pdf) for the distance estimate to M51).
$$P \propto A T^{4}
$$
Both we get from the spectrum.

And now, that we have a ruler for absolute sizes in the system, we can simply fit the shape of the dip with the above model that depends on the absolute size and velocity of your eclipser.

After we know the velocity of the source, we use Kepler's law to get the semi-major axis and period.

$$
\alpha = 45 \mathrm{au} \,\,(\dfrac{M\_\mathrm{tot}}{20 M\_\circ}) (\dfrac{20 \mathrm{km s}^{-1}}{v\_\mathrm{pl}})^2 \\
P\_\mathrm{orb} = 68 \mathrm{yr} \,\, (\frac{\alpha}{45 \mathrm{au}})^{3/2}(\dfrac{20 M\_\circ }{M\_\mathrm{tot}})^{1/2}$$
Probably assuming a circular orbit, as orbits with high eccentricity would not be stable in a system with a central binary star.


---

