I understand that multiple transit detections of an exoplanet are almost always used to derive its period, but is there a way to do it with only one transit detection? As in, with only one available light curve? I'm assuming we already know the masses/radii of the exoplanet and host star. I've searched online for an answer on this, and everyone seems to agree that you can at least get a rough estimate of the exoplanet's orbital period with only one transit detection, but nobody seems to explain how. The best equation I've gotten so far is from using Kepler's third law, where $$T=\frac{2π }{\sqrt{G(M+m)}}a^\frac{3}{2}$$ (with a being the semi-major axis) Of course, I don't think I can find the semi-major axis with just one transit observation. Is there any other way to do this?


1 Answer 1


If you assume that

  • the line of sight to the Earth lies exactly in the plane of the exoplanet's orbit, and
  • the orbit is roughly circular (negligible eccentricity),

then you can do this. Start with the fact that under these assumptions, the transit will take place over a segment of the exoplanet's orbit that is approximately the same length as the star's diameter $d$. Draw out some diagrams of light rays to see why this is so.

Moreover, a planet in a circular orbit moves at a constant velocity throughout its orbit. This means that the ratio of the transit time to the orbital period will be the same as the ratio of the star's diameter (the segment of the orbit during which transit takes place) to the orbital circumference. In other words, $$ \frac{t_\text{transit}}{T} = \frac{d}{2 \pi a} . $$ These facts coupled with Kepler's Third Law would let you solve for $a$ give the star's diameter, its mass, and the transit time.

That said, the above two assumptions are frequently violated by exoplanets, and would lead to errors in this estimate. So any such estimate should be taken with a grain of salt.

  • $\begingroup$ With exquisite photometry and a good model, I think the shape of the transit itself can give you the impact parameter and hence orbital inclination. But the second assumption seems unavoidable. $\endgroup$
    – ProfRob
    Commented Mar 28, 2023 at 6:29

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