I'm trying to calculate the transit probability for a planet orbiting a star. Here's what I know:

Mass of planet: 1.8 Jupiter masses
Period: 2.2047 days (0.0332 years)
inclination, i: 80.8 degrees
Eccentricity, e: 0.0
Radius of star: $2.05R_\odot$
Radius of planet: $1.51R_{Jupiter}$

Here's what I've tried so far, but the result is too high for me to believe it's correct :')

$p_{tr} = \frac{R_{st}+R_{pl}}{a(1-e^2)}$ (this is a formula from the book)
finding a from $P^2=a^3$ to be 0.0331659 (I think this is where I might be wrong)
$P_{tr}=\frac {2.05*R_\odot+1.51*R_{Jupiter}}{a*(1-e^2)}=4.64E10$

It's assumed that the orbit is circular.

I've already looked through similar questions asked in here without success

  • 3
    $\begingroup$ First: probability must be in interval 0.0-1.0. Second: to determine size of orbit you need mass of the parent star. $\endgroup$
    – Leos Ondra
    Dec 29, 2022 at 20:41
  • 1
    $\begingroup$ As pointed out in the comment to James K's answer, if you know the orbit fully (as in this case), you can work out if it is eclipsing or not, hence no probability is needed. Only if you don't know the orbit sufficiently, in particular if you don't know the inclination, you can define the probability as the fraction of possible orbits which are eclipsing. $\endgroup$
    – Walter
    Dec 30, 2022 at 13:16
  • $\begingroup$ @LeosOndra Why does the interval need to be 0.0-1.0 seconds? I don't know the mass of the parent star though :/ I don't know if it makes a difference, but in the previous question it is shown that $m(x)=\frac{75 M} {7} (\frac {x^3}{3}-\frac {x^4}{5}-\frac {x^5}{25})$ $\endgroup$
    – C H
    Jan 2, 2023 at 17:45
  • $\begingroup$ No. The probability is pure number. Second is adjectivum in the second sentence :-) $\endgroup$
    – Leos Ondra
    Jan 2, 2023 at 21:23

1 Answer 1


First we need the mass of the star. If it is a main sequence star, one can estimate using M = R^(1.25) (source) Units are in solar mass/radius

This gives an estimated mass of 2.45 Solar masses. (Looking at some real stars, this might be a bit too high, but this doesn't change the final conclusion)

With this one can calculate the distance of the star from the planet. Using Kepler's laws: 6.7 million km or 9.6 solar radii

With an inclination of 80.8, the planet will pass the star at $9.6\times\cos(80.8)$ or 1.5 solar radii.

Since the star is 2.05 solar radii, that is well within the stars disc, so transit is certain.

Here is a simulated picture of the planet transiting
enter image description here

There is some uncertainty resulting from the initial estimate of the star's mass. If the mass is lower, the planet would have to orbit more closely, and so the transit is still certain to occur. Also the star is assumed to be spherical, and that isn't the case for some stars. But for any reasonable estimate of the star's properties, a transit is all but certain.

  • 1
    $\begingroup$ The real point is that if you know the inclination you probably can decide certainly if a transit occurs. Transit probability calculations occur when you don't know the inclination. $\endgroup$
    – James K
    Dec 29, 2022 at 21:23
  • $\begingroup$ Also, this planet is fried, it is orbiting an A0 star at less than 5 stellar radii... it is going to be seriously hot, at least on the side facing the star. $\endgroup$
    – James K
    Dec 30, 2022 at 13:25

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