It's a miracle we find anything in the Universe at all

We are bounded in a nutshell of Infinite Space: Week 8: Worksheet #14: Problem #1: It's a miracle we find anything in the Universe at all

1. Draw a planet passing in front of its star, with the star on the left and much larger than the planet on the right, with the observer far to the right of the planet. The planet’s semimajor axis is a.

(a) Show that the probability that a planet transits its star is \(R_\star / a\), assuming \(R_P \ll R_\star \ll a\). What types of planets are most likely to transit their stars?

(b) If 1% of Sun-like stars in the Galaxy have a Jupiter-sized planet in a 3-day orbit, what fraction of Sun-like stars have a transiting planet? How many stars would I need to monitor for transits if I want to detect 10 transiting planets?

a. From the images, we know the planet is transiting in front of the star from our perception of it. Furthermore, we can see that the area the planet could be in in order for us to properly see it is actually smaller than the area of the star, it being farther away, however we can disregard this because of the extreme distance this system is from the observer. Therefore, the chance of observing the planet in front of the star from our point of view is the small cross section of the star we are able to observe directly, and this entire path as a belt around the Sun indicating the possibility of the planet being in our line of sight at one point of its orbit, over the area of the sphere of radius a where the planet could conceivably be: \[ \frac{Area ~of~belt}{Surface ~area ~of ~the ~sphere}\] and now defining the areas of these surfaces: \[ \frac{\sigma_A}{A_{Sphere}} = \frac{2\pi a (2R_\star)}{4\pi a^2},\] which gives an elegant proportion the question asked for: \[ \frac{\sigma_A}{A_{Sphere}} = \frac{R_\star}{a}\]

b. Now assuming we want to find 10 planets, how many stars do we need to look at randomly? Starting off with the equation we just derived, we know the percentage is equivalent to the star and planet’s: \[R_\odot / a = \% ,\] and also using the simplified version of Kepler’s third law relating years to AU, so as to find the A from the information given: \[ P^2 = a^3 ,\] writing the time in years: \[ 3 days~\to \frac{3~days}{365 days/year} \approx \frac{1}{122} ~years,\] Now we plug this into the equation and find the semi-major axis: \[ \left(\frac{1}{122}\right)^2 = a^3,\] \[ a = 0.04 AU.\]  

Next, we plug these into the equation we found earlier, plugging in the radius of the Sun in cm and the semi-major axis which is now turned into the equivalent in cm: \[ 1 AU = 1.5 \times 10^{13} cm,\] \[\frac{R_\odot}{a} = \frac{7\times 10^{10} cm}{1.5 \times 10^{13} cm \cdot 4 \times 10^{-2}} ,\] \[\frac{R_\odot}{a} = \frac{7\times 10^{10} cm}{6 \times 10^{11} cm }\] \[ \frac{R_\odot}{a} = 1.1 \times 10^{-1} .\] This represents the percentage chance of the planet being in an orbit observable to us. Next we multiply this percentage by the percentage of stars with a mass/radius comparable to that of the Sun \(1\%\): \[ 1.1 \times 10^{-1} \cdot 0.01 = 0.0011 = 0.11 \% ,\]  which is the percentage of Sun-like stars with a visible transiting planet. Finally we now compute how many stars we need to observe to see the 10 planets multiplying 10 times the whole of stars divided by the percentage of Sun-like stars with orbiting planets: \[ 10 \cdot \frac{100}{0.1} = 10 \times 10^3 = 10^4 = 10,000, \] meaning you would need to look at a lot of stars to see the 10 transiting planets.