Blog #19 - Transiting Exoplanet pt. 2

Transiting Exoplanet pt. 2

2. Now draw the star projected on the sky, with a dark planet passing in front of the star along the star’s equator.

The picture above depicts a planet transiting in front of its host star as observed from Earth

(a) How does the depth of the transit depend on the physical properties of the star and planet? What is the depth of a Jupiter-sized planet transiting a Sun-like star?

The optical depth or, \(\delta F\), is solely dependent on the cross sectional areas of the planet and the star, thus the radius is the only physical property we care about when considering depth of transit. \[\delta F = \frac{A_p}{A_{\star}} = \frac{\pi R_p^2}{\pi R_{\star}^2} = \frac{R_p^2}{R_{\star}^2} \] From the previous problem we know for a Jupiter-sized planet and a Sun-sized star, \[R_{\star} \approx 10*R_p\]Substituting into the equation for optical depth, we get \[\delta F = \frac{R_p}{10 R_p} = 10 \% \space transit \space depth\]

(b) In terms of the physical properties of the planetary system, what is the transit duration, defined as the time for the planet’s center to pass from one limb of the star to the other?

We know that for an full transit to occur, the planet's center must travel a distance of \(2 R_{\star}\) if observing the planet as traveling along a line at the equator of the star. We can use the simple relation between distance, velocity, and time to get the total transit time. \[t = \frac{d}{v}\]and \[v = \frac{2 \pi a}{p} \]This gives us \[t = \frac{2 R_{\star} P}{2 \pi a}\]To get the answer in terms of solely physical properties we can convert the orbit period, P, to \[P = 2 \pi \sqrt{\frac{a^3}{MG}}\] leading to a final answer of \[t = 2 R_{\star} \sqrt{\frac{a}{M_{\star}G}} \]Where \(a\) is the semi-major axis, M is the mass of the star, and G is the universal gravity constant.

(c) What is the duration of “ingress” (from no transit to full transit) and “egress” (from full transit to no transit) in terms of the physical parameters of the planetary system?


We know that given a spherical star and planet, the time of ingress must equal the time of egress\[t_{ingress} = t_{egress} \]Following the same method as part (b) we can solve for the time for the planet to move across the star's surface the distance of one planetary radius equating to the ingress transit time as seen in the diagram below.

From this diagram we can solve the time of transit using the same relation of distance, velocity, and time. \[t = \frac{d}{v} = \frac{R_p}{\frac{2 \pi a}{P}} = R_p \sqrt{\frac{a}{M_{\star}G}}\]