Worksheet 13.2, Problem 2: Exoplanet Transit

Now draw the star projected on the sky, with a dark planet passing in front of the star along the star’s equator.
(a)  How does the depth of the transit depend on the stellar and planetary physical properties? What is the depth of a Jupiter-sized planet transiting a Sun-like star?
(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?
(c)  What is the duration of “ingress” and “egress” in terms of the physical parameters of the planetary system? 

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

The transit depth is the relative decrease in measured luminosity of a star when a planet passes in front of it. Since the seat and planet are very far way, we can approximate them as disks.  Intuitively, the relative luminosity of a star is the amount of surface area that is uncovered thus the transit depth would be a ratio of the covered surface area to the total surface area.

\[\Delta L=\frac{A_{planet}}{A_{star}}=\frac{\pi R_p^2}{\pi R_*^2}=\frac{R_p^2}{R_*^2}\]

\[\boxed{\Delta L=\frac{R_p^2}{R_*^2}}\]

We can now apply this to a Jupiter sized planet, given that \(10R_{Jup}=R_*\)

\[\Delta L=\frac{R_{Jup}^2}{100R_{Jup}^2}=\frac{1}{100}=\boxed{1\%}\]

That's a small decrease, but it is still observable.

(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?

Since the system is so far away, we can approximate again.  We assume that the semi major axis a of the planet is much larger than the solar radius, making this much simpler.  

So, for the planet to pass fully across the star, it needs to travel a distance \(2R_*\).

Since we do not know \(a\), we should use angular diameter instead.  

\[S=\theta a\]

Since arc length is about \(2R_*\), 

\[2R_*=\theta a\]

\[\theta = \frac{2R_*}{a}\]

The planet needs to pass through \(\theta\).  We can get the time for this through a proportion.

\[\frac{P}{2\pi}=\frac{t}{\theta}\]

\[t=\frac{P\theta}{2\pi}\]

\[t=\frac{P\frac{2R_*}{a}}{2\pi}\]

\[t=\frac{PR_*}{\pi a}\]

It's time to revisit Kepler 3rd Law.  

\[P^2=\frac{4\pi^2 a^3}{GM}\]

\[P=2\pi a \sqrt{\frac{a}{GM_*}}\]

Let's substitute this in for P.

\[t=\frac{2\pi a \sqrt{\frac{a}{GM_*}}R_*}{\pi a}\]
\[\boxed{t=2 R_*\sqrt{\frac{a}{GM_*}}}\]

There is our answer, dependent only upon measurable fundamental quantities of the system.

(c)  What is the duration of “ingress” and “egress” in terms of the physical parameters of the planetary system? 

This is pretty much a modification of the prior part.  For an ingress or egress the planet has to move through its own diameter to enter or exit the transit of the star.  Thus instead of moving through \(2R_*\) it now moves through \(2R_p\).  This may seem cheap, but its totally valid: we can simply substitute \(R_*\) with \(R_p\) in our previous answer to get our new answer.  Crafty eh?

\[\boxed{t=2 R_p\sqrt{\frac{a}{GM_*}}}\]

I worked with G. Grell and S. Morrison on this problem.