Background: As a planet orbits around a star, the star's brightness periodically dims as shown in the following picture. By measuring the variation of brightness, we are able to deduce information about the system.

enter image description here

Source: http://blog.timesunion.com/weather/kepler62/2770/

Problem description: To calculate the maximum variation of brightness of a star with an orbiting planet.

Attempted solution:

Let $\ell_\text{max}$ be the brightness of the star (from the earth's POV) when the planet doesn't cross the star and $\ell_\text{min}$ when the planet crosses in its entirety the star. This is the max variation scenario and the variation is given by: \begin{align*} \left(\Delta \ell \right)_\text{max} = \frac{\ell_\text{min} - \ell_\text{max}}{\ell_\text{max}} \end{align*}

When the planet doesn't cross the star: \begin{align*} \ell_\text{max} = \frac{L_\text{star}}{4\pi r_\text{es}^2} \end{align*}

where $r_\text{es}$ is the distance between earth and star (from center to center).

When the planet crosses the star, then the power per $m^2$ that reaches the planet's surface is: \begin{align*} \frac{L_\text{star}}{4\pi r_\text{ps}^2} \end{align*}

where $r_\text{ps}$ is the distance between planet and star. Therefore, the power that is blocked by the entire surface of the planet is: \begin{align*} \frac{L_\text{star}}{4\pi r_\text{ps}^2} \pi R_\text{planet}^2 = \frac{L_\text{star}}{4} \left( \frac{R_\text{planet}}{r_\text{ps}} \right)^2 \end{align*}

The apparent brightness of the star, as seen from earth, will be (EDIT: the way we spread the power loss to the surface of a sphere with radius $r_\text{es}$ is most likely wrong): \begin{align*} \ell_\text{min} &= \ell_\text{max} - \frac{L_\text{star}}{4} \left( \frac{R_\text{planet}}{r_\text{ps}} \right)^2 \frac{1}{4\pi r_\text{es}^2} \\ &= \ell_\text{max} - \frac{\ell_\text{max}}{4} \left( \frac{R_\text{planet}}{r_\text{ps}} \right)^2 \end{align*}

\begin{align*} \left( \Delta \ell \right)_\text{max} = \frac{\ell_\text{min} - \ell_\text{max}}{\ell_\text{max}} = -\frac{1}{4} \left( \frac{R_\text{planet}}{r_\text{ps}} \right)^2 \end{align*}

Question 1: I've come accross some sites on the Internet and they give a result that is $\left(R_\text{planet}/R_\text{star}\right)^2$. Basically they take the ratio of planet's area to star's area without any more detailed calculations.

Which makes me think that my calculations are wrong. Any insights ?

Qestion 2 (added afterwards): Why doesn't the transit depth depend on the distance between the planet and its star (or between us and the planet) ? If the planet was very very close to earth, wouldn't it cause an eclipse of the star ?

  • $\begingroup$ If an Earth-sized, solar-mass white dwarf were orbited by a Jupiter-sized planet, it could be completely eclipsed, so the answer is clearly that the brightness can vary by 100%. $\endgroup$
    – rob
    Jun 14, 2014 at 23:41

1 Answer 1


The inconsistency comes from assuming the planet has a greater-than-infinitesimal size while leaving the star as a point source.

Usually in transit diagrams we think of the star as a disk of radius $R_\mathrm{star}$ emitting parallel rays perpendicular to its surface. The $\pi R_\mathrm{star}^2$ area of the star is partially blocked by the $\pi R_\mathrm{planet}^2$ area of the planet. This is because, to first order, all $\pi R_\mathrm{star}^2$ of the star is emitting the same amount of light in our direction (each point on the surface emits isotropically in all $2\pi$ steradians outward).

If you want to look at the energy intercepted by the planet, things get more complicated. You found the total reduction in luminosity (turned into a brightness via the Earth-star distance) if we were to capture starlight from all $4\pi$ steradians around the star. But of course there is no reduction in the brightness seen by a small detector when the planet is not blocking the line of sight to the star, and at the same time the reduction is greater when the planet is actually transiting.

Take a look at this eclipse diagram from Wikipedia:


Imagine drawing a large sphere around the star. If I define $L_\mathrm{min/max} = 4\pi r_\mathrm{es}^2 \ell_\mathrm{min/max}$, then the $L_\mathrm{max} - L_\mathrm{min}$ you calculated is the change in the total power reaching this sphere between the cases of not having a planet and having a planet. But you can't take that reduction in power and distribute it evenly over the sphere, since during a transit we are in either the penumbra or antumbra, and the amount of light blocked is viewing angle-dependent.

The question you answered correctly is "how much power and power per unit area escapes the star in all directions given there is a planet in orbit?" The problem is during the transit we have a special line of sight. You can't take total power (equivalently, power per unit area averaged over $4\pi r^2$) and propagate it out to our location.

  • $\begingroup$ Hi Chris and thank you very much for taking the time to help me! I understand the first paragraph and basically it answers my question why the solutions I found across the net used the ratio of the cross sections. What I still miss is why I calculated the starlight across all $4\pi$ since I divide by $4\pi r_j^2$ everytime to get the power me $m^2$ and then multiply with the maximum cross section. $\endgroup$
    – stathisk
    Jun 14, 2014 at 12:25
  • $\begingroup$ @Zet see if the added explanation helps $\endgroup$
    – user10851
    Jun 14, 2014 at 12:54

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