# What is the derivation of this formula for eclipsed flux?

From my lecture notes:

Planetary Transit Searches

If a planetary system’s orbital plane lies along our line of sight, planets will from time to time pass in front of their star, absorbing some of the light from the star that would otherwise reach us. This kind of thing can be seen in our own Solar System where Venus or Mercury can be seen to pass in front of the Sun. The last transit of Venus was in June 2012. Planetary transits will cause a small, but potentially measurable, dip in the brightness of a distant star observed from Earth. What flux decrease will a planetary transit produce?

If the uneclipsed flux of the star is $$F_s$$, the eclipsed flux $$F_t$$, the flux of the planet is $$F_p$$, the radius of the star is $$R_s$$ and of the planet is $$R_p$$ then: $$F_t=F_s-\left(\frac{R_p}{R_s}\right)^2F_s+F_p\tag{1}$$

But no derivation is given for eqn $$(1)$$.

So using the flux-luminosity definition $$F=\frac{L}{4\pi d^2}$$ to try to get a relation between the distances $$R_s$$ and $$R_p$$ and specializing to the cases of $$F_s$$ and $$F_p$$, hence, $$F_s=\frac{L_s}{4\pi {R_s}^2}\qquad \text{&}\qquad F_p=\frac{L_s}{4\pi {R_p}^2}$$ where $$L_s$$ is the luminosity of the star. $$\frac{F_s}{F_p}=\left(\frac{R_p}{R_s}\right)^2\implies F_s=\left(\frac{R_p}{R_s}\right)^2F_p$$

So, I can't even verify the formula is correct (let alone derive it), I have no idea where this formula, $$(1)$$, even comes from. I searched the internet but couldn't find a derivation for it. Does anyone know of a derivation for $$(1)$$ or point me towards some resource that goes through the derivation of it?

Quotation used in this post is from ICL dept. of Physics

## 1 Answer

The uneclipsed flux of a star, as seen on Earth, is given by $$$$F_s=\frac{L_s}{4\pi d^2}\,,$$$$ where $$d$$ is the distance from Earth to the star.

A planet eclipsing the star blocks the flux coming from part of the star's surface. How much it blocks depends on the cross-sectional area of the planet: $$$$F_s^\mathrm{blocked} = \frac{A_p}{A_s}\,F_s\,.$$$$ For spherical planet (and thus circular cross-sectional area), $$A_p=\pi R_p^2$$ and similarly for $$A_s$$, so the ratio reduces to $$R_p^2/R_s^2$$. (I'm assuming that the distance between the Earth and star/planet is much greater than the distance between star and planet.)

Finally, one has to take into account the flux of the planet, $$F_p$$.

Combining the above, the eclipsed flux of the star can be written: $$$$F_t = F_s - \frac{R_p^2}{R_s^2}\,F_s + F_p\,.$$$$

• That's great, but might I ask why you are not using the surface area of the hemisphere ($2 \pi R^2$)? It makes no difference to the end result, but, mathematically it is the hemisphere of the planet that intercepts the flux from the star (even if the star and planet are vastly separated). Thanks for your answer. Apr 26, 2020 at 10:07
• @Electra Yes, you're right. I was visualising the star and planet as overlapping circles on a 2D plane when I wrote that. Though as you say, their ratios will be the same regardless. Apr 26, 2020 at 11:18