How to measure the ratio of a planet's radius to a star?

Question

I was reading a physics problem related to astronomy, and upon re-reading it, I realized that it could be really indicated to extrapolate some really interesting physics-related information. One of these is:

How could we measure the ratio of a planet's radius to a star, for example using transits? The only idea I have is to compare them when the planet passes exactly in front of the star (i.e. they are aligned with our view), but this only makes sense if the distance between the two is much smaller than the distance between us and that star system (which I think is true enough for every system except the Solar system) and if it is possible to obtain such high resolutions (and I already had my doubts about the distance between the two, which should be much greater than their radii anyway).

Answer 1

As for the question as it stands now, this is a fundamental tenet in exoplanet transit observations, see also simply wikipedia:

A planet in front of its star will block an amount of stellar flux roughly proportional to its surface area (modulo limb darkening of the star). Hence the drop in the star's flux $\Delta F$ is proportional to the ratio of planet to stellar radius squared, i.e. $$ \Delta F/F \propto (r_P/r_s)^2 $$

Answer 2

My answer is inspired by an exercise I did recently:

The densities of the Sun and Jupiter are approximately equal, and their masses are in the ratio of $1000:1$. What variation in the apparent magnitude of the Sun would an observer outside the Solar System detect when Jupiter transited in front of the Sun's disk?

So knowing $\Delta m$, we'd get $$\Delta m=m_{\ast p}-m_{\ast}=2.5\log\left(\dfrac{f_{\ast}}{f_{\ast p}}\right)\overset{(\ast)}{=}2.5\log\left(\dfrac{I\Omega_{\ast}}{I\Omega_{\ast p}}\right)=2.5\log\left(\dfrac{\pi\left(\dfrac{R_\ast}{r_{\ast}}\right)^2}{\pi\left(\dfrac{R_\ast}{r_{\ast}}\right)^2-\pi\left(\dfrac{R_p}{r_p}\right)^2}\right).$$

And so,

$$\dfrac{R_p}{R_\ast}=\sqrt{1-10^{-\frac{\Delta m}{2.5}}}\ \dfrac{r_p}{r_\ast}.$$

Where $m\equiv$ apparent magnitude, $f\equiv$ apparent flux, $I_\nu\equiv$ spectral radiance, $\Omega\equiv$ solid angle subtended by the body and the observer, and $r\equiv$ distance between Earth and body.

($\ast$) Note we assumed $$I=\int_0^{\infty}I_{\nu}d\nu$$ is isotropic, thus, $I_\nu$ depends solely on $\nu$ and not any angle: $$f=\iint I_\nu \underbrace{\cos\theta}_{\text{Lambert}} d\nu d\Omega\overset{I_{\nu}=I_{\nu}(\nu)}{=}\int I_\nu d\nu\int\cos\theta d\Omega=I\cdot\Omega$$