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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).

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  • $\begingroup$ This is a wall of text, not a question. It needs significant reworking to be a question (particularly, it needs a question mark somewhere) $\endgroup$ Commented Nov 8, 2023 at 21:51
  • $\begingroup$ Thank you for the warning. I have rearranged the questions, parceling out the content, despite the fact that I was unable to delete anything in my text. If you would like, please provide me with more advice on how to word the question better. $\endgroup$
    – Bml
    Commented Nov 8, 2023 at 21:59
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    $\begingroup$ There is a one question per post limit. You've five here. I'm also pretty sure some of these questions have answers already, if not here then on Astronomy. Please do a search for questions like these on the sites. $\endgroup$ Commented Nov 8, 2023 at 23:07
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    $\begingroup$ Sorry for posting all these questions. I'm reducing the number of questions to one by keeping one of the ones I proposed. $\endgroup$
    – Bml
    Commented Nov 8, 2023 at 23:26
  • $\begingroup$ Related (but not a duplicate): Is there a way to calculate/estimate the orbital period of an exoplanet from only one transit detection? $\endgroup$ Commented Nov 9, 2023 at 12:20

2 Answers 2

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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. $$ \Delta F/F \propto (r_P/r_s)^2 $$

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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: $$f=\iint I_\nu \underbrace{\cos\theta}_{\text{Lambert}} d\nu d\Omega\overset{I_{\nu}=I_{\nu}(\nu)}{=}\int I_\nu d\nu\int\cos\theta d\Omega=I\cdot\Omega$$

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  • $\begingroup$ Let me know whether this has any flaws. $\endgroup$
    – Conreu
    Commented Nov 9, 2023 at 0:47
  • $\begingroup$ Thank you for your reply. Could you clarify a few more steps? For example, where does $2.5 come from in your operations, etc...? $\endgroup$
    – Bml
    Commented Nov 10, 2023 at 19:43
  • $\begingroup$ It is defined like this so that a body with magnitude m is 100 brighter than a m+5 magnitude one. $\endgroup$
    – Conreu
    Commented Nov 11, 2023 at 15:58
  • $\begingroup$ Thanks, I think I got it. Maybe you meant that the brighter an object is, the lower its magnitude number: a difference of $1.0$ in magnitude corresponds to a brightness ratio of ${\displaystyle {\sqrt[{5}]{100}}} \approx 2.512$. Right? $\endgroup$
    – Bml
    Commented Nov 11, 2023 at 16:09
  • $\begingroup$ Yes, like pH scale: lower the number, higher the acidity (brightness). $\endgroup$
    – Conreu
    Commented Nov 11, 2023 at 18:07

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