IOAA 2023, Теория — Задача 9. Second eclipse

Figure 1: Observed lightcurve for system Bolek.

Figure 2: Observed lightcurve for system Lolek.

Answer Sheet

Figure 3: Predicted lightcurve for the second eclipse for system Bolek.

Figure 4: Predicted lightcurve for the second eclipse for system Lolek.

Solution

For each of the systems, since the eclipses are short we can assume that the angle swept along the orbit during the eclipse is negligibly small, which means that the tangential velocity in relative orbital motion is constant.

System A

A sharp peak minimum in a central eclipse implies that $R_1=R_2$. Therefore, both eclipses will be total, with minima of brightnesses $m_1$ and $m_2$, respectively.

Define the magnitude at minimum as $m_{\mathrm{min}}=m_1$ and the baseline magnitude as $m_{1,2}=m_1+m_2$. From Pogson’s law applied to the two cases: $m_1-m_{1,2}=-2.5\log_{10}(L_1/(L_1+L_2))$ and $m_2-m_1=-2.5\log_{10}(L_2/L_1)$. For the first case, with simple operations we obtain: $L_2/L_1=10^{(m_1-m_{1,2})/2.5}-1$. Plugging this into the second case: $m_2=m_1-2.5\log_{10}(10^{(m_1-m_{1,2})/2.5}-1)$.

As the orbits are near-circular, timescales will be the same for both eclipses. Therefore it is sufficient to ‘copy’ the shape of the primary one while scaling down the depth to $|m_{1,2}-m_2|$. Reading off $m_{1,2}$ and $m_{\mathrm{min}}$ from the plot, we arrive at $L_2/L_1=2$, $m_2=11.44$, and the following figure:

Figure 5: Predicted lightcurve for the second eclipse for system A – correct answer.

Note about marking: Reproducing the exact shape of the eclipse (concave/convex, smooth descent, etc.) is not subject to grading as it requires mm-level precision, and the focus is on the thought process instead.

System B

For the second system, we follow a generally similar scheme, but we see that the minimum is flat. $t_{\mathrm{fall}}$ is the time of the descent from the baseline brightness to the lowest point, i.e., between contacts I and II, while $t_{\mathrm{flat}}$ is the time between contacts II and III during which the lightcurve remains at its lowest point:

From this, we obtain $t_{\mathrm{flat}}/t_{\mathrm{fall}}=2(R_2-R_1)/2R_1$ and, after rearranging, $R_2/R_1=t_{\mathrm{flat}}/t_{\mathrm{fall}}+1$.

Here $R_1$ is the radius of the smaller star and there are two solutions – it could either be (partially) eclipsing or (entirely) eclipsed. Let us first assume that it was eclipsed. After measuring $m_{\mathrm{min}}$ for the eclipse and $m_{1,2}$ for the baseline we again (just like for system A) assume $m_{\mathrm{min}}=m_1$ and write down: $L_1/L_2=10^{(m_1-m_{1,2})/2.5}-1$.

Reading off $m_{1,2}$ and $m_{\mathrm{min}}$ from the plot, we arrive at $L_2/L_1=4$, $R_2/R_1=2$.

$L_i=S_i\pi R_i^2$, where $S_i$ stands for the surface brightness of each star.

$\Longrightarrow S_1=S_2=S$ (or $T_{\mathrm{eff},1}=T_{\mathrm{eff},2}=T_{\mathrm{eff}}$).

In such a case, it does not matter if star 2 is eclipsing or eclipsed. During the baseline, we always see two disks with a total brightness of $S(\pi R_1^2+\pi R_2^2)$, while during the eclipse, we see one disk of radius $R_2$ and total brightness $S\pi R_2^2$.

Note about marking: This solution can be independently verified by assuming that the smaller component was eclipsing and cutting out a disk of surface $\pi R_1^2$ and brightness contribution $S_1\pi R_1^2$ in the star of surface $\pi R_2^2$ and surface brightness $S_2$. Instead of $m_{\mathrm{min}}=m_1$ of star 1, we would have measured $m_{\mathrm{min}}=m_{\mathrm{ecl},1}$ of a combination of the two, with a brightness $L_{ecl1}=S_1\pi R_1^2+S_2\pi(R_2^2-R_1^2)$. The same conclusion $S_1=S_2=S$ will follow. Any solution correctly arriving at this conclusion and the correct figure – no matter the order of steps taken – should be scored equally. For the maximum number of points, however, the student should make a convincing point that this is the only solution, i.e. lifting the previously made assumption (or exploring both cases).

To draw the second eclipse, one must simply copy the shape of the first one:

Figure 6: Predicted lightcurve for the second eclipse for system B – correct answer.