OWAO 2022, Теория — Задача 8. Perfect Nebula

• Only an upper limit: with τ > 1/8, lg r < 4.03, i.e. r ≲ 11 kpc.

Let us assume that the radius of nebula $l$ is much more than radius of planet’s orbit $R$. The star luminosity is $B$, optical depth of nebula is $\tau$. The planet is inside the nebula, so the energy flux from the star near the planet is

The energy scattered by the nebula is

Angular element $d\Omega$ of the nebula scatters

and that results in illuminance near the planet

Energy flux on the planet surface from an angular element is

Thus, total amount from sky hemisphere is

Here the fact that average $\langle\cos\theta\rangle = \frac{1}{2}$ over the hemisphere is used.

The subsolar point of the planet receives $J_S + J_N$, while the opposite point receives $J_N$. Recalling Stefan–Boltzmann’s law, we conclude that

Therefore,

The latter equation has no solutions for $\tau$, unless we consider the “close model” $l \gtrsim R$ with rough approximation. In this case the nebula fragment near the planet can be considered as a flat layer with optical depth $\tau < 1$, illuminated with the energy flux $J_S$. It scatters $\tau$-part of the incident flux to the outer and inner side of the nebula:

To meet the temperature relation,

this gives the lower limit value of $\tau_{\min} = \dfrac{1}{8}$. Generally speaking, we have to write the final condition on the optical depth:

The Sun has the absolute magnitude $M_\odot = +4.7^{\mathrm{m}}$ and apparent magnitude $m = +20^{\mathrm{m}}$. Taking the logarithm of Beer’s law (1), we obtain the equation for the distance $r$:

Finally,

Marking Scheme:

• Distance determination, without the absorption — 2 pt.
• Estimation of the ratio of energy flows — 2 pt.
• Understanding that the condition can be fulfilled at different thicknesses, there is only a lower limit — 2 pt.
• Adequate energy transfer model (not necessarily the same as above) — 2 pt.
• Distance determination, taking absorption into account:

  • Inequality — 2 pt,
  • Equality only — 1 pt. in particular.