IAO 2002, Теория, Група α, български — Задача 2

• $S_E\approx40$ km²; $h\approx750$ km.

The drawing enables to realise that the light from the satellite actually spreads in a cone with angle $\alpha=0.5^\circ$, so that it covers area on the ground

Using the relation $m_1-m_2=-2.5\log(b_1/b_2)$ and taking into account stellar magnitude of the Sun $m_1=-26.8^m$ and magnitude of the flare $m_2=-8^m$, we can find that $(b_1/b_2)\approx3.3\cdot10^7$, and so we see $1/(3.3\cdot10^7)$ part of the Sun through the reflecting plate.

It means that the visible projection of the square of the plate $188\times86$ cm ($S=1.62$ m²) is $1/(3.3\cdot10^7)$ part of $\pi(\alpha h/2)^2$. Projection gives some factor about $2^{1/2}$ which should be taken into account since as usual we see the plate under the angle of $45^\circ$ (see figure):

Taking into account that $\alpha=9.1\cdot10^{-3}$ rad, one can find that $h\approx750$ km and the area on ground covered by the beams from one antenna

Бележка за източника: споменатата рисунка не е приложена в нито един от двата ключа. Руският ключ обсъжда същия проекционен множител, но го пропуска в отпечатаната формула за $S$. Английската формула и отпечатаните резултати са запазени.