IAO 2003, Теория — Задача 1

• $m\approx3^m$: възможно е наблюдение с невъоръжено око.

Official English sketches for solutions — theosol.doc.

Let us compare the light fluxes observed by our "naked eye" from Sputnik I (in zenith at perigee) and from the full Moon.

If $W$ = light flux from the Sun near Earth, $\alpha_*$ = albedo of the observed objects, $S_* = \pi r_*^2 = \pi d_*^2/4$, where $r_*$ and $d_*$ – radius and diameter of the observed objects, $R_*$ = distances from the observed objects to the observer:

Flux from Moon to observer: $I_M = W\cdot\alpha_M\cdot S_M / (2\pi R_{M-O}^2)$.

Flux from Sputnik I to observer: $I_I = W\cdot\alpha_I\cdot S_I / (4\pi R_{I-O}^2)$.

(We do not take into account effects related to diagrams of scattering and random/mirror scattering. In the student's solutions it's also not really necessary to understand difference between the 2 and the 4 in $2\pi R_{M-O}^2$ and $4\pi R_{I-O}^2$.)

Since Sputnik I was made of highly polished aluminium alloy, we may assume $\alpha_I = 1$.

So $I_M / I_I = \alpha_M\cdot(S_M / S_I)\cdot(4\pi R_{I-O}^2 / 2\pi R_{M-O}^2) = 2\alpha_M\cdot(d_M/d_I)^2\cdot(R_{I-O}/R_{M-O})^2$.

Necessary data may be taken from the Table of planetary data. Calculations may then give us:

$1.87\cdot10^6 = 1.87\cdot100^3$, i.e. the difference in stellar magnitudes: $0.7 + 3\cdot5 = 15.7$.

Taking into account that magnitude of the full Moon is $-12.7^m$, the magnitude of Sputnik I (in zenith at perigee) $m = -12.7^m + 15.7^m = 3^m$. One may see objects of up to magnitude $6^m$ with the naked eye, so Sputnik may be visible.