IAO 2003, Теория — Задача 4

• $T\approx1.07$ години, $a\approx1.047$ au, $m\approx-8.0^m$.

Official English sketches for solutions — theosol.doc.

If the next opposition of a hypothetical «Mars-2» will be in summer 2018, it means that the synodic period of «Mars-2» is equal approximately to $T_S = 15$ years. So the sidereal period of «Mars-2» $T_{M2}$ may be found from the formula:

where $T_E$ is sidereal period of Earth,

To find semiaxis $a_{M2}$ of the orbit of «Mars-2» we can use Kepler's 3rd Law:

so $a_{M2} = a_E\cdot T_E\cdot T_S / (T_S - T_E)\}^{3/2} = 1.047\ \text{a.u.}$

Assuming that the orbit of «Mars-2» is circular, the minimum distance to Earth at opposition of this planet will be at 0.047 a.u., i.e. with an apparent size $0.524/0.047 = 11.1$ times smaller than in mean opposition of Mars when its magnitude is equal to $-2.0^m$. «Mars-2» is also closer to Sun with an apparent size $1.524/1.047 = 1.46$ times. $5\cdot\lg(11.1 \times 1.46) = 6.0$, $-2.0^m - 6.0^m = -8.0^m$! So the hypothetical «Mars-2» might be as bright as $-8.0^m$ at opposition.

Бележка за източника: последната формула за голямата полуос е отпечатана с несдвоена скоба и степен 3/2, въпреки предходния трети закон на Кеплер. Изразът е запазен дословно. Запазено е и „times smaller“, въпреки числото 11.1 и получения по-голям блясък.