OWAO 2022, Теория — Задача 3. Anti-Earth

• The Earth crosses the minor axis earlier, by Δτ = (e/π)T⊕ = 1.94 d ≈ 2 d; Anti-Earth is then visible next to the Sun (c = ea = 2.5·10⁶ km = 3.6 R☉).

Using Kepler’s second law and observing the symmetry, one may conclude that

as an area of a quarter of the ellipse is $A_1 = \frac{A}{4} = \frac{\pi}{4}ab$, and the area of the highlighted triangle is $A_2 = \frac{1}{2}bc = \frac{e}{2}ab = \frac{e}{2\pi}A$.

The Earth crosses the minor axis earlier,

The distance between the center of the Sun and the center of the Earth’s orbit is

Therefore, Anti-Earth is definitely visible next to the Sun at that moment. The fact that it did not reach the minor axis only improves the conditions for its visibility.

Marking Scheme:

• Correct understanding of the situation, providing the correct diagram — 2 pt.

• Estimation of the time interval.

  In case of using Kepler’s second law:

  • The law is properly mentioned or formulated — 1 pt,
  • Difference of areas and time intervals — 2 pt,
  • Correct value of the time interval — 1 pt.

  In case of using Kepler’s equation:

  • The equation is correct — 1 pt,
  • Difference of mean anomalies is calculated — 2 pt,
  • Correct value of the time interval — 1 pt.

• Describing the visibility:

  • Full justification of visibility is given — 3 pt: The linear eccentricity is calculated and compared with the position of the Sun,
    OR the eccentric anomaly is calculated and it is shown that it exceeds $90^{\circ}$ by more than the angular radius of the Sun,
    OR the true anomaly is calculated from the eccentric and the cosine theorem is used.
    Particular case: just the fact that the Earth moves $2^{\circ}$ in 2 days — 1 pt.
  • Correct answer — 1 pt.