# Let us assume that in the solar system there is a planet “Counter-Earth” of small mass, moving exactly along the orbit

**Let us assume that in the solar system there is a planet “Counter-Earth” of small mass, moving exactly along the orbit of the Earth with a lag of six months. Would it be possible to register this planet using ground-based observatories?**

By the condition of the problem, the mass of the planet is small, and we will not be able to register it by gravitational perturbations on the movements of other celestial bodies. However, oddly enough, we can just see it! Recall that the Earth’s orbit is elliptical with an eccentricity e = 0.017. According to Kepler’s II law, the radius vector directed from the center of the Sun to the center of the planet describes equal areas at equal intervals. Let’s depict the orbit of the Earth (and Counter-Earth) around the Sun in the figure. The sun is at point S, the perihelion of the orbit is at point P, and the aphelion is at point A. We denote the center of the ellipse by O, then the length of the segment OS will be equal to ae, where a is the semi-major axis of the Earth’s orbit. Let the Earth (point E1) be in the middle of the time interval between aphelion and perihelion, and Counter-Earth (E2) between perihelion and aphelion. Then the areas of the parts of the ellipse bounded by the segments SE1 and SE2 coincide. If we assume that the eccentricity of the Earth’s orbit e is small, then from the equality of the areas up to e2, the length of the segment SE0 will be equal to 2ae. This means that when viewed from the Earth, the Counter-Earth will be at an angular distance of 2e rad or 1.95 ° from the Sun. Of course, this is not enough to see Counter-Earth at sunrise or sunset, but it would give us the ability to easily find it during the total phase of solar eclipses.