# Visibility of a planet on the other side of the Sun

Context: I've been having a discussion with my vectors and mechanics professor (course Mathematics BSc) about a problem on a recent coursework. The following is the model I came up with. I'm not an astronomer, so bear with me.

Define:

1. $$S$$ the Sun;
2. $$E$$ the Earth;
3. $$P$$ some other planet in the solar system;
4. $$O$$ the position of an observer somewhere on $$E$$;
5. $$0 the latitude of $$O$$;
6. $$\Pi_E$$ the orbital plane of $$E$$, the $$xy$$ plane;
7. $$\Pi_P$$ the orbital plane of $$P$$;
8. $$\Pi_O$$ the plane tangent to $$E$$ at $$O$$;
9. $$\alpha$$ the acute angle between $$\Pi_E$$ and $$\Pi_P$$;
10. $$\beta=90-l$$ the acute angle between $$\Pi_E$$ and $$\Pi_O$$.

Assume:

1. The local time at $$O$$ is midnight;
2. The date is the Summer solstice;
3. The visual sizes of $$P$$ and $$S$$ are both non-zero but otherwise negligible;
4. The projections of $$E$$, $$S$$ and $$P$$ onto the $$xy$$ plane are roughly but not exactly collinear;
5. The solar system is Euclidean;
6. Light travels in infinite straight lines.

Given these assumptions, I think that $$P$$ is visible from $$O$$ if and only if $$\alpha>\beta$$. Am I right?

• I've down-voted because this seems to have no value to anyone other than the person who is taking this quiz. It seems needlessly difficult to read, what's wrong with writing your question in plain English? – uhoh Dec 22 '19 at 9:39
• I'm voting to close this question as off-topic because it is a geometry puzzle but not a question about Astronomy as defined in the help center. – uhoh Dec 22 '19 at 9:40
• @uhoh I'll let you be the judge of relevance, since I'm new here. I don't know why you find it difficult to read. – mjc Dec 22 '19 at 19:02
• How do you find a stellar system where the planets are bigger than their stars? It is possible only if the star is extraordinarily small (like a white dwarf or a neutron star). It is impossible in ordinary star systems. – peterh - Reinstate Monica Dec 23 '19 at 10:25
• @peterh-ReinstateMonica Negligible visual size of $S$ is a simplifying assumption; removing that assumption would just mean accounting for the arc length of the heavens, as seen from $O$, covered by $S$, which is not that much work but more than I wanted to do. – mjc Dec 23 '19 at 14:05

Lets put our observer on the equator. So the angle between his tangent plane and the Earth's orbit (at local midnight on the solstice) is 90 - 23.5 = 66.5 degrees. This is your $$\beta$$

Let's put our planet in orbit very close to the ecliptic, so the angle between the earth's and the planet's orbit is small (say Jupiter, with an inclination of 1.3 degrees) This is your $$\alpha$$

Let us further suppose that Jupiter is at opposition. It will culminate at midnight and be very much visible (indeed it will be about 23.5 +-1.3 degrees from the zenith), but $$\beta>\alpha$$.

It is not surprising. Nearly all the planets orbit close to the ecliptic. so $$\alpha$$ is typically very small (a few degrees) If a planet were only visible when $$\alpha>\beta$$, then planets would rarely be visible.

In more conventional astronomical terms:

• ΠE is the ecliptic
• ΠO is the observer's horizon
• α is the inclination of P's orbit, usually denoted i
• β = 90° - l is the angle between the horizon and the celestial equator

If the order of roughly collinear bodies (assumption 4) is Sun-Earth-P, then P is at opposition and easiest to observe around midnight. If the order is P-Sun-Earth or Sun-P-Earth, then P is in conjunction with the Sun and difficult to observe at any time. In general the angle Sun-Earth-P is the elongation of P. The projection of that angle in the ecliptic plane is the difference between the Sun's and P's ecliptic longitudes.

For a ground-based observer, the Sun brightens the whole daytime sky, making any planet hard to observe unless you have a telescope and are very clever at pointing it. Even for a space telescope, the minimum safe elongation is much greater than the angular diameter of the Sun.

To determine ground-based visibility, the key quantities are the elevation angles of P and the Sun relative to the horizon (your ΠO). The absolute value of either angle is insufficient; P must be above the horizon and the Sun must be below it. Time and date (assumptions 1 and 2) will be inputs to this calculation, but a direct comparison to i (your α) is not a reliable test. Dot products of the observer's zenith vector (normal to the horizon, away from the Earth) with vectors ES and EP might be useful.

• Thanks for the interesting and useful answer. Mapping astronomical nomenclature to my symbols helps. Since for present purposes I'm more interested in the Euclidean geometry, perhaps I can remove your requirement for the Sun to be below the observer's visual horizon (alluded to in my assumption 3). – mjc Dec 24 '19 at 15:59
• @mjc This would only save you a dot product operation; to get EP, you'd have to find both ES and SP anyway. In practice there's a trick to finding even Venus in the daytime. – Mike G Dec 24 '19 at 22:15