[SPACE] Solar Eclipses on Other Worlds

[Trachmyr]

How would one go about determining the frequency, totality and duration of solar eclipses on other planets?

In the setting I run, the solar eclipse on a particular world has been stated to have a major effect of the natives way of life and activity cycles, but I’ve never actually went in to details (as I didn’t really have any other than a fuzzy idea of the effects). Now I will be running a game where the effect really needs to be known, so I was wondering how to calculate it.

Pertinent Information:

• The Star is Alpha Centauri B {K1V, Mass 0.907, Luminosity 0.50, Radius 0.865}

• The Planet is Altea (aka Thalassia) {Diameter 3.9K Miles, Apparent Rotation 540 Hours*, Sidereal Rotation 486 Hours, Orbital radii 0.65 AU (0.64 to 0.66 due to Eccentricity), Orbital period 0.55 (4860 Hours), Axial Tilt is negligible}

• The Satellite is Celestia (actually a Binary Planet) {Diameter 3.4K Miles, Orbital period is 486 Hours*, Orbital separation is 114K Miles, Axial Tilt is 32%)}

• [I]*Altea and Celestia are Tide-locked to each other, and are in fact a binary system. They are in geosynchronous orbit, and the same face of Altea always faces the same face of Celestia. This region on Altea is called the Moonsea (no connection to the D&D FR region of the same name).

I was wondering what the appearance of the Solar eclipse would be like if one were in the “Moonsea”, and how large the ‘moon‘ appears to be in the sky. Anyone know how to figure this out? (Well I’m sure there is… anyone willing to tell me how to figure it out?)

Thanks,
-Trachmyr

[Diomedes]

In the setting I run, the solar eclipse on a particular world has been stated to have a major effect of the natives way of life and activity cycles, but I’ve never actually went in to details (as I didn’t really have any other than a fuzzy idea of the effects). Now I will be running a game where the effect really needs to be known, so I was wondering how to calculate it.

As a guess, it would be when "day" at the Moonsea matches up with a vernal or autumnal equinox.

I was wondering what the appearance of the Solar eclipse would be like if one were in the “Moonsea”, and how large the ‘moon‘ appears to be in the sky. Anyone know how to figure this out? (Well I’m sure there is… anyone willing to tell me how to figure it out?)

The angular width of the moon would be double the arctangent of the lunar radius divided by the lunar distance. The Moon is about half a degree. By an amazing coincidence, the sun is also half a degree, which results in the sun being blocked, but not the corona. So Earth's solar eclipses are more interesting than they might otherwise be.

[Trachmyr]

Originally Posted by Diomedes

The angular width of the moon would be double the arctangent of the lunar radius divided by the lunar distance. The Moon is about half a degree. By an amazing coincidence, the sun is also half a degree, which results in the sun being blocked, but not the corona. So Earth's solar eclipses are more interesting than they might otherwise be.

Trig was never my strong math, but I get 1.7 degrees... making it appear 3.4times as wide as our own moon. While the sun would be 0.53 degrees, making it appear just a fraction larger than our own. Is that about right?

So if the planet has a 540 hour apparent rotation... thats one and a half hours per degree. If the moon was not inclined to the planet's orbit, then that mean a eclipse would only last about 3 hours and 20 minutes, being a total eclipse for just 1 hour and 45 minutes. Is that right? Much shorter than I was hoping for.

[Agemegos]

Originally Posted by Trachmyr

Trig was never my strong math, but I get 1.7 degrees... making it appear 3.4times as wide as our own moon. While the sun would be 0.53 degrees, making it appear just a fraction larger than our own. Is that about right?

So if the planet has a 540 hour apparent rotation... thats one and a half hours per degree. If the moon was not inclined to the planet's orbit, then that mean a eclipse would only last about 3 hours and 20 minutes, being a total eclipse for just 1 hour and 45 minutes. Is that right? Much shorter than I was hoping for.

You've overestimated totality. You have calculated the time it takes for the moon to traverse an apparent longitude, i.e. the time from first contact to third contact or from second contact to fourth. But totality lasts only from second contact to third contact. You have to subtract the apparent width of the sun from that of the moon, i.e. 1.17 degrees. Which gives 1 hour 45 minutes maximum duration of totality. (Totality lasts a shorter time towards the northern and southern edges of the totality zone.)

Also, don't forget that a solar eclipse is visible only from a narrow arc on the planet's surface: totality is observable from a strip as wide as the moon's apparent diameter minus the sun's apparent diameter, and at least partial eclipse from a strip as wide as the sum of their apparent diameters. Two-and-a quarter degrees of latitude see any particular eclipse, and about half that see any part of totality. If the inclination of the moon's orbit to the planet's equator is close to zero (likely: tidal forces will tend to reduce obliquity as they increase separation and synchronise rotations) the solar eclipses will be seen only close to the equator. The maximum latitude of the centre-line of the path of totality is arccos {(radius of moon's orbit / radius of planet) * cos (obliquity of moon's orbit)}. If that comes out undefined some eclipses will miss the planet to the north or south, and eclipses will only be visible around ascending node and descending node, i.e. if the moon is near to the points where it crosses the sun's latitude at the time when it happens to be passing the sun's longitude.