What kind of lunar orbit would cause a total solar eclipse to happen once a day?

Question

The eclipse needs to be visible from at least one region on the planet every revelation and last about one hour.

Answer 1

Simple. The moon doesn't block the sun.

The habitable moon you're on orbits a larger planet, say a gas giant. Since it would likely be tidally locked to the planet, the moon's "day" would equal the orbital period around the planet. During part of the orbit, the sun is blocked by the planet.

Note: for the following, we'll assume the center of the moon's (Earth's) orbit is coincident with the center of Saturn, just to make things a little easier, and that both bodies are perfect spheres.

If the larger planet were Saturn, it has a mean density of 0.687, and Earth 5.51. The Roche Limit is calculated as:

2.423 x (planet radius) x ∛(planet density/moon density)

Subbing in the appropriate numbers, we get:

2.423 x 58,232km x ∛(0.687/5.51)

= 2.423 x 58,232km x ∛(.12)

= 69,137 km

So there's the Roche Limit.

Now, how far out from the planet would the moon have to be to get an orbital period of 24 hours? There's lots of orbital period calculators online, so I won't bother with the equations, but an orbital radius of 193,500 km is almost spot on to 24 hours. And, well outside the Roche Limit calculated previously, so we're good there.

Now, how much of the sky is covered by the planet from the moon? Well, Saturn is 116,464 km in diameter. Earth's radius is 6356 km, so someone with Saturn directly overhead would see it covering 34.568 degrees of the sky. Which, since the sun would take 12 hours to cross it, would conceal the sun for over two hours.

So, it works. By tweaking the size of the planet (making it denser, thus smaller), you can reduce the amount of sky it covers, thus reducing the length of the eclipse. You could also alter the inclination of the moon's orbit, meaning the sun isn't blocked by the full diameter of the planet. I won't bother doing more, but there's how you get a total solar eclipse once a day, every day.

Answer 2

Slow down the planet's spin

A lunar eclipse happens when the moon is directly between the planet and the sun. Because the orbit of Earth's moon isn't on the same plane as the orbit of the Earth around the sun, it doesn't happen every month. But, if you were to reorient the planes to match, then you would have a lunar eclipse every month.

In order to advance that to 'once a day', which you specify is a 'revolution', and have a 'total lunar eclipse' take an hour, you need very specific conditions. For starters, a total lunar eclipse takes an instant because the moon is constantly in motion and only occurs on the direct line of the Earth, Moon, and Sun. It's possible for a partial lunar eclipse to take longer, but not if you want a 24 moon rotation. In orbital mechanics, the closer an orbit is to the focus of the orbit, the faster the orbit is. To have a 24-hour lunar orbit, you would need the moon to be very close to the Earth, close enough to start causing problems, and there's no way a partial eclipse could last anywhere near an hour under those conditions.

So the solution is quite simple - adjust the moon's orbital plane to match the planet and then slow down the planet's rotation to give it a 708 hour day. That should work.

Answer 3

Setting the moon at geosynchronous orbit with zero obliquity would do the trick. By zero obliquity I mean the moon's orbit would not be inclined with regard to the earth's orbit about the sun.

This way a viewer would see the moon staying in the same spot along the path the sun takes across the sky (the ecliptic). The sun would pass behind the moon the same time each day.

Geosynchronous is about nine times closer than the present lunar orbit. So a moon with 1/9 the moon's radius would still subtend half a degree as the moon and sun presently do. Which would give it 1/9^3 or 1/729 of the moon's volume. As a SF writer I would use this tinier moon unless you wanted extreme tidal effects.

Although a smaller moon could mean no eclipse seen from northern or southern latitudes.

Answer 4

The higher an orbit is, the longer it takes. The Moon currently takes ~28 days to make a complete orbit. So the orbit needs to be much lower.

It needs to be at or very near 0 inclination with respect to the plane of the ecliptic, or else it will miss blocking the sun on many orbits, just like happens with our moon in real life.

The orbit needs to be circular or else it will miss during some times of the year.

I don't think you can have both hour-long eclipses and year-round daily eclipses. The lower an object orbits, the faster it moves. So it will pass the sun quickly. You might be able to get an hour long eclipse with a properly scaled elliptical orbit, the eclipse happening when the moon is at apoapsis, but then you can't have daily eclipses all year. It will only work for a portion of the year where the moon's apoapsis and the sun are on the same side of the planet.

Answer 5

Another possibility is having a planet with many moons of almost totally identical size which are equally spaced in a ring around the planet. The moons all share the same orbit, equally spaced.

According to this post: https://planetplanet.net/2017/05/03/the-ultimate-engineered-solar-system/1 at the PlanetPlanet blog, a bunch of identical mass objects can share the same orbit if equally spaced, according to a paper by Smith and Lissauer in 2010. They calculated that at least 7 identical objects would be needed, and calculated examples with numbers of objects up to 42.

https://ui.adsabs.harvard.edu/abs/2010CeMDA.107..487S/abstract2

So assume that your planet has between 7 and 42 identical objects equally spaced in a ring about it. What distance for would be necessary to ensure one solar eclipse each day?

Well, if each successive eclipse is by the next moon in the ring, the spacing between moons would be equal to the time it took one moon to orbit during one day of the orbit. So with 7 to 42 moons orbiting, the total time for one moon to orbit would be 7 to 42 days.

At the lowest satellite orbits, a single orbit of Earth can last for about 90 minutes.

At a geosynchronus orbit:

Geosynchronous orbit (GSO) and geostationary orbit (GEO) are orbits around Earth matching Earth's sidereal rotation period. Although terms are often used interchangeably, technically a geosynchronous orbit matches the Earth's rotational period, but the definition does not require it to have zero orbital inclination to the equator, and thus is not stationary above a given point on the equator, but may oscillate north and south during the course of a day Thus, a geostationary orbit is defined as a geosynchronous orbit at zero inclination. Geosynchronous (and geostationary) orbits have a semi-major axis of 42,164 km (26,199 mi).3 This works out to an altitude of 35,786 km (22,236 mi). Both complete one full orbit of Earth per sidereal day (relative to the stars, not the Sun).

https://en.wikipedia.org/wiki/List_of_orbits4

The Moon orbits Earth with a semi major axis of 384399 kilometers, and has an orbital period of 27.321661 Earth days. Because the Earth travels several degrees along its orbit around the Sun during that period, the synodic period of the Moon, the time it takes to return to the same position relative to the Sun and to have the same phase, is a bit longer, 29.530589 Earth days.

So if it took 7 to 42 Earth days for the ring of moons to complete one orbit around an planet with the mass of Earth, the orbit with seven moons would be a lot higher than geostationary orbit at one sidereal Earth Day and a lot lower than lunar orbit at 27.321661 Earth days, and the orbit with forty two moons would be a lot higher than lunar orbit - possibly too high to be within Earth's Hills sphere and have stable orbits.

If your alien planet has Earth-like temperatures resulting from receiving Earth-like amounts of radiation from a Sun-like star that it orbits at about one Astronomical Unit distance, the angular diameter of the star will be about the same as that of the Sun as seen from Earth. Because the Earth's orbit around the Sun is elliptical, the angular diameter of the Sun as seen from Earth varies between 31.6 and 32.7 minutes of arc5, or 0.526666 to 0.545 degrees of arc. Since the Moon's orbit around Earth is also elliptical, the Moon's angular diameter varies between 29.3 and 34.1 arc minutes3

https://en.wikipedia.org/wiki/Sun3

https://en.wikipedia.org/wiki/Moon5

When the angular diameter of the Moon is less than that of the Sun during a solar eclipse, the eclipse is annular, with a ring of the Sun's surface showing around the Moon. A total solar eclipse only happens when the angular diameter of the Moon is equal to or greater than that of the Sun. So if you want all the stellar eclipses to be total the angular diameters of the moons should always be greater than that of the star.

If it is necessary or desirable for the moons to all be spheroids to avoid any oddly shaped moons affecting the duration of eclipses, the moons will have to be large and massive enough for their gravity to squeeze them into approximately spherical shapes (unless they are gigantic hollow spherical artificial satellites). Moons made of rock could retain irregular shapes when much larger than moons made of ice, and ice moons might evaporate if within the habitable zone of their star.

For these igneous planetesimals, the diameter needed to overcome rigid body forces and become round is about 620 miles (1,000km). The main belt asteroid Vesta is 326 miles (525km) in diameter. In its early history, Vesta’s interior was at least partially molten and may at one time have been in hydrostatic equilibrium; however, after cooling, Vesta was battered out of round by large impacts.

Farther from the Sun, ice condensed along with silicate grains to form planetesimals consisting of a fragile mixture of ice and rock. The minimum diameter to achieve hydrostatic equilibrium for an icy body is comparatively small, about 310 miles (500km).

https://astronomy.com/magazine/ask-astro/2017/08/the-diameter-of-spherical-bodies6

So the minimum diameter for roughly spherical moons may be about 500 kilometers if icey or about 1,000 kilometers if rocky.

The Moon has an average radius of 1,737.4 kilometers and thus an average diameter of about 3,474.8 kilometers. 3,474.8 kilometers is 3.4748 times the minimum diameter of 1,000 kilometers for a spherical rocky moon and 6.9496 times the minimum diameter of 500 kilometers for a spherical icey moon.

So a minimum size spherical rocky moon would have to be 3.4748 times as close as the Moon to totally eclipse the Sun, and thus have an orbit with a semi major axis of about 110,624.78 kilometers or less, while a minimum size spherical icey moon would have to be 6.9496 times as close as the Moon to totally eclipse the Sun, and thus have an orbit with a semi major axis of about 55,312.392 kilometers or less.

Larger spherical moons could orbit the planet in higher orbits than those minimum sized spherical moons.

The mass of the Moon is 0.012300 of Earth's mass, so 7 Moon-sized moons of the alien planet would have a total mass of about 0.0861 of Earth's mass and 42 Moon-sized moons of the alien planet would have a total mass of about 0.5166 of the Earth. And I am rather uneasy about the mathematical stability of a ring of Moon-sized bodies orbiting a planet with about the mass of Earth. There might be too much mass in the moons for long term orbital stability.

So I would want to make the moons as small and low mass as possible, and use as few moons as possible, to make the total mass of the moons as low as possible, thus putting them in as low orbits as possible.

I hope this suggestion will be useful.

Answer 6

Make 3753 Cruithne much larger. Placing the Moon in L1 isn't gonna work on long terms, L1 is an unstable equilibrium point (move the body from there and it won't return)

Answer 7

It is actually very simple!

Luna's orbit plane is 5.5° inclined compared to Earth-Sun plane. Make it 0° and you will have eclipses everyday. Source: https://www.ifa.hawaii.edu/~barnes/ASTR110L_S03/lunarorbit.html