160 Year Solar Orbital Period & 1.9 Year Moon Orbital Period - Is it possible?

Question

A defining factor of my world is the hardships inhabitants have to face when surviving long seasons. Due to the extreme length of a year, time is largely measured by moon cycles. I need to make sure though that the solar/luna cycles are a) Generally possible, and b) able to sustain life.

Here is the general premise of my solar/luna cycle:

• The Planet's sun is many times larger than the Earth's sun (no exact size determined yet)
• The Planet takes 120* Earth-Years to orbit
• The Planet is ~1.5x the size of the Earth, and ~1.5x the mass.
• The Planet has an axial tilt, giving it Summers & Winters (and Spring & Autumn)
• The Moon takes 1.9 Earth-Years to orbit the Planet
• The Moon is visible in the Planet's sky, similar in size to Earth's Moon (or even appearing slightly bigger)

So my questions are:

• Is it possible to have a habitable planet that takes 160 Earth-Years to orbit the sun if the sun is sufficiently large enough, and the planet's orbital distance is sufficiently far away?

• Is it possible to have a moon that takes 1.9 Earth-Years to orbit a planet, and if so: how big would the moon have to be for it to be visible in the sky (as described above) when considering the (yet-uncalculated) distance its orbit would have to be for it to take 1.9 Earth-Years?

Any help would be greatly appreciated, even if it's pointing me in the direction of tools and resources that could help me calculate this by myself.

EDIT: I made a mistake in my original post. The length of the planet's orbit is 120 Earth-Years, not 160. I've also changed it so that the moon's cycle is now only ~48.8 Earth Days. I imagine this is a lot more realistic.

Thanks for all your replies so far!

Answer 1

The planet's orbit.

Neptune orbits the Sun in 165 years. The long dimension of its orbital ellipse is 30 times that of the Earth. This is an example of the square-cube law: All other things being equal, the square of the orbital time is proportional to the cube of the long dimension of the orbital ellipse. So for a 160 year orbit, the long dimension of the orbital ellipse would be 29.5 times that of Earth's orbit. This distance needs to be further increased by the cube root of the increase in the star's mass. Multiplying by the cube root of 10.9 makes this distance be 65 times that of Earth's orbit.

(distance) is proportional to (period)^(2/3) * (central mass)^(1/3)

but if the central mass is large enough to give the same power per unit area,

(distance) is proportional to (period)^(42/51)
so the factor of 65.3 = 160^(42/51)
or 51.5 = 120^(42/51)

By the way, all of the numbers in this answer are absurdly precise. Rounding them off is probably a good idea. If you want to calculate similar factors, but for a different orbital period, you can copy-and-paste these formulas (on the right-hand sides of the equals signs, but before the "for" notes) into WolframAlpha. Then change the input variable values, and press the compute button (the = sign). The biggest mistakes will be in my choices of assumptions and formulas, not in how the calculations are done.

The star's size

All other things being equal, the light per unit area received from a star is inversely proportional to the square of the distance from the star. For 29.5 times the Earth's distance, that is a factor of 868. For stars on the main sequence, a star about 6.9 times as massive as the Sun would be about 868 times more powerful than the Sun.

Combining the inverse-square law, the square-cube law, the effect of stellar mass on orbital period, and the relationship between stellar mass and luminosity means that the 6.9 factor needs to be raised to a power of 21/17, or to 10.9. For 65 times the Earth's distance, the star's power output needs to be 4,200 times the Sun's output, which corresponds to the 10.9-fold increase in the star's mass.

Logan Kearsley is correct that if we are trying to match the power per unit area using a star on the main sequence (just like the Sun, but bigger, hotter, and bluer),

(star mass) is proportional to (orbit time)^(8/17),
so the factor of 10.9 = (160)^(8/17)
or 9.5 = (120)^(8/17)

(power needed) is proportional to (distance)^2
so 4,200 = 65^2 for 160-year orbit
or 2,700 = 52^2 for 120-year orbit

In general, the more powerful the star, the faster it runs out of fuel. A star that puts out 868 times the power of the Sun would have a short life. If the Sun can be expected to have a total life of 10,000,000,000 years, a star that puts out 868 times the power might last 75 million years. A star that puts out 4,200 times the power of the Sun might last 25 million years.

(lifespan) is proportional to (star mass) / (power output)
so 25 million years = 10,000 million years * 10.9 / 4,200 for 160-year orbit
or 35 million years = 10,000 million years * 9.5 / 2,700 for 120-year orbit

The moon's orbit

1.9 years is about 24 times as long as Luna's orbital period, so the proposed moon orbits about 8.32 times as far from the planet as Luna is from Earth.

If the moon moves from east to west, and has an apparent cycle of 48.8 days, its orbital period is 47.8 / 27.3 = 1.75 times that of Luna. If the planet is 1.5 times the mass of Earth, the moon has

(distance) is proportional to (period)^(2/3) * (central mass)^(1/3)
or a factor of 1.66 = (1.75)^(2/3) * (1.5)^(1/3) times that of Luna

The revised moon's size is plausible

To take up the same amount of space in the sky, the revised moon would need to be 1.66 times the diameter of Luna, or 4.6 times the volume of Luna. As long as the revised moon is not much denser than Luna, its mass is still much smaller than that of the planet.

(diameter) is proportional to (distance).

(volume) is proportional to (diameter)^3,
so the factor of 4.6 = (1.66)^3

Mitigation

Have you considered the solution that Piers Anthony used for providing enough power per unit area for human colonization of the Jupiter and Saturn systems? The equivalent of giant lenses that focus light on the colony worlds. He used this in his Bio of a Space Tyrant series. More practical -- but explicitly artificial and expensive -- versions include reflector satellites around your planet (a la Bujold's "soletta array" around Komarr), and beaming power from satellites in the inner solar system (a la Ringo's Very Scary Array in the Troy Rising series).

Answer 2

In order to get a longer orbital period, you need a larger orbital radius--and to keep the planet warm enough at a larger radius, you need a larger star.

The luminosity equation for stars between 2 and 55 solar masses is approximately $L = 1.4M^{3.5}$. The necessary radius scales with the square root of luminosity, and orbital period scales with star mass and orbital radius as $T = \sqrt{R^3/M}$ With a little algebra, we can solve for stellar mass M and radius R in given T = 160 years:

$M = (\frac{T}{\sqrt{1.372}})^{8/17} = 10.114$

$R = \sqrt{1.4}(\frac{T}{\sqrt{1.372}})^{14/17} = 67.868$

So, your star is a little over 10 times as massive as the sun, and your planet orbits at nearly 68 AUs--considerably farther out than Pluto!

Such a star will only live for a little over 3 million years, so life did not originate naturally on this planet. It was colonized and terraformed--or magic did it. The sun will not have a noticeable disk in the sky; it will appear as an actinic blue point source, producing sharp shadows and extremely rapid sunsets and sunrises.

Now, we have the "moon" to worry about. Basically, you can't have a moon like that--but that doesn't mean you can't have the effect of a moon like that. It just won't actually be a moon--rather, your planet will be a moon of a gas giant (with a 1.9 year orbital period, and a system that's only a couple million years old at best, you don't need to worry about tidal locking).

Gas giants don't get much bigger than Jupiter--when you add more mass, they just get more dense. So, we'll assume that the gas giant has Jupiter's radius. To make it appear the same size in the sky as Earth's moon (about half a degree across), your planet will need to orbit the gas giant at 16,384,786 km (or slightly less, to make it slightly bigger), or a little over 1/10th of an AU. Given that distance and the required orbital period of 1.9 years, we can calculate a mass of 7.238e26kg; that's a little over a third Jupiter's actual mass, and a little bit more than Saturn's. So, assuming Jupiter's radius may be a little optimistic, but it definitely seems plausible that there's a range of gas giant masses and orbital distances which would make that all work out. Additionally, while an orbital distance of a tenth of an AU may seem overly large for the gas giant to retain your planet as a moon, it turns out that given that mass it would have a Hill sphere of about 1.5 AU, so the system should be perfectly stable and well-behaved.

Answer 3

The equation which defines orbital periods is Kepler's third:

$$ \frac{T^2}{R^3} = \frac{4\pi^2}{GM} $$

Where:

$T$ = The orbital period; You want it to be 1.9 years = $5.992 \times 10^7$ seconds
$R$ = Mean distance from the center of mass.
$G$ = The gravitational constant ($6.674 \times 10^{-8} cm^3 g^{−1}s^{−2}$)
$M$ = The sum of the masses involved

Let's plug some numbers in there.

$$ \frac{(5.992 \times 10^7)^2}{R^3} = \frac{4\pi^2}{6.674\times10^{-8}\times (8.958\times10^{24} + 7.347\times10^{22})} $$

The numbers in the parenthesis are the mass of 1.5 earths plus the mass of a moon, by the way.

Solving it gives us approximatelly $1.587 \times 10^{18}$ meters, or a quadrillion and a half of kilometers.

For comparison, our Moon is about 384,400 kilometers away from us, which is slightly over a light second. The moon in your world, though, would be further than 150 light years from the planet. The only way for this to work so that your moon orbits your planet is if both are deep in intergalactic space, away from any other significant mass. Maybe even not then; I haven't bothered to calculate the escape speed for your planet at that altitude, which is probably less than a fraction of an atomic radius per second.

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The orbital period for your planet involves less mathing around. Unfortunately, contrary to what I said in the comments, now I think it is not possible.

A sol-like star will eventually die and expel much of its outer material, which may form a planetary nebula; The remains become a white dwarf, which is both less mass massive than the original star and brighter. This is important.

Look at the equation at the beginning of my answer again. Let's twist it a little, while keeping it the same:

$$T^2GM = 4\pi^2R^3$$

This means that, for a fixed mean distance ($R$), time and total mass are inversely proportional. Reduce the mass and you have longer periods.

A white dwarf may be hotter than the sun for billions of years, which is long enough for life to form on a captured planet, or to rise again on a planet devastated by the original star's death. This means it will output more energy. Meanwhile, since it'll be less massive, planet periods will be longer. For a remaining mass of half a sun, the orbital period you want would be found at the distance Jupiter is from the sun. But even though the star might be up to 40% hotter than the sun, the distance involved and the fact that the star is smaller means less radiation. Your planet's surface temperature would be something between Titan's and Europa's, both of which are really cold.

One alternative to heat it up is to give it a hellish atmosphere like Venus's, but that is not compatible with life as we know it.

You may have to handwave it, or accept smaller numbers. Orbiting around a half-solar mass white dwarf would mean one orbit for every four real Earth years.

Answer 4

•Is it possible to have a habitable planet that takes 160 Earth-Years to orbit the sun if the sun is sufficiently large enough, and the planet's orbital distance is sufficiently far away?

The habitable zone depends on the luminosity of the star. The brighter the star, the further away from it is the habitable zone. The further away the planet is, the longer it takes to orbit the star. As a 0th order approximation, it's an assumption that holds true.

•Is it possible to have a moon that takes 1.9 Earth-Years to orbit a planet, and if so: how big would the moon have to be for it to be visible in the sky (as described above) when considering the (yet-uncalculated) distance its orbit would have to be for it to take 1.9 Earth-Years?

The Earth takes one year to orbit the Sun. The Sun has in the sky the same apparent size of the Moon. I have the feeling, again, a 0th order approximation, that such a long lasting orbital period is not suitable for a moon together with the other constrains you give. I suspect the moon would either be captured by some other attractor, or simply be so big to not be a moon.

Answer 5

Your moon's orbit is not possible, it exceeds the Hill Sphere Radius by a large margin.

Jasper did most of the required background legwork already.

The most stable moon orbit is a circular orbit (zero eccentricity), it which case the Hill Sphere radius is expressed as HR = A * (m/3M)^(1/3)

Where A is the orbital radius of the planet, m is the mass of the planet, and M is the mass of the star.

Using Alpha = 1 AU

Substituting using earth & sun parameters with Jaspers calculations.

HR = ((1.5*5.97e24 kg) / (3*6.65*1.88e30 kg))^(1/3) HR = 29 * 6.09e-3 AU = 0.29 AU

No moon orbit larger than 0.29 AU is stable. I.e., 24 million miles / 43.5 million km.

This is smaller than the required 593 * lunar orbit radius (239,000 miles / 385,000 km) calculated by Jasper.