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This question already has an answer here:

I would like to know what distance a moon would have to be from an near earth size planet to appear twice as big in the night sky.

I would also like this moon to essentialy double Earth's high and low tides...so about a 12 foot variation.

Could I keep the moon cycle at or near 28 days? What would be a way to keep the above conditions but make the moon cylce 50-60 days? ---would this affect the seasons?

Please, if possible keep the formulas light. I am trying to create a world for a novel.

Thank You!

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marked as duplicate by Philipp, Azuaron, L.Dutch, kingledion, Mołot Mar 17 '17 at 14:36

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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By twice as large I assume you mean twice the arc not twice the area. That's the standard way to measure celstial objects but not earthly objects. A 2x2 sheet of paper is 1/4th the size of a 4x4 sheet of paper for example, but for objects in the sky, twice as far across is generally referred to as twice the size. I point that out just for clarity.

Twice the size is simple. For the same size object in orbit it's half the distance. If you want the moon to be 1.5 moon diameters, then 2 divided by 1.5 or 1.3333 times closer. Our moon is 388,400 km from Earth on average ,so 288,300 km. (distance is measured from the center to center not surface to surface, so that's a couple hundred km off if you want to get technical but close enough).

Tidal forces scale but the cube or 3rd power of the distance, so at 1.333 times closer, the tides would be about 2.37 times larger, everything else like the plane size and the moon's density remaining the same so that isn't too far off. 12 foot tides could easily be explained by smaller oceans, the multiplication doesn't have to be exact.

As for the 28 day orbital period, 1.333 times closer the period is to the power of 1.5, so around an Earth mass, 28/1.54 or about 18 days. The lunar period is actually 27.3 days but it appears to be 29.5 days due to the Earth's orbit around the sun. You can fudge that number by making the planet lighter and it should have lower gravity then too. If you want Earth gravity and Earth mass, a 28 day lunar period isn't accurate in your scenario. 18-19 range day would be better.

There's also some wiggle room in how long a planetary year is. See here on how the Earth year affects the Lunar month.

As long as you don't provide every specific, like you don't identify the planet's precise mass, then you have some wiggle room on the orbital period, but if you have the object 1.33 times closer and an orbital period of 28 days, that's only possible if the mass of the planet a fair bit lower, like 60%-70% Earth mass. Or, some other method of cheating like a smaller star and a year that's only 60 days long, or some combination of the two. It might be easier to compromise on the 28 days and make it an 18-19 or 17-20 day lunar period.

If you want the lunar cycle to be 50-60 days, that runs into problems. You need the planet to be really small, like the mass of Mars or you need to move the moon further out. A Mars size planet would have a hard time holding onto it's atmosphere and if you move the moon too far out, you run into hill-sphere problems where the planet might have a hard time holding onto the moon. In theory, if you have a larger-hotter sun, 1.5 times the mass of our sun for example, you could move the Earth like planet twice or a bit more than twice as far from the sun, increasing the Hill Sphere, but you'd still need to move the Moon further out, not 1.33 times closer to achieve a 50-60 day lunar period.

As a rule of thumb - closer moon, shorter lunar month. Further away moon, longer lunar month.

The moon doesn't affect seasons, that's the planetary tilt, or, in some cases of high eccentricity, the planet's orbit.

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  • $\begingroup$ shouldn't a lower density of this larger moon allow for a longer period? $\endgroup$ – Burki Mar 17 '17 at 13:12
  • $\begingroup$ Thank you very much for your detailed answer. That is helpful $\endgroup$ – P. Payton Mar 17 '17 at 15:54
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Answering the first part of your question:

You want to have a moon with 1.5 larger diameter and that looks twice bigger in the sky. Using the same formula I posted in my answer to this question, you can quickly get the distance from the center of the planet.

In short $angular amplitude = 2arctan(distance/2Diameter)$.

Based on the condition you set, solving the equations

$New Distance = ((1.5*Moon Diameter)/(tan(2*atan(moon diameter/2*Distance))))/2$

Will give you the new distance. It will hardly be the same as our present moon, therefore the orbital period will be different.

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