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The world I am envisioning is a rocky planet with oceans, plate tectonics, atmosphere, and several other similarities to our globe. The mass and size is roughly equal, but its moon is orbiting much closer around it than our moon revolves around Earth. For the sake of clarity: the moon has roughly the same size and mass as our moon.

I am wondering what the consequences would be if it were to orbit the planet at a distance of 1/20th the distance our moon orbits the Earth (let's say 20 000 km as an approximation).

The questions I have are:

  1. Would such a system be stable?
  2. If yes:

    a) what orbital speed would be necessary for the moon to stay in geocentric orbit?
    b) would an eccentricity close to zero be possible, or would the trajectory have to look different in order for the system to be stable (if yes: what would it look like)?
    c) what sidereal day would the moon have at the altitude of 20 000 km and the necessary speed?

  3. How many degrees of the sky would it take up when seen from the world's surface? (I would love to simulate for myself with the software from spaceengine.org, but I lack hardware spec for that and they lack a Linux version)

  4. What would the tidal effects on the oceans look like with such a close orbit (as an average, I might add; I am well aware that tidal patterns are greatly affected by local geography, depth of the sea, shape of the sea bottom, etc)?

    Side note: I read that a moon 20 times closer to the Earth would mean tidal effects 400 times stronger, but I do not have the mathematical knowledge to double check the numbers myself.

  5. Would a geocentric orbit by an object of this mass allow for an axial tilt of zero degrees for the planet, or would that be impossible?

  6. What would the moons pull on the planet mean in terms of plate tectonics (earthquakes, volcanoes, etc)? I assume it would be increased, but is it possible to calculate how much more increased it would be, or are there too many unknown factors involved for that?

I hope I have been specific enough, and I look forward to your answers and thoughts on the subject!

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    $\begingroup$ scienceline.ucsb.edu/getkey.php?key=373 is relevant. It mentions that the early moon was 1/10th of its current distance from the Earth. So is it a stable system at 1/20th? Don't know for sure, but it's plausible. $\endgroup$ – Rob Watts Jun 15 '15 at 22:43
  • $\begingroup$ Thanks, I read it. It essentially explains that the moon collided with Earth, and due to the uneven gravity of the Earth it has gradually spiraled outward to the altitude it is at today. The events that are described in answer 1 in the link you posted gives me no hints of whether the constellation I envision may be stable or not. $\endgroup$ – fantasia Jun 15 '15 at 23:08
  • $\begingroup$ @fantasia welcome to Worldbuilding, and thank you for this interesting and well-asked question! $\endgroup$ – Monica Cellio Jun 16 '15 at 2:48
  • $\begingroup$ @MonicaCellio Thanks, I have lurking for months, and finally found some time to get back to writing - hence the question (and the ones that will follow). Wonderful community! $\endgroup$ – fantasia Jun 16 '15 at 21:04
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It could happen.

  1. Would such a system be stable?

The Earth's moon is currently 41 times further out than its Roche limit with Earth. So the moon can be 20 times closer and not break up.

a) what orbital speed would be necessary for the moon to stay in geocentric orbit?

The velocity of any satellite orbiting in a circle is:

$$v_o \approx \sqrt{GM \over r}$$

Where G is the gravitational constant, M is the combined mass of the bodies in question, and r is the distance between their centers.

To orbit at 1/20th the distance, the moon would need to be orbiting at 4.58 km/second, which is about 4.48 times as fast as it does now.

b) would an eccentricity close to zero be possible, or would the trajectory have to look different in order for the system to be stable (if yes: what would it look like)?

I've calculated using eccentricity of zero, I think this would be fine.

c) what sidereal day would the moon have at the altitude of 20 000 km and the necessary speed?

I'm not sure this makes sense given the definition of sidereal time. If you want to know the orbital period of the moon, it would take about 7.34 hours to orbit the Earth.

How many degrees of the sky would it take up when seen from the world's surface?

The angular diameter is given by:

$$ \delta = 2\arctan\left({d \over 2D }\right) $$

For the moon's diameter of 3,476 km and new distance of 19,220 km, it would appear to be about 10.3 degrees in diameter. This is about the size of your fist when you arm is held out.

What would the tidal effects on the oceans look like with such a close orbit (as an average, I might add; I am well aware that tidal patterns are greatly affected by local geography, depth of the sea, shape of the sea bottom, etc)?

There is a brief article discussing the tidal effects of a moon 20 times closer, it says:

If the Moon got much closer, say 20 times closer, it would exert a gravitational force 400 times greater than what we are used to. A mighty tidal bulge would be created, hitting the land and causing great flooding, with cities such as London and New York disappearing under water.

This may have been what you were talking about. It's a valid estimate. The mathematics for tidal force are included on the Wikipedia page. Though the article is just talking about gravitational force, the two are obviously related. But, the tidal force increase would be significantly stronger with this increase in gravitational force. The lunar tidal acceleration at the Earth's surface along the Moon-Earth axis would be about 0.0718 g (around 8,000 times stronger than now).

So the article is accurate, it just doesn't provide the increase for tidal forces, only the gravitational increase.

Would a geocentric orbit by an object of this mass allow for an axial tilt of zero degrees for the planet, or would that be impossible?

Allow for it? Sure, things will change significantly for the planet. I'm not sure what you're asking here. The axial tilt with respect to the Sun?

What would the moons pull on the planet mean in terms of plate tectonics (earthquakes, volcanoes, etc)? I assume it would be increased, but is it possible to calculate how much more increased it would be, or are there too many unknown factors involved for that?

The tidal force could be calculated. What effect that would actually have on plate tectonics is far beyond my ability to calculate/simulate. My guess is you could get answers like "more earthquakes" but not much more specific than that. Especially if this planet is not Earth.

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  • $\begingroup$ Hmm. I wonder, would the Earth-Moon act (even more) like a Double planet or less? $\endgroup$ – Amziraro Jun 16 '15 at 0:31
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    $\begingroup$ @Samuel, nice answer. But tides are related to $\frac {1}{r^3}$ so 20x closer gives you tidal forces 8000x as large. $\endgroup$ – Jim2B Jun 16 '15 at 3:59
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    $\begingroup$ @Samuel I think you're confused between gravitational force/acceleration between two bodies as a whole (the 1/r^2 terms at the top of the section of the article you link to); and tidal acceleration (computed at the bottom of it) which is an 1/r^3 term. $\endgroup$ – Dan Neely Jun 16 '15 at 5:26
  • $\begingroup$ This answer on Physics.SE might be of help as well. Tidal effects are based on the gradient (slope) of the gravitational field; not its absolute value. Since the gradient is a derivative of the acceleration, you get an extra 1/r term in it for 1/r^3. $\endgroup$ – Dan Neely Jun 16 '15 at 5:33
  • $\begingroup$ Nice answer, it boggles my mind to think of a fist sized object zooming through the sky every 7 hours.... $\endgroup$ – Pureferret Jun 16 '15 at 9:10
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The massive tides would have one other effect, which would be to slow the Earth and cause the moon to gradually spiral outwards from the Earth to a more distant orbit due to tidal torques. The current rate of slowing can only be calculated using accurate atomic clocks (with the occasional leap second being added to the year to make up for the slowdown), but even 4 billion years from now the Earth will not be tidally locked to the Moon (and we will have a few bigger issues with the Sun at that time).

If the Moon were as close as you want it to be, the slowdown would be faster (relatively speaking, maybe a few seconds a year) and the Moon would not visibly retreat from the Earth in a human lifetime, but over the generations this will become noticeable. Edmund Halley, of Halley's comet fame, noticed the discrepancies in ancient astronomical records in 1695, so a civilization with even limited astronomical knowledge will figure this out eventually.

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  • $\begingroup$ Yea, if the moon started out that close, it would be far away now. That's only a factor of 2 different from what really happened. If the moon is that close now, how did it get there in recent history? $\endgroup$ – JDługosz Jun 16 '15 at 0:11
  • $\begingroup$ This isn't correct. If the moon is closer than the geosynchronous altitude (and still orbiting prograde), then it speeds up the earth and spirals inward, opposite what it does today. $\endgroup$ – BowlOfRed Jun 16 '15 at 2:40
  • $\begingroup$ @BowlOfRed Two interesting yet very different answers here from two different posters: one says the moon's altitude and the tidal effects would make it spiral outwards from its primary, and one says it would spiral inwards. Anyone care to back their statement up with equations or references? $\endgroup$ – fantasia Jun 16 '15 at 21:18
  • $\begingroup$ universetoday.com/14258/… $\endgroup$ – BowlOfRed Jun 16 '15 at 21:40
  • $\begingroup$ @BowlOfRed Thanks. That settles it for me, since it was corroborated by several other sources I trust. $\endgroup$ – fantasia Jun 16 '15 at 22:01
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Not stable over geological time. That's a system that, because of the massive tidal forces in effect, is going to bleed angular momentum extremely rapidly. Regardless of whether you could get a stable starting set up or not, and I'm inclined to think not, the increased tidal gravity is exerted at both ends of the Earth/Moon system. Tidal friction creates a net loss of angular momentum in the system but most of the energy is debited from the lower mass faster moving member of it meaning that, in our world, the Moon is constantly giving up orbital energy to the Earth. In a system that creates four hundred times the force we see today that's a process that's going to go much faster, I'm not sure what would happen to the system, whether the moon in question would back away into space or fall in towards it's primary and be torn apart when it passed it's Roche Limit. What I am sure of is it's not stable, our own Earth/Moon system isn't stable, but this is really not stable.

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It doesn't appear to be possible without deliberate engineering or some very special sequence of events.

If the moon started that close, after 4 billion years it would be far away. How do you get it close in geologicaly recient time, so it has not run off again?

An n-body interaction could capture it, even give it a perige as close as you wanted. But then how do you circularize the orbit, bringing down the (very distant) apoge to match?

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    $\begingroup$ If playing Osmos has taught me anything, it is that, when @ perigee, if you apply a decelerating force, in the direction of travel, you'll drop the apogee without affecting much else about the orbit. With use of the correct 'maths,' you should be able to achieve a near-perfect orbit with a single, well timed burn of appropriate force. I'm not sure how many Estes D-? engines it would take, but hey... :-) $\endgroup$ – MrWonderful Jun 16 '15 at 6:21
  • $\begingroup$ Right, the acceleration applied at the closest point will lower the farthest. So a natural n-body interaction would need (1) an enormous delta-v, and (2) do so while it's close to the primary as well: so how can the well-aimed black hole or whatever affect the moon but not the primary too? $\endgroup$ – JDługosz Jun 16 '15 at 8:04
  • $\begingroup$ Btw, 30 years before Osmos I learned that writing orbital simulations on an 8-bit computer that had 2K of RAM, as coisins of a "lunar lander" kind of game. No graphics, just 1 line of LCD text. $\endgroup$ – JDługosz Jun 16 '15 at 8:07
  • $\begingroup$ @JDługosz Thank you for taking the time to answer. Could you please elaborate why the moon in question could not maintain a stable orbit at an altitude of 20 000 km, but instead would have to move outwards? Your mentioning of the moon being captured is also of interest. Do you mean that the primary could be able of capturing a moon and give it a perigee of 20 000 km (or any given distance), but that it won't be able to give it a circular orbit? Thank you once again, and bear in mind my ignorance when it comes to celestial mechanics when you try to explain. $\endgroup$ – fantasia Jun 16 '15 at 21:16
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    $\begingroup$ Our moon formed close and moved away. It is due to friction and phase delay in tides, and conservation of angular momentum. That's in Wikipedia, second paragraph in that section. $\endgroup$ – JDługosz Jun 17 '15 at 1:58

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