View Full Version : Question About the Finer Details of Orbits

BlackInk

2013-Aug-26, 08:07 AM

Hi there. Since I was very young me and my dad have been developing this other world and this week I thought I'd take it a bit further astronomically. This particular planet has a mass of 6,26512x10^24 kg and has five moons. What I want to know is the time it takes for these moons to orbit the planet. I have the distances to all the moons and their individual masses, so I'll post that in another comment below if there is anyone here willing to help me.

If this then goes well I have an additional question about the planet's orbit, but for now I just need to focus on the moons.

Thanks in advance.

slang

2013-Aug-26, 08:18 AM

Hi there. Since I was very young me and my dad have been developing this other world and this week I thought I'd take it a bit further astronomically. This particular planet has a mass of 6,26512x10^24 kg and has five moons. What I want to know is the time it takes for these moons to orbit the planet. I have the distances to all the moons and their individual masses, so I'll post that in another comment below if there is anyone here willing to help me.

If this then goes well I have an additional question about the planet's orbit, but for now I just need to focus on the moons.

Thanks in advance.

Hi BlackInk. I'm not sure if you want to try to figure it out for yourself with a little help, or just want to get the answers. If you care to try for yourself, start with Kepler's laws (http://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion).

You might also want to get a copy of Wold Building by Stephen L Gillett (http://www.amazon.com/World-Building-Science-Fiction-Writing-Stephen/dp/158297134X)

Just be warned that a few of the formulae in the book (at least in my edition) were off by a factor that exactly matched the metric to SI conversions.

BlackInk

2013-Aug-26, 09:15 AM

Hi BlackInk. I'm not sure if you want to try to figure it out for yourself with a little help, or just want to get the answers. If you care to try for yourself, start with Kepler's laws (http://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion).

I'd like to just get the answers. Even though astronomy intrests me I have never had the time to really get into and truly understand it for myself. So most of the stuff in that link means little to me.. Hehe.

slang

2013-Aug-26, 09:22 AM

Well, then list your masses and distances and hope someone is kind enough to get you some answers :)

BlackInk

2013-Aug-26, 10:01 AM

"Distance" is from center to center.

"Orbit" is the length of the orbit.

Planet mass - 6,26512x10^24 kg

First moon:

Distance - 234’519,6547145228 km

Orbit - 736’765,2243735597 km

Mass - 5,29757 x 10^23 kg

Second moon:

Distance - 284’612,376364245 km

Orbit - 894’136,1507066454 km

Mass - 8,23698 x 10^23 kg

Third moon:

Distance - 500’695,9996986944 km

Orbit - 1’572’982,874335216 km

Mass - 10,44218 x 10^23

Fourth moon:

Distance - 602’874,9428188817 km

Orbit - 1’893’987,491393165 km

Mass - 6,19733 x 10^23

Fifth moon:

Distance - 821’848,3754268496 km

Orbit - 2’581’912,818605697 km

Mass - 3,48293 x 10^23 km

These orbits are made to be a perfect circle. I know that's not really how perfectly realistic orbits looks like and maybe I'll dive into that even deeper in the future. For now I have just made them that way.

antoniseb

2013-Aug-26, 11:41 AM

You can get circular orbits for these bodies, but since they are very massive moons, the orbits will not be stable for very long, as the gravitation of the other moons will perturb each moon's orbit quite a bit.

BlackInk

2013-Aug-26, 12:05 PM

Yeah, I thought that might be a problem, but it's okay, for now I'd just like to know how long it would take for them to orbit. I'll tackle issues like that and other ones later.

NEOWatcher

2013-Aug-26, 12:15 PM

Have you tried searching the net for "orbital calculator"?

BlackInk

2013-Aug-26, 12:51 PM

Have you tried searching the net for "orbital calculator"?

That actually helped. Didn't think that existed. Thanks!

caveman1917

2013-Aug-26, 11:43 PM

That actually helped. Didn't think that existed. Thanks!

Assuming you're using this one (http://www.1728.org/kepler3a.htm), and you probably know this, but just to make sure, the mass that you'll have to provide is the mass of the planet - not the mass of the moons.

grapes

2013-Aug-27, 07:38 AM

You can get circular orbits for these bodies, but since they are very massive moons, the orbits will not be stable for very long, as the gravitation of the other moons will perturb each moon's orbit quite a bit.

Even worse, the total mass of the five moons appears to be more than 50% of the mass of the planet--one by itself is a sixth of the mass of the planet. That's going to affect the orbit times somewhat, the orbit calculators may not take that into account.

caveman1917

2013-Aug-27, 10:17 AM

Even worse, the total mass of the five moons appears to be more than 50% of the mass of the planet--one by itself is a sixth of the mass of the planet. That's going to affect the orbit times somewhat, the orbit calculators may not take that into account.

The one i linked to mentions to use the combined mass if the moon's mass is high relative to the planet's, but for a 5-moon system in which all moons are pretty massive this may even get you a worse result since it only takes account of one moon at a time. I suppose it's best to ignore the moon's masses for a first approximation, as anything better will require a numerical simulation to do it right.

grapes

2013-Aug-27, 06:19 PM

The one i linked to mentions to use the combined mass if the moon's mass is high relative to the planet's, but for a 5-moon system in which all moons are pretty massive this may even get you a worse result since it only takes account of one moon at a time. I suppose it's best to ignore the moon's masses for a first approximation, as anything better will require a numerical simulation to do it right.

I see it mentions it in the discussion of the earth/moon system. The moon is only 1/81 the mass of the earth, the moons in the OP are much larger. Also, when the sum of the masses is used, the radius is the semi-major axis, not the center-to-center distance. For a moon that is 1/6 the mass of the primary, that will be a significant difference, no?

caveman1917

2013-Aug-27, 06:33 PM

I see it mentions it in the discussion of the earth/moon system. The moon is only 1/81 the mass of the earth, the moons in the OP are much larger. Also, when the sum of the masses is used, the radius is the semi-major axis, not the center-to-center distance. For a moon that is 1/6 the mass of the primary, that will be a significant difference, no?

True. The main problem here is that we have a system of 5 such massive moons all with relatively equal orbital radii, if there were only one we could use the semi-major axis and sum of masses to get a good result, but with a system of 5 such moons any such correction for each moon seperately is likely to result in nothing more than noise. Hence my proposal to just ignore the moon's masses altogether, unless one wants to do an actual simulation.

tony873004

2013-Aug-27, 06:48 PM

These moons do not even complete a single orbit before becoming unstable.

http://orbitsimulator.com/BA/fsscr003.GIF

grapes

2013-Aug-27, 08:13 PM

Looks about right :)

slang

2013-Aug-28, 06:58 PM

These moons do not even complete a single orbit before becoming unstable.

http://orbitsimulator.com/BA/fsscr003.GIF

Wow, what beautiful tides they must get there :)

Jeff Root

2013-Aug-28, 09:46 PM

Hello, BlackInk!

The numbers you posted for "length of the orbit" are not

quite right. You multiplied the orbit radius by pi to get them,

but forgot to multiply by 2. The circumference is 2 pi times

the orbit radius.

I checked them all. :)

The orbit radii you posted were clearly also the results of

some calculation, because the number of significant digits

is unrealistic. You show radii to tenths of a micrometre.

It is good to keep all the significant digits which result

from a calculation while there are still more calculations

being done, but whenever you publish the results, you

should round them to a reasonable number of significant

digits. In this case, rounding to either the nearest metre

or the nearest kilometre would probably make sense.

-- Jeff, in Minneapolis

Jeff Root

2013-Aug-28, 10:03 PM

Also, when the sum of the masses is used, the radius

is the semi-major axis, not the center-to-center distance.

The semi-major axis *IS* the center-to-center distance.

If the orbit is circular.

For a moon that is 1/6 the mass of the primary, that will

be a significant difference, no?

Not if the orbits are circular. The fact that the primary

moves around almost as much as the moon doesn't mean

they aren't both in circular orbits. The problem is when

other moons have to be accounted for. You can have one

moon of any mass in circular orbit indefinitely. More than

one, and you lose circularity real fast if any of the moons

have large mass.

This also ignores perturbations from the planet's sun or

other nearby planets.

-- Jeff, in Minneapolis

grapes

2013-Aug-28, 11:04 PM

The semi-major axis *IS* the center-to-center distance.

If the orbit is circular.

Not if the orbits are circular. The fact that the primary

moves around almost as much as the moon doesn't mean

they aren't both in circular orbits.

The semi-major axis is the distance from the orbit (center of the moon) to the center of the orbit. If both orbits are circular, the center of the orbit is not the center of the other body, no?

Jeff Root

2013-Aug-29, 12:00 AM

Hmmm! You may be right. But your view of the center of

the orbit seems to depend on absolute positions in space.

I'm imagining the view from the primary, which doesn't

notice or care that it is moving, too.

-- Jeff, in Minneapolis

grapes

2013-Aug-29, 03:36 AM

Hmmm! You may be right. But your view of the center of

the orbit seems to depend on absolute positions in space.

I'm imagining the view from the primary, which doesn't

notice or care that it is moving, too.

Both ways of looking at the situation may be valid, but the distance in each is the same.

In a two-body ststem, a moon doesn't orbit the center of the primary, it orbits the focus, which is at their combined center of mass.

Jeff Root

2013-Aug-29, 04:00 AM

I think this is leading toward a relativity-based discussion

of what rotation is. I'll just agree with you and assume

that I was wrong.

-- Jeff, in Minneapolis

Hornblower

2013-Aug-29, 02:52 PM

A recurring misconception is the idea that each satellite body moves in a Kepler ellipse with its focus at the barycenter of the system. This is an excellent approximation for a body whose orbital radius is large in proportion to the mutual separations of the other bodies, but it breaks down if that body's orbital radius is of the same order of magnitude as the others and the satellites are not of vanishingly small mass compared to the primary. An extreme case was an Apollo spacecraft on a free-return trajectory to the Moon. It followed a figure-eight path that was retrograde around the Moon and would have returned to a perigee near Earth's surface had it not been placed in a low orbit upon arrival at the Moon. That trajectory was good for only one circuit. Had the perigee been a bit too high so the spacecraft would miss the atmosphere, its next outward path would have been nowhere near the Moon and it would have gone heaven only knows where.

I don't know the details, but my educated guess is that Tony's simulator does numerical integrations for multibody problems that have no analytic solution.

Grey

2013-Aug-29, 06:12 PM

I don't know the details, but my educated guess is that Tony's simulator does numerical integrations for multibody problems that have no analytic solution.He's using Gravity Simulator (http://www.orbitsimulator.com/gravity/articles/what.html), which appears to do precisely that. For a situation like this, where the masses of the moons are large enough relative to the primary that no stable orbits are possible, it's the only option for having any hope of predicting what happens. It would be interesting to know an order-of-magnitude estimate for how large the moons can be relative to the planet before it's no longer realistic to expect even moderately stable orbits.

Rocky1775

2013-Sep-18, 06:40 PM

These moons do not even complete a single orbit before becoming unstable.

http://orbitsimulator.com/BA/fsscr003.GIF

You motivated me to search for an orbit simulator, and I found this:

http://phet.colorado.edu/sims/my-solar-system/my-solar-system_en.html

This is going to keep me entertained for hours.....

Powered by vBulletin® Version 4.2.3 Copyright © 2019 vBulletin Solutions, Inc. All rights reserved.