I've been working on a system over the past week or so, and I think I have one figured out, but due to my lack of knowledge regarding these things, I would like someone to check my figures to see whether the system is stable and whether my world would actually be able to support life.
System info:
The star is a G5 class with three planets and an asteroid belt, the distances are:
Planet 1 - 0.525 AU (Small, terrestrial)
Planet 2 - 0.89 AU (Earth-like but smaller - This is the world I want to support life)
Asteroid belt - 1.775 - 1.95 AU
Planet 3 - 4.057 AU (This planet is a gas giant of a similar size to Neptune and has an elliptical orbit. It's closest approach to the star is about 2.7 AU and it's farthest is about 5.425 AU)
Assuming the system itself is stable, I want to make sure that the world can support human life, here's some info on it:
Radius: 4,782 km (.75 times Earth)
Circumference: 30,046 km
Surface Area: 287,361,782 km2
Volume: 4.580546798*1011 km3
Density: 6.48 g/cm3 or 6480 kg/m3
Mass: 2.96819*1024 kg (.50 times Earth)
Gravity: 0.88 times Earth
Day Length: 18 Hours
Year Length: 408.61 Days
Number of Moons: One, plus a partial ring
Orbital Period of Moon: 1 Day 14 Hours 16 Minutes 5 Seconds
Orbital Period of Ring: 5 Hours 9 Minutes 15 Seconds
The ring is about 2 radii away from the planet and the moon about 6.8 radii. The moon is outside the Roche limit.
Now, assuming no problems with all of that, either, I was thinking of having a axial tilt where the south pole is always tilted towards the star. Would this be possible while still having a stable orbit/rotation/environment to support human life?
Any feedback or suggestions will be greatly appreciated.
A Theoretical Planetary System - Need Checking
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Spaceman Spiff
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Everything seems in order, except:
1) You don't say if the moon is 'large' like our Earth's moon. I'd say if the moon is more massive than 1/75,000th the planet's mass, then I'd call it 'large' because it'll raise tides more strongly than our Earth's moon does here. At identical density to its planet, such a moon would have a diameter of about 225km.
2) The moon's period and distance do not square up with the planet's mass. It's ambiguous whether you mean the moon is 6.8 radii from the planet's surface or the planet's centre. Assuming either, I can't get the period and mass to fit. Assuming 6.8 radii from the centre, I calculate (assuming the moon's mass is negligible to the planet's mass) the period should have been about 22 hours 48.5 minutes.
3) Looks like you've set the star mass as exactly equal to the sun, judging from the planet's distance and period, if you're happy with that...
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Given all that, the main problem is that that it isn't possible to keep one pole of a planet facing the sun all year round.
If you think about it, that's two rotations: one for the 18 hour period about the poles, and a second that precesses the poles exactly once a year (your 408.61 days). That would require a very strong torque, for example the Earth's moon can only get the Earth's poles to precess once every 26,000 years, let alone one year.
A large moon can speed that up, but not much less than 100's of years (?).
You can have a 90?° axial tilt, but the poles will swap facing the sun each half year.
Spiff.
1) You don't say if the moon is 'large' like our Earth's moon. I'd say if the moon is more massive than 1/75,000th the planet's mass, then I'd call it 'large' because it'll raise tides more strongly than our Earth's moon does here. At identical density to its planet, such a moon would have a diameter of about 225km.
2) The moon's period and distance do not square up with the planet's mass. It's ambiguous whether you mean the moon is 6.8 radii from the planet's surface or the planet's centre. Assuming either, I can't get the period and mass to fit. Assuming 6.8 radii from the centre, I calculate (assuming the moon's mass is negligible to the planet's mass) the period should have been about 22 hours 48.5 minutes.
3) Looks like you've set the star mass as exactly equal to the sun, judging from the planet's distance and period, if you're happy with that...
.
Given all that, the main problem is that that it isn't possible to keep one pole of a planet facing the sun all year round.
If you think about it, that's two rotations: one for the 18 hour period about the poles, and a second that precesses the poles exactly once a year (your 408.61 days). That would require a very strong torque, for example the Earth's moon can only get the Earth's poles to precess once every 26,000 years, let alone one year.
A large moon can speed that up, but not much less than 100's of years (?).
You can have a 90?° axial tilt, but the poles will swap facing the sun each half year.
Spiff.
Also relevant to consider with 90 degree orientations is that if the planet's moons/rings orbit its equator (which is usually the case), they might get caught up in the Kozai mechanism. This results in orbital inclination and eccentricity getting traded back and forth, and with a ninety degree inclination the maximum eccentricity for an initially circular satellite would probably smash it into the planet.
The satellite could be saved if tidal circularisation of the orbit occurs on a faster timescale than Kozai eccentricity pumping occurs, or possibly if the planet is oblate then it would get pushed back into the equatorial plane which would prevent the eccentricity buildup. A combination of these effects, plus satellite-satellite interactions keeps Uranus's satellite system stable, but Uranus is more massive and more oblate than a terrestrial planet would be, which would enhance the satellite-saving effects. It is also further from the Sun, which I'd guess would increase the Kozai timescale.
The satellite could be saved if tidal circularisation of the orbit occurs on a faster timescale than Kozai eccentricity pumping occurs, or possibly if the planet is oblate then it would get pushed back into the equatorial plane which would prevent the eccentricity buildup. A combination of these effects, plus satellite-satellite interactions keeps Uranus's satellite system stable, but Uranus is more massive and more oblate than a terrestrial planet would be, which would enhance the satellite-saving effects. It is also further from the Sun, which I'd guess would increase the Kozai timescale.
Spaceman Spiff wrote:1) You don't say if the moon is 'large' like our Earth's moon. I'd say if the moon is more massive than 1/75,000th the planet's mass, then I'd call it 'large' because it'll raise tides more strongly than our Earth's moon does here. At identical density to its planet, such a moon would have a diameter of about 225km.
225 km seems very small for a 'large' moon. For the sake of familiarity, I think I'll have the moon at the same size as ours.
Spaceman Spiff wrote:2) The moon's period and distance do not square up with the planet's mass. It's ambiguous whether you mean the moon is 6.8 radii from the planet's surface or the planet's centre. Assuming either, I can't get the period and mass to fit. Assuming 6.8 radii from the centre, I calculate (assuming the moon's mass is negligible to the planet's mass) the period should have been about 22 hours 48.5 minutes.
I've put the moon at 32517.6 km from the surface, if we take that as 5.1 Earth radii and the mass of the planet as .5 times Earth, then I get:
Code: Select all
T = 1.4 * ???(5.1^3)/0.5
= 32.25 hours
= 1 day 14 hours 15 minutesA little longer than my original value, but still close.
Spaceman Spiff wrote:3) Looks like you've set the star mass as exactly equal to the sun, judging from the planet's distance and period, if you're happy with that...
It's slightly smaller than our sun:
Mass: 1.8293602*1030 kg
Diameter: 1,243,056 km
It's a shame about the axial tilt, but I'm sure I can come up with some other nasty things for the climate.
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Spaceman Spiff
- Posts: 420
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- With us: 22 years 9 months
- Location: Darmstadt, Germany.
Level10 wrote:225 km seems very small for a 'large' moon. For the sake of familiarity, I think I'll have the moon at the same size as ours.
'Large' depends on by what standard:
- physically large like Titan or Ganymede;
- apparently large as in subtended angle in the planet's sky (e.g., 0.5?° for Earth's moon);
- or (as I then realised in a fit of inspiration) causes tides larger than our moon - that's where I got 225 km from.
For your new distance, a 225 km moon would appear 80% the size of our moon in the sky. If you adopt a moon the size of Earth's ( about 3,300 km), then it would appear 5.8?° across in your planet's sky - almost 12 times to size of our moon.
From a tidal effect standard, that's huge! Tidal forces are proportional to the mass of the moon, the mass of the planet, the radius of the planet and the inverse cube of the distance between them (see the Wikipedia on "Tidal Forces" at http://en.wikipedia.org/wiki/Tidal_force ).
Comparing this system to our Earth-Moon system, the tidal force is (assuming a moon mass equal to our moon's) 1 ?— 0.5 ?— 0.75 ?— ( 1 / ( 32,517.6 km / 384,400 km ) )?? = 620 times stronger. That would also lock the moon and planet into mutual synchronous rotation in no time (see the Wikipedia on "Tidal Locking" at http://en.wikipedia.org/wiki/Tidal_locking ).
I might then say who needs an extreme climate when you have extreme tides! - except the tides won't move about a tidally locked planet, libration excluded.
This large, close moon would probably also mess up the rings. Well, I'm sure you can play about with a moon mass and diameter in the range 3,300 km to 225 km.
Level10 wrote:I've put the moon at 32517.6 km from the surface, if we take that as 5.1 Earth radii and the mass of the planet as .5 times Earth, then I get:Code: Select all
T = 1.4 * ???(5.1^3)/0.5
= 32.25 hours
= 1 day 14 hours 15 minutes
Ah. I see 'day' here is 18 hours , not 24 hours. Hmm, I still get 28 hours 1.7 minutes from:
27.321 days ?— ??? ( ( ( 32,517.6 km + 4,782 km ) / 384,400 km )?? / 0.5 )
That's 27.321 days for the Moon's orbital period, 384,400 km for the mean Earth-Moon semi-major axis.
Level10 wrote:It's slightly smaller than our sun:
Mass: 1.8293602*1030 kg
Diameter: 1,243,056 km
Ah, then the star is really 0.92 solar masses (see the Wikipedia on "Solar Mass" at http://en.wikipedia.org/wiki/Solar_mass ). In that case, the planets' periods should be 1 / ??? 0.92 longer than you have them.
I can't comment on chaos syndrome's Kozai mechanism. He's beaten me there
. Tidal locking could make it irrelevant though, tidal forces damp oscillations.
So, why not have fun with a near 90?° obliquity for planet and moon?
Spiff.
Last edited by Spaceman Spiff on 26.01.2007, 16:16, edited 1 time in total.
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Spaceman Spiff
- Posts: 420
- Joined: 21.02.2002
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The equation is Kepler's third law, scaled to the Earth-Moon system. I just noticed a slight boo-boo I made before about the moon's period: I should have added the planet's radius, not the Earth's. I've edited the calculation in it's previous post context above (it's now even less than what you first calculated, but that's in the past now).
Also, one should properly add the planet and moon masses together. Taking the Earth's mass as 5.9742?—10^24 kg, your planet mass is 0.49683 Earth masses, and your moon mass is now 0.00151 Earth masses, so the total system mass is 0.49834 Earth masses.
The orbital period is now:
27.321 days ?— ??? ( ( ( 69,817.2 km + 4,782 km ) / 384,400 km )?? / 0.49834)
or 79 hours 24.6 minutes. I hope you see how the numbers fit in.
Simple trigonometry, don't y'know!:
2 ?— arcsin ( 798.68 km / 69,817.2 km )
or 1.31?°.
Looks like the tides are about 6.3 times as strong as our moon's: ( 384,400 km / ( 69,817.2 km + 4,782 km ) ) ?? ?— ( 0.00151 / 0.0123 ) ?— 0.75 ?— 0.5. (Our moon is 0.0123 Earth masses.)
I hope the arithmetic is simple for you to repeat if you tinker some more....
Will you be submitting an add-on soon?
Spiff.
Also, one should properly add the planet and moon masses together. Taking the Earth's mass as 5.9742?—10^24 kg, your planet mass is 0.49683 Earth masses, and your moon mass is now 0.00151 Earth masses, so the total system mass is 0.49834 Earth masses.
Level10 wrote:Moon distance = 14.6 radii from surface (69817.2 km)
The orbital period is now:
27.321 days ?— ??? ( ( ( 69,817.2 km + 4,782 km ) / 384,400 km )?? / 0.49834)
or 79 hours 24.6 minutes. I hope you see how the numbers fit in.
Level10 wrote:Moon distance = 14.6 radii from surface (69817.2 km)
Moon radius = 798.68 km
Simple trigonometry, don't y'know!:
2 ?— arcsin ( 798.68 km / 69,817.2 km )
or 1.31?°.
Looks like the tides are about 6.3 times as strong as our moon's: ( 384,400 km / ( 69,817.2 km + 4,782 km ) ) ?? ?— ( 0.00151 / 0.0123 ) ?— 0.75 ?— 0.5. (Our moon is 0.0123 Earth masses.)
I hope the arithmetic is simple for you to repeat if you tinker some more...
.
Will you be submitting an add-on soon?
Spiff.