Ultimate space simulation software

 
User avatar
PlutonianEmpire
Pioneer
Pioneer
Posts: 524
Joined: 02 Nov 2016 18:13
Location: Planet Meabh
Contact:

Science and Astronomy Questions

04 Jun 2021 21:05

Hi all, I'm in need of a plausible detail for some of my fiction I write. I have a rogue planet with a set of orbital parameters as follows, looping around Sol:
Orbit {
     Epoch 2458860.74038
     RefPlane "Ecliptic"
     PericenterDist 0.9
     Period 106.73035377261823
     Eccentricity 1.04
     Inclination 32.8
     AscendingNode 256.56103
     ArgOfPericenter 306.41318
}
So, the assumed SMA would be -22.5, with a Perihelion date of January 12, 2020.

Rogue Planet has two moons, with an external barycenter with the outer one, all with the following parameters:

Planet:
Radius 5469.6462 km
Mass 0.60543 M_earth
AlbedoBond 0.829
AlbedoGeom 0.935477

Inner Moon:
Radius 1326.212 km
Mass 0.005454 M_Earth
AlbedoBond 0.119
AlbedoGeom 0.1428
SemiMajorAxis 98304.460 km
Period 4.542 Days

Outer Moon:
Radius 2258.295 km
Mass 0.027646 M_earth
AlbedoBond 0.147
AlbedoGeom 0.1764
SemiMajorAxis 302025.266 km
Period 25.626

With that in mind, the Rogue Planet reaches apparent Magnitude 15 at 88 AU from Earth on Halloween of 1976, slightly dimmer than Pluto's AppMag of 14.5 that same year, when Charon was discovered.

Now, take into accout all the sky surveys and exploration that were operating throughout the 20th century. Between 1976 and 2020, what might be decently realistic discovery dates for Rogue Planet and its moons?
Specs: Dell Inspiron 5547 (Laptop); 8 gigabytes of RAM; Processor: Intel® Core™ i5-4210U CPU @ 1.70GHz (4 CPUs), ~2.4GHz; Operating System: Windows 7 Home Premium 64-bit; Graphics: Intel® HD Graphics 4400 (That's all there is :( )
 
A-L-E-X
Galaxy Architect
Galaxy Architect
Posts: 3340
Joined: 06 Mar 2017 20:19

Science and Astronomy Questions

29 Jul 2021 08:21

Got into a bit of an argument with a TV personality on Twitter who kept repeating "energy cannot be created or destroyed" repeatedly.  I know there are special cases where this law is violated and I really hate it when TV personalities simplify science to make it (according to them) more understandable.  So I used the following links, with articles written by physicists, to prove my point that it is violated under specialized conditions.

https://interestingengineering.com/is-i ... -of-energy

https://www.physlink.com/education/askexperts/ae605.cfm

https://www.quora.com/How-does-an-infin ... ite-energy
 
User avatar
pzampella
Space Tourist
Space Tourist
Posts: 33
Joined: 12 Dec 2016 07:44
Location: Madrid
Contact:

Science and Astronomy Questions

06 Nov 2021 07:06

I was checking the latest version of SE today, after almost 2 years, and while checking Betelgeuse, I noticed that the closest procedural planet is only 2.53 AU from the star's center (0.27 AU from its surface). Being so close to its parent star, it should be improbable for it to have a satellite, and in this case, it has 8 moons!

I understand that gravity is tricky in this sense, and that when mass is distributed in a larger space, the effects are no as strong. However, I have heard several times that the absent of satellites in Venus is related to its close distance to the sun.

What do you think?
 
User avatar
Watsisname
Science Officer
Science Officer
Posts: 2247
Joined: 06 Sep 2016 02:33
Location: Bellingham, WA

Science and Astronomy Questions

06 Nov 2021 20:23

I was checking the latest version of SE today, after almost 2 years, and while checking Betelgeuse, I noticed that the closest procedural planet is only 2.53 AU from the star's center (0.27 AU from its surface). Being so close to its parent star, it should be improbable for it to have a satellite, and in this case, it has 8 moons!
The key idea for whether a planet can support moons, and how many or how far out they can be, is the "Hill sphere", a roughly spherical region of space in which the gravity of the planet dominates over the gravity of the star (while also accounting for other forces due to the planet's motion around the star). Your intuition is right that the size of the Hill sphere is smaller for a planet closer to its star (the size is proportional to the orbital distance). It is also smaller if the planet's mass is smaller (or the star's mass is greater), and goes as the cube root of that ratio.

Betelgeuse is about 11 times more massive than the Sun, which all else being equal would make Hill spheres about cuberoot(11) ≈ 2.2 times smaller than in the solar system. But 2.53 AU is still a pretty good distance from the star. It means the size of the Hill sphere for an Earth-mass planet at that distance from Betelgeuse would be about 1.7 million kilometers. This is actually a bit larger than Earth's Hill sphere at 1AU from the Sun (1.5 million kilometers). For a Jupiter-mass planet at 2.53 AU, the Hill sphere would extend to about 12 million kilometers. 

So depending on the mass of the procedural planet in question and what the moon system looks like (how big and spread out the moons are), having 8 of them might not be that unrealistic. :)
 
A-L-E-X
Galaxy Architect
Galaxy Architect
Posts: 3340
Joined: 06 Mar 2017 20:19

Science and Astronomy Questions

06 Nov 2021 22:55

I was checking the latest version of SE today, after almost 2 years, and while checking Betelgeuse, I noticed that the closest procedural planet is only 2.53 AU from the star's center (0.27 AU from its surface). Being so close to its parent star, it should be improbable for it to have a satellite, and in this case, it has 8 moons!
The key idea for whether a planet can support moons, and how many or how far out they can be, is the "Hill sphere", a roughly spherical region of space in which the gravity of the planet dominates over the gravity of the star (while also accounting for other forces due to the planet's motion around the star). Your intuition is right that the size of the Hill sphere is smaller for a planet closer to its star (the size is proportional to the orbital distance). It is also smaller if the planet's mass is smaller (or the star's mass is greater), and goes as the cube root of that ratio.

Betelgeuse is about 11 times more massive than the Sun, which all else being equal would make Hill spheres about cuberoot(11) ≈ 2.2 times smaller than in the solar system. But 2.53 AU is still a pretty good distance from the star. It means the size of the Hill sphere for an Earth-mass planet at that distance from Betelgeuse would be about 1.7 million kilometers. This is actually a bit larger than Earth's Hill sphere at 1AU from the Sun (1.5 million kilometers). For a Jupiter-mass planet at 2.53 AU, the Hill sphere would extend to about 12 million kilometers. 

So depending on the mass of the procedural planet in question and what the moon system looks like (how big and spread out the moons are), having 8 of them might not be that unrealistic. :)
Wat do you think there are planets around Betelgeuse and we just haven't found them yet?  Or has Betelgeuse absorbed any that used to be there.  I wonder if we've found planets around any red super giants....maybe Antares?
 
User avatar
Watsisname
Science Officer
Science Officer
Posts: 2247
Joined: 06 Sep 2016 02:33
Location: Bellingham, WA

Science and Astronomy Questions

07 Nov 2021 01:39

As far as I know we have neither confirmed nor excluded planets around Betelgeuse, nor found any around any other supergiant star. Using exoplanets.org (for instance by plotting star radius vs temperature, and another good check is the star's surface gravity), the largest and most luminous star I can find that has a known planet is HD 220074. The star is a red giant about 55 times the radius of the Sun (or about 0.26 AU), with about 700 to 800 times the luminosity of the Sun. The planet is 11.2 ± 1.9 Jupiter masses and orbits at 1.60 ± 0.13 AU.

I'm pretty sure we have found evidence of evolved stars that have swallowed their planets from the pollution of heavy metals in their atmospheres. A planet can also survive (and continue orbiting!) inside of an evolved giant star for a surprisingly long time, because the density of the star's outer layers is quite low.
 
A-L-E-X
Galaxy Architect
Galaxy Architect
Posts: 3340
Joined: 06 Mar 2017 20:19

Science and Astronomy Questions

07 Nov 2021 01:51

Wow thats something Wat!  it would be fun to see something like that (a planet orbiting inside a giant star), I want to see what the atmosphere of such a planet would look like.  One of the really neat things about SE is seeing the colors of different atmospheres.  I envision its atmosphere would be mixed in with the star's gaseous outer shell and would glow orange or red (depending on the color of the star.)  Maybe these even exist inside blue giants, though the shorter lifetime of those stars probably makes planets much less likely?
 
User avatar
midtskogen
Star Engineer
Star Engineer
Posts: 1468
Joined: 11 Dec 2016 12:57
Location: Oslo, Norway
Contact:

Science and Astronomy Questions

08 Nov 2021 04:42

A planet can also survive (and continue orbiting!) inside of an evolved giant star for a surprisingly long time, because the density of the star's outer layers is quite low.
Sure, but how tenuous? Google tells me that Betelgeuse's density is 0.000012 kg/m³.  That corresponds to the density of Earth's atmosphere at rough 80 km altitude, which would make a very short "orbit".  Obviously, giant stars have a steep density gradient (otherwise a planet near the photosphere would hardly orbit anyway...)
NIL DIFFICILE VOLENTI
 
User avatar
Watsisname
Science Officer
Science Officer
Posts: 2247
Joined: 06 Sep 2016 02:33
Location: Bellingham, WA

Science and Astronomy Questions

08 Nov 2021 11:34

Sure, but how tenuous? Google tells me that Betelgeuse's density is 0.000012 kg/m³.  That corresponds to the density of Earth's atmosphere at rough 80 km altitude, which would make a very short "orbit".
Yes, 10[sup]-5[/sup] kg/m[sup]3[/sup] or 10[sup]-8[/sup] g/cm[sup]3[/sup] is approximately right (allow a few orders of magnitude in either direction depending on the particular star and how far into it we go). For more details, see for example these stellar structure models of giant stars, and especially figure 4:

Image


Anyway, how long could a planet's orbit inside such a star last? Something like a small stone or satellite certainly does not last long (much less than one orbit) at similar densities 80km above Earth. However, a planet inside a star is a very different beast. Think of it this way: the inertia of a solid sphere goes as the cube of its radius, while the drag force is proportional to its cross sectional area which goes only as the square of its radius. This means a rocky planet 10,000 km across will last hundreds of millions of times longer than a 10 cm diameter stone.

Of course, the circumference of a giant star is also bigger than the circumference of the Earth. But not by as much -- a factor of a few thousands, to hundreds of thousands. So the net result is that a planet can actually make up to tens of thousands of orbits below the photosphere of a giant star before ultimately giving out to increasing drag and vaporization.

If this still sounds too insane to believe (and yes is a pretty crazy thing and one of my favorite tidbits about giant stars), it's well worth doing a more careful calculation or simulation. Last year I had written an orbital drag model with Earth's exponential atmosphere to study satellite reentry and the Apollo reentry descent profiles, and now we can apply it to a simple model of the star HD 220074 from earlier. For starters I set the density everywhere equal to the average density, which is about 10[sup]-4[/sup] kg/m[sup]3[/sup], or about 10 times denser than Betelgeuse. Then I set a spherical object with a density of 5500 kg/m[sup]3[/sup] in a circular orbit just below the surface. How long for the orbit to drop to 0.95 stellar radii, as a function of the object's radius?

100km: less than half an orbit
500km: 1.5 orbits
1000km: 3 orbits
2000km: 6 orbits
4000km: 12 orbits
6000km: 18 orbits

A nice match with the earlier physical intuition: doubling the size of an object doubles how far it can go against drag. It gets even better if we give the star a realistic density profile (can easily get hundreds to thousands of orbits). One final factor I haven't treated here is the heating. But evolved stars have cooler surfaces, and it turns out it's not too big of a problem for a planet. They can last beneath the photosphere for thousands, in some cases even millions of years!
 
User avatar
midtskogen
Star Engineer
Star Engineer
Posts: 1468
Joined: 11 Dec 2016 12:57
Location: Oslo, Norway
Contact:

Science and Astronomy Questions

08 Nov 2021 12:51

the inertia of a solid sphere goes as the cube of its radius, while the drag force is proportional to its cross sectional area which goes only as the square of its radius
That's a good point, and also explains why a mouse can jump off a tall building and survive whilst an elephant certainly wont.

One other thing to consider.  Even a few thousand orbits is a short moment in planetary and stellar timescales, so a planet skimming a stellar surface could still be extremely rare.
NIL DIFFICILE VOLENTI
 
User avatar
Watsisname
Science Officer
Science Officer
Posts: 2247
Joined: 06 Sep 2016 02:33
Location: Bellingham, WA

Science and Astronomy Questions

08 Nov 2021 13:41

Yes, a short period relative to the life of the star. However, combined with the number of stars and that the vast majority of them have planets, it happens quite often -- estimates are for a planet to be engulfed by its host star evolving off the main sequence about every 1 to 10 years throughout our galaxy. 

This may actually be important for the stars, too. One effect is that the planet pollutes the envelope of the star with heavier elements, and there is some evidence that we have seen stars where that has happened (not yet in the act, as far as we know). But another consideration (and the main topic of the paper in the above link) is that the orbital energy being dragged away doesn't simply vanish, but goes into heating the envelope of the star. In some cases this may be small enough to just be a hiccup, but in others it could be significant and produce a clear signature of the process in action. This would be easiest to see for planets engulfed by smaller giant stars, since their envelopes have higher densities and lower luminosities.
 
A-L-E-X
Galaxy Architect
Galaxy Architect
Posts: 3340
Joined: 06 Mar 2017 20:19

Science and Astronomy Questions

09 Nov 2021 23:54

Sure, but how tenuous? Google tells me that Betelgeuse's density is 0.000012 kg/m³.  That corresponds to the density of Earth's atmosphere at rough 80 km altitude, which would make a very short "orbit".
Yes, 10[sup]-5[/sup] kg/m[sup]3[/sup] or 10[sup]-8[/sup] g/cm[sup]3[/sup] is approximately right (allow a few orders of magnitude in either direction depending on the particular star and how far into it we go). For more details, see for example these stellar structure models of giant stars, and especially figure 4:

Image


Anyway, how long could a planet's orbit inside such a star last? Something like a small stone or satellite certainly does not last long (much less than one orbit) at similar densities 80km above Earth. However, a planet inside a star is a very different beast. Think of it this way: the inertia of a solid sphere goes as the cube of its radius, while the drag force is proportional to its cross sectional area which goes only as the square of its radius. This means a rocky planet 10,000 km across will last hundreds of millions of times longer than a 10 cm diameter stone.

Of course, the circumference of a giant star is also bigger than the circumference of the Earth. But not by as much -- a factor of a few thousands, to hundreds of thousands. So the net result is that a planet can actually make up to tens of thousands of orbits below the photosphere of a giant star before ultimately giving out to increasing drag and vaporization.

If this still sounds too insane to believe (and yes is a pretty crazy thing and one of my favorite tidbits about giant stars), it's well worth doing a more careful calculation or simulation. Last year I had written an orbital drag model with Earth's exponential atmosphere to study satellite reentry and the Apollo reentry descent profiles, and now we can apply it to a simple model of the star HD 220074 from earlier. For starters I set the density everywhere equal to the average density, which is about 10[sup]-4[/sup] kg/m[sup]3[/sup], or about 10 times denser than Betelgeuse. Then I set a spherical object with a density of 5500 kg/m[sup]3[/sup] in a circular orbit just below the surface. How long for the orbit to drop to 0.95 stellar radii, as a function of the object's radius?

100km: less than half an orbit
500km: 1.5 orbits
1000km: 3 orbits
2000km: 6 orbits
4000km: 12 orbits
6000km: 18 orbits

A nice match with the earlier physical intuition: doubling the size of an object doubles how far it can go against drag. It gets even better if we give the star a realistic density profile (can easily get hundreds to thousands of orbits). One final factor I haven't treated here is the heating. But evolved stars have cooler surfaces, and it turns out it's not too big of a problem for a planet. They can last beneath the photosphere for thousands, in some cases even millions of years!
Wow, thats amazing Wat.  Intuitively, I had originally thought a smaller object would actually last longer because it's less affected by drag, but it makes sense that a larger object should last longer because of how much more material it has.  Would the composition of a planet also make a difference?  That is, planet density?  Some systems actually have gas giants in the inner solar system!
 
A-L-E-X
Galaxy Architect
Galaxy Architect
Posts: 3340
Joined: 06 Mar 2017 20:19

Science and Astronomy Questions

09 Nov 2021 23:56

the inertia of a solid sphere goes as the cube of its radius, while the drag force is proportional to its cross sectional area which goes only as the square of its radius
That's a good point, and also explains why a mouse can jump off a tall building and survive whilst an elephant certainly wont.

One other thing to consider.  Even a few thousand orbits is a short moment in planetary and stellar timescales, so a planet skimming a stellar surface could still be extremely rare.
But in that case the effect is the reverse, as the smaller creature lasts longer.  That is what I originally thought would be the result with the planet, but the opposite seems to be true.  I also want to know how much planetary composition and density factors in to longevity.
 
A-L-E-X
Galaxy Architect
Galaxy Architect
Posts: 3340
Joined: 06 Mar 2017 20:19

Science and Astronomy Questions

09 Nov 2021 23:58

Yes, a short period relative to the life of the star. However, combined with the number of stars and that the vast majority of them have planets, it happens quite often -- estimates are for a planet to be engulfed by its host star evolving off the main sequence about every 1 to 10 years throughout our galaxy. 

This may actually be important for the stars, too. One effect is that the planet pollutes the envelope of the star with heavier elements, and there is some evidence that we have seen stars where that has happened (not yet in the act, as far as we know). But another consideration (and the main topic of the paper in the above link) is that the orbital energy being dragged away doesn't simply vanish, but goes into heating the envelope of the star. In some cases this may be small enough to just be a hiccup, but in others it could be significant and produce a clear signature of the process in action. This would be easiest to see for planets engulfed by smaller giant stars, since their envelopes have higher densities and lower luminosities.
Wat, should spectroscopic analysis be able to find evidence of this?  Would a planet large enough be able to slow down the star's evolutionary process and perhaps even reignite it's hydrogen fusion "engines"?  I guess for that to happen it would have to be a big gas giant with a lot of hydrogen?
 
User avatar
Watsisname
Science Officer
Science Officer
Posts: 2247
Joined: 06 Sep 2016 02:33
Location: Bellingham, WA

Science and Astronomy Questions

10 Nov 2021 02:25

Would the composition of a planet also make a difference?  That is, planet density?  Some systems actually have gas giants in the inner solar system!
Yes, absolutely. A planet twice as dense has twice the inertia but the same drag force, so its orbit will last twice as long. Terrestrial planets fare better than gas giants for another reason, which is that the bulk of gas giants is evaporated away more rapidly. They also expand due to the heating (increasing the drag force even further).
But in that case the effect is the reverse, as the smaller creature lasts longer.
Midtskogen means the mouse will have a smaller terminal velocity than the elephant and thus more likely to survive the fall. Which is indeed an excellent intuition for why a smaller object's orbit decays more quickly. Objects with the same density but different sizes have different terminal velocities -- the smaller, the slower. The decay of a planet's orbit inside a star, or of a satellite in Earth's upper atmosphere, can also be compared to throwing something sideways in air. The slower the object's terminal velocity, the more rapidly it will be slowed down by drag, and the less distance it will travel horizontally before reaching the ground. 
Would a planet large enough be able to slow down the star's evolutionary process and perhaps even reignite it's hydrogen fusion "engines"?
No. :) But what it can do besides brighten the star slightly is also increase its rotation by transferring the orbital angular momentum, which could explain some unusually rapidly rotating red giants.

Who is online

Users browsing this forum: Semrush [Bot] and 0 guests