Space Travel harder in M-Class starsystems

[Widdekind]

Stellar Habitable Zones are defined by:

 

L

_*

/

D

_HZ

²

~

const

 

And, stellar Luminosities scale approximately as the fourth-power of the star Mass (Bowers & Deeming. Astrophysics I: Stars, pg. ~28).

 

Thus, the Habitable Zones of smaller & dimmer stars are deeper down into their parent stars' Gravity Wells:

 

U

_HZ

=

- G M

_*

/ D

_HZ

~

M

_*

^-1

 

And so, Interplanetary Space Travel in such systems would require considerably greater (~several times more) Energy expenditures.

 

Moreover, most M-Class stars' potentially Habitable Planets, b/c they orbit so closely to their parent star, are probably Tidally Locked to those stars*, and probably possess no Moons**. Thus, those potentially Habitable Planets will spin much more slowly, making Rocket Launches even more Energetically costly, and they will possess no nearby Planetoids for easily accessible off-world Colonies.

*

BBC / Channel 4

Alien Worlds

(DVD)

**

http://www.astronomyforum.net/planetary-forums/82391-most-habitable-exo-planets-moon-less.html

Moreover still, M-Class stars are often violent Flare Stars*. Thus, Interplanetary Space Travel in such systems would pose greater Radiation Hazards to space-farers.

*

BBC / Channel 4

Alien Worlds

(DVD)

However, most M-Class stars' potentially Habitable Planets must, apparently, be comparatively small*. They would, thusly, have lower Surface Gravities, making them comparatively easier to escape.

*

http://en.wikipedia.org/wiki/Extrasolar_planet#Stellar_characteristics

**

http://www.physicsforums.com/showthread.php?t=278022

Edited May 2, 2009 by Widdekind

[ydoaPs]

Why would a Minshara class planet need to be in a smaller/dimmer star? IIRC, Vulcan orbits a Red Giant.

Edited May 2, 2009 by ydoaPs

[Widdekind]

To date, most known Extra-Solar Planets orbit K, G, & F-Class stars*.

*

http://en.wikipedia.org/wiki/Extrasolar_planet#Stellar_characteristics

[VedekPako]

To date, most known Extra-Solar Planets orbit K, G, & F-Class stars*.

*

http://en.wikipedia.org/wiki/Extrasolar_planet#Stellar_characteristics

 

That's probably more to do with astronomers only looking at K,G & F stars because they are looking for the holy grail of planet hunting, an Earth-like world. After all the possible K, G & F stars are examined, astronomers will look at M stars and probably find more planets around them than K, G, & F stars.

[Martin]

I'm in partial agreement with Vedek. The planet hunters have tended to focus on sun-like stars. Sol is type G, so they have tended to study F, G, and K.

And so more planets have been found around F, G, K because they were looking more at those types of stars.

 

But M type stars are quite numerous, and I believe planets have been found at some of them. So in the end it might turn out that there are more planets in total that are orbiting M stars.

 

For anyone joining the discussion late, astronomers have this curious classification of stars which, going from massive hot stars down to smaller cooler stars, is the "spectral sequence"

O, B, A, F, G, K, M

 

And we were taught a mnemonic to remember the sequence: "Oh Be A Fine Girl, Kiss Me."

 

The sequence and the quaint mnemonic go back to an earlier and more innocent time. According to Wikipedia the classification scheme was invented around 1900 by Annie Jump Cannon

http://en.wikipedia.org/wiki/Annie_Jump_Cannon

She is also credited with making up the mnemonic to help remember that peculiar sequence of letters.

Edited June 4, 2009 by Martin

[Reaper]

 

But M type stars are quite numerous, and I believe planets have been found at some of them. So in the end it might turn out that there are more planets in total that are orbiting M stars.

 

We know of at least one vaguely Earth-like planet around a red dwarf, perhaps a planet you are already familiar with?

 

http://en.wikipedia.org/wiki/Gliese_581_d

 

 

Furthermore, it has been recently found to be well in it's habitable zone, so there is the potential for life.

[SH3RL0CK]

The difficulty of space travel depends not necessarily on the gravity well you are climbing, but how high you are climbing it. It has been said "that once you are in orbit around the earth, you are halfway to anywhere" and the intent of this statement is that the earth is the deepest gravity well that most space probes have to leave.

 

Thus, space travel is highly dependant on the size/gravity of the specific planet or moon it is launched from. I suspect this will be highly variable, and I don't see any reason why the stellar class would influence the planet size. (Edit: The reference provided in the wiki link isn't an internet address, so I couldn't read it; but the use of "suggests" in the wiki article means it isn't assured that the planets of M-type stars are smaller...Maybe the stellar type here has an influence, but I don't understand any reason why that would be the case so I will maintain the planets could be any size until either there is more data, or a scientific reason for habitable planets/moons being less massive...)

 

The second variable is how much climbing of the stars gravity field to get to the planet/moon of interest. While a civilization might be deeper in the gravity well, so might the other planet of interest. Again, I am not aware of any reason why relative distances between planets and moons should be influenced by the stellar mass, so I suspect this too would be highly variable.

 

Now as far as leaving the star itself, it could be harder for certain star types. But because the immense distances between stars (and the time required to travel this far) is a much, much greater obstacle; I don't see the stellar mass as being significant here either.

 

For the above reasons, I would state that the stellar type would have much less influence than the variables above and therefore it is my opinion that space travel is not necessarily harder in any particular star type.

 

Please correct me, however, if my understanding regarding my assumptions above (stellar type does not influence planetary mass or relative distances between planets) are not true.

Edited June 4, 2009 by SH3RL0CK

[Martin]

Reaper, thanks for pointing that out. The Gliese 581 system is a good example.

 

Sherlock, one reason I like Widdekind's exercises is they add some zest to the astrophysics by deriving consequences for ET (or future humans and our robot surrogates) from basic formulas. In a sense it makes the formulas come alive, by putting some imagined beings into the picture. It is the kind of thing that could eventually lead someone to a creative online teaching tool or beginning astro workbook.

 

In this case Widdy has a valid point! And it is based on some very simple algebra!

 

Check this out. I'll repeat what Widdy is doing.

 

The wattage of the star varies as M^4 (he gives a reference)

So the habitable zone is a belt a few percent around some ideal distance D which is where the wattage hitting a square meter is ideal for water. OK?

So M^4/D^2 = constant

 

So D is proportional to M^2 by a kind of universal proportionality the same for all stars and planetary systems (just approximate but fair enough)

 

So for starters imagine you are a colonizer evaluating prospects for settlement and you see a star with an earth-mass planet in circular orbit at the distance

D from the star (or a few percent from the ideal distance D).

 

Is this a good prospect? Well Widdy can calculate for you how much potential energy you have to sacrifice to get down to that planet.

 

He calls that energy U, and he points out that by standard Newton it is proportional to M/D.

But D is proportional to M^2! We already said.

So the energy you have to blow off is proportional to M/M^2 which is 1/M.

 

So the energy cost to the colonizer of getting down into the habitable, where he can do a flyby of the planet, is reciprocal to the mass of the star.

 

Habitable planets of low-mass stars are more costly to visit. Because a low M means a high 1/M.

 

=====================

 

This type of thing is going to carry over to pretty much any scenario you imagine. it applies not only a colonizer coming from a long distance away but also it applies to the locals who want to migrate from planet to planet within the habitable zone.

The energy difference between the inner and outer boundary of the habitable is going to be a fixed percent of 1/M. So the smaller M is, the larger that will be.

 

=====================

 

Basically what I see Widdekind doing is experimenting with a new kind of highschool or freshman college physics textbook, which integrates and motivates

some classic basics by imagining these principles and laws applied to the job of spreading life thru the galaxy.

It is not a bad idea for a pedagogical approach.

 

And whenever you have a physics textbook example there are bound to be some simplifying assumptions. Like "other things being equal". Or like "the colonizer's main concern is the energy cost of getting down to the level of the planet in the star's gravity well." Obviously you can imagine other concerns, but in a physics problem, to illustrate something, you narrow down.

Edited June 4, 2009 by Martin

[SH3RL0CK]

Oh please don't misunderstand, I really enjoyed the thought exercise. I've thought it through and for the reasons outlined above, I disagree with the assertion that space travel is more difficult for certain stars than for others; based simply on the fact that I think the variables I pointed out will dominate the variables Widdekind suggested.

 

But I freely admit I could be totally wrong (and would welcome being proved wrong BTW) and in no way intended to throw cold water on these ideas. Its very clever, IMO, to try to deduce the consequences of stellar type with regards to space travel; I would never have thought of it! And his logic is indeed compelling.

[Martin]

... And his logic is indeed compelling.

 

Yes, within the narrow confines of how he set the problem up. I'm glad you enjoyed it too. I get a mild kick out of scenarios involving some simple algebra.

 

Actually, if I (or my robot surrogate) were the Colonizer Captain, I would avoid type M stars because I just don't like the color! Yuck! too reddish and too much infrared compared with visible. I (and my surrogate) like cool days with bright golden sunlight.

 

But I might use Widdekind's algebra to justify my decision to skip that star, to give an excuse to the board of directors of the Foundation that was paying me.

Edited June 4, 2009 by Martin

[swansont]

I'm in partial agreement with Vedek. The planet hunters have tended to focus on sun-like stars. Sol is type G, so they have tended to study F, G, and K.

And so more planets have been found around F, G, K because they were looking more at those types of stars.

 

Right. The spectrum of the kind of signal they can detect is by no means complete. Earth-like planets in earth-like orbits around sun-like stars are not yet detectable, though we are getting close.

 

One tool in this improvement is an optical-frequency comb used (the "astro-comb") to improve the calibration of telescopes as they measure spectral line Doppler shifts.

http://www.photonics.com/Content/ReadArticle.aspx?ArticleID=33353

[VedekPako]

Re: HZ

 

The habitable zone around a M star is small, smaller than the HZ around an F star, so, M stars will have less chance of having a planet in the HZ? True, but, gliese 581 has planets in the HZ, so, does this mean that M class stars are more likely to have planets in the HZ because the smaller the star, the more closely planets will form?

[Reaper]

Re: HZ

 

The habitable zone around a M star is small, smaller than the HZ around an F star, so, M stars will have less chance of having a planet in the HZ? True, but, gliese 581 has planets in the HZ, so, does this mean that M class stars are more likely to have planets in the HZ because the smaller the star, the more closely planets will form?

 

I don't think we really know all that much about how planets form around various types of stars to be able to say with certainty.

 

But, there is no reason that the process should be any different than the one that formed our solar system (or other solar systems); after the accretion disk phase, the size and mass of the planets that form at various orbits really depend on the distribution of mass around the stars and not so much it's proximity.

[Widdekind]

Smaller, dimmer stars typically produce far fewer Terrestrial-type planets, but make more Jupiters & Kuiper Belt Objects:

http://www.physicsforums.com/showthread.php?t=278029&page=2

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...So the energy you have to blow off is proportional to M/M^2 which is 1/M.

 

So the energy cost to the colonizer of getting down into the habitable, where he can do a flyby of the planet, is reciprocal to the mass of the star.

 

Habitable planets of low-mass stars are more costly to visit. Because a low M means a high 1/M..

 

That is an excellent extension of the reasoning !

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That's probably more to do with astronomers only looking at K,G & F stars because they are looking for the holy grail of planet hunting, an Earth-like world. After all the possible K, G & F stars are examined, astronomers will look at M stars and probably find more planets around them than K, G, & F stars.

 

The cited Wikipedia article accounts for that:

Most known Exoplanets orbit stars roughly similar to our own Sun, that is, main-sequence stars of spectral categories F, G, or K. One reason is simply that planet search programs have tended to concentrate on such stars. But even after taking this into account, statistical analysis suggests that lower-mass stars (red dwarfs, of spectral category M) are either less likely to have planets, or have planets that are themselves of lower mass and hence harder to detect.

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Oh please don't misunderstand, I really enjoyed the thought exercise. I've thought it through and for the reasons outlined above, I disagree with the assertion that space travel is more difficult for certain stars than for others; based simply on the fact that I think the variables I pointed out will dominate the variables Widdekind suggested.

 

But I freely admit I could be totally wrong (and would welcome being proved wrong BTW) and in no way intended to throw cold water on these ideas. Its very clever, IMO, to try to deduce the consequences of stellar type with regards to space travel; I would never have thought of it! And his logic is indeed compelling.

 

You raise a valid concern. However, how does the Escape Velocity of Earth compare with that of the Sun, from Earth's Orbit ??

 

Isn't the latter basically Earth's Orbital Velocity ? And, isn't Earth's Orbital Velocity ~30 km/sec, or around 3x that of its Escape Velocity ?

 

Moreover, don't Gravitational Potential Wells "add" or "super-impose" ? Thus, if you put the same planet, down deeper into a low-mass stars' Habitable Zone, havn't you actually twice-increased the Energy costs ?

[SH3RL0CK]

You raise a valid concern. However, how does the Escape Velocity of Earth compare with that of the Sun, from Earth's Orbit ??

 

Isn't the latter basically Earth's Orbital Velocity ? And, isn't Earth's Orbital Velocity ~30 km/sec, or around 3x that of its Escape Velocity ?

 

Moreover, don't Gravitational Potential Wells "add" or "super-impose" ? Thus, if you put the same planet, down deeper into a low-mass stars' Habitable Zone, havn't you actually twice-increased the Energy costs ?

 

Interesting perspective, I'll have to give this some more thought. I was under the impression that as the spacecraft (once in earth orbit) is already travelling at 30 km/sec. So to go to mars, for example it needs to attain approximately the orbital velocity for mars, or about 24 km/sec.

 

http://nssdc.gsfc.nasa.gov/planetary/factsheet/marsfact.html

 

Thus, wouldn't it only need an additional 6 km/sec to go from Earth to Mars? You've already done 11 km/sec to get off the earth. You aren't trying to escape from the sun, just to alter your orbit around the sun so you can go from one planet to the next. This isn't my best subject so please help me if I am missing something here?

Edited June 7, 2009 by SH3RL0CK
correction of typo