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Author Topic: Krypton's core is made of uranium? (mathy types wanted)  (Read 26069 times)
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Uncle Mxy
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« Reply #40 on: October 15, 2005, 03:24:36 PM »

Quote from: "llozymandias"
In the comics krypton's sun was a red giant.  It just strikes me as weird to think that the kryptonians would give their sun the same name as the "creator" of the universe.

Is it any less weird that we humans generally refer to the name of our planet as Earth, the same name that we would label some lesser plot of land (but usually without a capital "E")?
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Great Rao
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« Reply #41 on: October 15, 2005, 04:11:15 PM »

Quote from: "llozymandias"
Is it any less weird that we humans generally refer to the name of our planet as Earth, the same name that we would label some lesser plot of land (but usually without a capital "E")?

I agree, let's rename it "Hooston."

S!
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"The bottom line involves choices.  Neither gods nor humans have ever stood calmly in a minefield forever.  Good or evil, they are bound to choose.  And when they do, you will see the truth of all that motivates us.  As a thinking being, you have the obligation to choose.  If the fate of all mankind were in your hands, what would your decision be?  As a writer and an artist, I've drawn my answer."   - Jack Kirby
Captain Kal
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« Reply #42 on: October 17, 2005, 02:34:31 PM »

The problem I discovered when figuring out Krypton's orbit for DakotaSmith was that 25:18 Earth year:Krypton year ratio pretty much made it very difficult if not outright impossible for me to put the planet in a habitable zone around any normal red star -- red dwarf or red giant.

It actually came closest with a red dwarf star verging on an orange dwarf at 0.33 solar masses and an orbit about 0.86 of our own distance from our sun.  But that would give a very feeble 2.7% of the solar energy that our own world receives from our sun.  If we took a red giant then we get the opposite problem that an orbit consistent with that year ratio give a planet that should have its oceans and atmosphere burned off by being too close to that darned red giant sun.  If we accept the red dwarf option, then not only were Kryptonians exposed to solar radiation that had an overall lower energy spectrum, but the actual intensity was a tiny fraction of ours.  The difference in solar energy coming to our world would be a lot more dramatic than just exchanging a red lamp for a yellow one.

One compromise is that the red dwarf figures I gave are augmented in three ways.  Krypton had a higher carbon dioxide content in its atmosphere hence having a much more pronounced greenhouse effect.  Al Schroeder speculated as much in one of his internet articles (though until now I really leaned towards the red giant option).  This is consistent with Krypton's red sky (carbon dioxide scatters red light more giving mars a red sky) and "The Sun of Superman" story where we learn without Krypton artificially affecting their sun that its natural state is an orange dwarf star.  The other factor is Krypton's own internal heat from that natural uranium reactor in its core was helping keep the temperature habitable on the planet surface.  Krypton was almost a failed star afterall and it likely had a lot of trapped heat from planetary formation and fizzled fusion reactions.

Another option is their sun was still a red giant star but the artificial presence of the Sun-Thrivers feeding on that solar energy reduced the output sufficiently to let Krypton hold onto its water and air.  For a 2 solar mass red giant star Krypton's orbit would be 1.56 x ours from our sun, FYI.

Thoughts?

(Oh, BTW, if anyone wants, I can post the calculations and formula this is based on so they can play with it to see what other possibilities they can come up with.)
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Gary
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« Reply #43 on: October 19, 2005, 05:08:30 PM »

Okay, so I'm somewhat slower than a speeding bullet.... Smiley

Quote from: "Captain Kal"
If we maintain the same proportionate density, then to double the gravity means eight times the mass with twice the radius.  Though the planet has 8 x more mass the distance to the core is doubled, and gravity obeys the inverse square law so the relative gravity is reduced by a factor of four.  8 / 4 gives us 2.


So far so good....

Quote from: "Captain Kal"
Following that same reasoning, a 35 G world with the same density as Earth would have 35 x the radius and 42,875 x the mass.  Clearly, this cannot be so since this would give a planet well beyond the brown dwarf range and it would be a small red dwarf star.  Hence, I adjusted my old figures based on the idea that a brown dwarf would be the max value for a planetary mass.

But we've specified a uranium core that suggests an overall density that's about 2.42 x greater than Earth's. (Uranium density = 19.1 g/cm^3 vs Iron density = 7.87 g/cm^3.)  That reduces the mass by that 2.42 factor to 17,666 x Earth's.


The Earth's not solid iron, and has a density of 5.515 g/cm^2 according to Wiki, but I think you're assuming that the other stuff has the same density ratio as well.

Quote from: "Captain Kal"
To achieve that 35 G gravity with that mass we divide 17,666 by 35 and take the square root of that.  That gives us a radius factor 22.46 x Earth's.  That's nearly 0.21 x the Sun's radius.

So, a uranium core implies a mass of 17,666 x Earth's, or 0.053 x the Sun's mass, and a radius 0.21 x the Sun's.


Here's where I had a problem with this. Your mass calculation above was based on the premise that you were changing the mass but not the radius.  If you now change the radius, the mass is going to change proportionally, and you don't have 35x Earth gravity any more.

Or something. Actually, I'm kinda scratching my head trying to figure out what you did here. So I'll just tell you how I would answer this, and maybe you can figure out where we disagree.

The gravitational attraction of a planet is proportional to (mass) / (radius) ^ 2. Since the mass goes like (density) x (radius) ^ 3, in terms of density and radius the gravitational attraction is proportional to (density) x (radius).

This is why, as any college physics teacher will tell you, it's better to do algebra and not put in actual numbers until the very end. Smiley

Anyway, you want an overall ratio of 35 for the gravitational attraction. The densities are in a ratio of 2.42. So you would need the radii in a ratio of 35 / 2.42 ~= 14.5. Krypton would need a radius of 14.5 Earths, or about 0.13 Suns.
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JulianPerez
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« Reply #44 on: October 20, 2005, 10:45:24 PM »

Here's a possibility concerning the destruction of Krypton, and why it was so catastrophic:

Earth, with its comparatively low gravity, has almost none of the lighter gases, which when heated achieve escape velocity from earth. This accounts for why Helium was found on the sun before it was discovered on the earth, but why gas giant planets, like Jupiter, with far greater gravity than our planet, have helium and hydrogen in their atmospheres. Krypton, with its far greater gravity, might be able to keep on to these kinds of "light" gases.

Thus, explosions would be far more destructive on Krypton than on earth, as they would cause the hydrogen in the atmosphere to ignite.
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