EPT2. What happens if you heat up wood to very high temperatures inside a chamber without oxygen (edit: not necessarily a vacuum, perhaps a chamber with helium or something)?
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airtown
United States410 Posts
2. What happens if you heat up wood to very high temperatures inside a chamber without oxygen (edit: not necessarily a vacuum, perhaps a chamber with helium or something)? | ||
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Adeny
Norway1233 Posts
1 - I'd think the pull would stay the same but distributed in other directions as you're leaving some of the mass behind you. 2 - IDK. | ||
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ZeaL.
United States5955 Posts
2) ?? | ||
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Archas
United States6531 Posts
Question 1 intrigues me greatly; I never thought about it before. As for question 2, heating wood in a vacuum is very difficult in the first place, since wood is not a good conductor of heat, and the only way to heat wood in a vacuum is through conducting heat radiation through the wood. Assuming you could get it done, there'd be no fire, since fire requires oxygen. The wood would just degrade steadily and release gases; once the process completes, you'd have a very high-grade chunk of charcoal remaining. | ||
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quirinus
Croatia2489 Posts
2. Edit: 1. There's a theorem (forgot the name, but some famous physicist showed it) that shows that only the mass under the height you're standing should be used to calculate the pull, so you'd have less mass that pulls you, but you'd be closer to the center. (I think the same theorem says that the mass under your height can be assumed to be located at the center - if it's uniformly spherically distributed) The net result would be that it pulls you less, as obviously, if you were at the center, you'd have to feel no force, as the same force "pulls" you from all directiones. It's not that hard to show/prove it too. 2. http://en.wikipedia.org/wiki/Charcoal, happened naturally a long time ago... | ||
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airtown
United States410 Posts
On November 28 2010 08:41 Adeny wrote: Homework? Nope. | ||
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hacklebeast
United States5090 Posts
2. it melts, assuming that "very high" means something hotter than humans can make. the pull decreaces because as you move into the center because all of the matter that you pass starts pulling you backwards. At the center the pull of any matter is perfectly countered by matter on the opposite side, so there is no prevailing foce (ignoring the fact that the moon is not a perfect sphere.) | ||
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MetalFace
United States75 Posts
1) I think as you get closer to something, the gravitation attraction you'll feel is stronger, so as you get closer to more of the moon, the pull of gravity would increase. 2) This one I'm not so sure of, but I think it might be called gasification. Pretty much the natural gas in wood can be extracted from it through a process involving heating up wood in a sealed container. I'm not sure on either of those though. Good luck with whatever project you're doing where you heat up wood in the middle of the moon. | ||
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Archas
United States6531 Posts
On November 28 2010 08:45 MetalFace wrote: I'm not sure on either of those though. Good luck with whatever project you're doing where you heat up wood in the middle of the moon. It's clearly a terrorist plot to turn the Moon into a flaming meteor and crash it into Washington D.C. | ||
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bbq ftw
United States139 Posts
1. If a reinforced hole was drilled straight through the moon and a ladder was added, would the pull of gravity increase or decrease as you climbed down to the center? Pretty sure it decreases. Logic is a bit iffy but: Gravity at surface of moon is some finite value. Gravity at the middle of the moon is zero (there's no mass within the sphere defined by r=0) Assuming that the gravity vs radius is continuous (no abrupt breaks) one can surmise gravity is decreasing. In addition, bad over simplified math says: the sphere defined by r=x where x is your current position has volume proportional to x^3. Sphere volume assuming constant density, therefore volume proportional to mass, mass proportional to x^3 the denominator of gravity term is x^2 -- x is also the distance from moon's center of mass (center) from current position gravity proportional to x^3/x^2 = x lower x, lower gravity = decreases 2. interesting question, i have no idea. | ||
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Slayer91
Ireland23335 Posts
On November 28 2010 08:45 MetalFace wrote: I haven't taken a physics class in years, but from what I remember: 1) I think as you get closer to something, the gravitation attraction you'll feel is stronger, so as you get closer to more of the moon, the pull of gravity would increase. As you reach the dead centre, the gravitational affects of each part of mass will all cancel out, and you'd feel no force. It is stronger with respect to r^2 only if the mass pulling you is all in one direction. I assume the wood would vapourise eventually, even without combustion. | ||
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AcrossFiveJulys
United States3612 Posts
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Blisse
Canada3710 Posts
FG = (G x m1 x m2) / r^2 have anything to do with this? Cause I don't know if UND means infinite or zero... 2. Do you mean if say you placed it on an electric heater or something similar? | ||
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sheaRZerg
United States613 Posts
On November 28 2010 08:42 quirinus wrote: There's a theorem (forgot the name, but some famous physicist showed it) that shows that only the mass under the height you're standing should be used to calculate the pull, so you'd have less mass that pulls you, but you'd be closer to the center. You can use Gauss' law for gravity, much like you would for the electric charge. That is what you are describing for part 1. So the gravitational acceleration would be less on the interior of the sphere than at the surface. Edit: http://en.wikipedia.org/wiki/Gauss'_law_for_gravity | ||
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airtown
United States410 Posts
On November 28 2010 09:01 vica wrote: 1. Would need to ask which pull of gravity, Earth's, or the Moon's. Questioning why the same question could not have been Earth instead of the Moon. 2. Do you mean if say you placed it on an electric heater or something similar? 1. Ignore any outside pulls of gravity 2. Um, idk. Does it matter? | ||
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airtown
United States410 Posts
Can someone put it into laymans terms? I'm only in algebra 2. | ||
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MoltkeWarding
5195 Posts
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MetalFace
United States75 Posts
-from the wiki page you linked. Pretty much, the forces of gravity inside a sphere will cancel each other out, like everyone's pretty much been saying. So as you go inside the moon, the forces will be stronger, but the directions will start to cancel each other out, until you get to the middle where all of the forces will be cancelled. | ||
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SpiritoftheTunA
United States20903 Posts
if 'a' is the radius of the moon, and the gravitational pull at the surface is F = G*m1*m2/a^2, you just have to multiply the whole thing by a factor of your radius from the center in order to get the gravitational force for anything under the surface. so ends up working out as F = r*G*m1*m2*/a^2 (where r is the only variable in this case, your radial distance from the center of the moon) | ||
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SpiritoftheTunA
United States20903 Posts
On November 28 2010 09:16 MoltkeWarding wrote: The pull of gravity would increase, but be more evenly distributed in different directions, so there would be a net decrease, assuming that the mass of the moon were equally distributed throughout its body. If we can imagine a moon with its mass heavily concentrated at the core, it's possible that the net pull would increase. This actually isn't the case, even if 90% of the moons' mass were within its innermost 10% sphere, the gravitational pull would decrease as you go towards the center because you still will no longer feel the downward pull of all of the ground you're digging under. as long as the mass is distributed radially symmetrically, your gravitational attraction towards the very center should never increase as you dig into the mass. (actually it might not increase towards the center no matter what) EDIT: actually that was severely wrong see post below me for details | ||
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