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Forum » SpaceEngine » Science and Astronomy Discussions » Science and Astronomy Questions
Science and Astronomy Questions
midtskogenDate: Friday, 11.10.2013, 22:08 | Message # 196
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Triple integral? This you solve with basic geometry and the risk of making errors is less. So the tube is πd long, i.e. 9425 meters. Let's say you want a 5 meter inner radius and a 1 meter thickness (6 meter outer radius). The curvature doesn't matter here. The volume is then π×((6m)²-(5m)²)×9425m=325,704m³. Multiply with that 8000 kg/m³ and you get 2,605,636 tonnes.

If we assume a current price of $4500/kg for sending stuff into orbit (the price of steel itself is a couple of $ per kg), you need to spend about 11.7 trillion dollars to launch this, or pretty much precisely the US debt. So, Americans, you owe us a 3 km diameter Stanford torus. Get to work!





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Edited by midtskogen - Friday, 11.10.2013, 22:41
 
WatsisnameDate: Saturday, 12.10.2013, 00:58 | Message # 197
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Sure, but I just prefer the integration for deriving the volumes of solids. It really is just an extension of geometrical thinking, and if you're no stranger to calculus then this integral is insanely easy to solve; you can pretty much almost do it in your head. smile




 
HarbingerDawnDate: Saturday, 12.10.2013, 01:39 | Message # 198
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Quote midtskogen ()
If we assume a current price of $4500/kg for sending stuff into orbit

Falcon Heavy could carry material to orbit for $2200/kg, not counting the savings that will be achieved by its reusability smile Including those savings, it could achieve $400/kg or less. So America owes you 10 Stanford tori biggrin





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midtskogenDate: Saturday, 12.10.2013, 07:07 | Message # 199
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Quote Watsisname ()
It really is just an extension of geometrical thinking, and if you're no stranger to calculus then this integral is insanely easy to solve

Sure, but I last did calculus and integrals at the university 20 years ago and it included many formulas that really couldn't be derived intuitively, whereas the need for πd or πr² pops up more frequently. smile





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WatsisnameDate: Saturday, 12.10.2013, 21:26 | Message # 200
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Fair enough; I'm sure that without regular use the tricks of calculus are forgotten very quickly. tongue

By the way, if anyone's wondering where those crazy integrals came from in the first place, it comes straight out of thinking about how to create the shape:



Take a point in space, extend it radially from r(a) to r(b) to get a segment, then sweep that segment through an angle 2π along r to get a hollow disk, then sweep that disk through another 2π along R to get a hollow torus. Mash all these bits together and you get the triple integral of r*R*dθ*dφ*dr, with θ from 0->2π, φ from 0->2π, and r from a->b. Put another way, this triple integral is quite literally a construction formula for the torus.

That's kind of why I like this stuff; I think it's really beautiful. smile It can also be applied to almost anything, even shapes for which the area or volume formulas are not at all obvious, or even in dimensions higher than 3.





 
SalvoDate: Saturday, 12.10.2013, 22:29 | Message # 201
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Quote midtskogen ()
If we assume a current price of $4500/kg for sending stuff into orbit (the price of steel itself is a couple of $ per kg), you need to spend about 11.7 trillion dollars to launch this, or pretty much precisely the US debt. So, Americans, you owe us a 3 km diameter Stanford torus. Get to work!


It's 2142 and they take materials from other planets biggrin





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themohawkninjaDate: Saturday, 26.10.2013, 02:36 | Message # 202
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If we find an gas giant that is 5-10 times larger than Jupiter orbiting around a star, and it is the only planet-sized body orbiting the star, is the system a star with a gas giant, or a brown dwarf and star binary?
 
HarbingerDawnDate: Saturday, 26.10.2013, 02:47 | Message # 203
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A brown dwarf is an object which is large enough to fuse deuterium in its core. 5-10 Jupiter masses is not large enough to do this (the normally accepted cutoff for brown dwarfs is 13 Jupiter masses). Therefore, it is simply a planet orbiting a star.




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WatsisnameDate: Saturday, 26.10.2013, 03:17 | Message # 204
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Yep. smile The presence or lack of additional planets in the system should also have no impact on what we call the 5-10MJ body, and given that range of masses it is considered a gas giant.




 
themohawkninjaDate: Friday, 08.11.2013, 02:22 | Message # 205
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I've been watching a video about the creation of the Earth, and as it talked about the massive impacts that the Earth sustained in it's development, especially with the Moon (it asserts the impact theory of the moon's creation), how come the Earth is in such a relatively circular orbit? Shouldn't all of those impacts put the Earth in a very eccentric orbit?
 
HarbingerDawnDate: Friday, 08.11.2013, 02:32 | Message # 206
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Quote themohawkninja ()
Shouldn't all of those impacts put the Earth in a very eccentric orbit?

Only if they'd all come from the same direction at the same point in the orbit. Keep in mind that most of the colliding objects would have had similar orbits to each other, so the effect each would have on the final orbit would be very small. And anyway, the forces caused by gravitational interactions between the planets later could circularize an eccentric orbit, or make a circular orbit eccentric.





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neutronium76Date: Sunday, 10.11.2013, 09:16 | Message # 207
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Cont'd from here:

So if, according to the above doc, the mass of a galaxy's central black hole is 0.5% the mass of the galaxy itself, how come and there is enough gravitational force to keep the stars in orbit around the galactic center? Especially those, like our sun, that reside in outer region of the galactic bulge? Unless astronomers prediction above is wayout wrong, it seems that the mechanisms involved in the formation and shape of a galaxy are much more complicated than those of a solar system (because a solar system's mass is 95-99% located in its central star) blink .





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WatsisnameDate: Sunday, 10.11.2013, 10:44 | Message # 208
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Orbits of stars (any object really) depends on the total mass enclosed within that orbit, and essentially ignores everything outside the orbit. You can plot the velocity of stars as a function of their distance from the galactic center and get a rotation curve (from which we also get this observation that there is more mass in and around galaxies than what we can see visually -- hence dark matter).

The supermassive black hole at the center is very important for the motions of stars very close to the center. But as you look at orbits of stars much farther out, the contribution from the black hole becomes much less important because there's all this additional material enclosed within the orbit.





 
neutronium76Date: Monday, 18.11.2013, 12:09 | Message # 209
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Cont'd from here

Quote Watsisname ()
Earlier in this thread, werdna asked if the Oort cloud would pose a risk to high-velocity spacecraft leaving the solar system. The answer 'not really' was given without much proof, but I recently got to thinking about it again and wanted to get a numerical answer. I was curious how much area of the sky is actually taken up by comets in the Oort cloud, if you added them all up. So here's a little math exercise and some interesting tidbits found out of it.

We don't really have a good model of comet size distribution or position, only some general estimates, so let's treat the Oort cloud as being of some number n of comets of typical radius rc and distance from the Sun, that is the average distance of the Oort cloud, RO. We will need to consider the total mass of the Oort cloud MO, and typical density of a comet ρ. Since the cloud is so diffusely populated, we may also ignore self-shadowing (no comets hiding behind other comets). If that was actually important, then, as midtskogen noted, it would dim the background stars.

So the ratio of solid angle taken up by comets to the whole sky is the solid angle of a comet times the number of comets divided by the solid angle of a sphere seen from within (4pi steradians)

Treating comet nuclei as spherical, their total number may be estimated as

Using small angle approximations, the angle subtended by a comet is

Combining expressions,

Estimating the Oort cloud to contain about 2 Earth masses1 in the form of comets of ~1km in radius, with a typical density2 of ~0.6kg*m-3, then this ratio is on the order of 10-8. That is, if you chose a line of sight at random, then the chance of it intersecting a comet in the Oort cloud is about one in a hundred million, give or take a couple orders of magnitude given the assumptions and uncertainties. Nevertheless, it is clearly extremely small, not even worth worrying about for interstellar spacecraft or probes.

Still, I don't actually have a good grasp of what the number means physically. Here's a good way to visualize it: If we take the total area of the sky covered by the comets, then what is the angular size of a single body that covers the same amount of sky?

Plugging in values, we find an equivalent angular diameter of about 9x10-5 radians, or about 20 arcseconds. That's comparable to the angular size of the planet Saturn as seen from Earth.

So with all the comets in the Oort cloud, indeed many trillions of them, even if you stuck all of them together in a flat disk face-on to the Earth, it would still only be a very tiny point in a huge sky, needing at least a good set of binoculars to resolve it.



That's pretty good research! I have another question: what is the risk of micometeoride impact on spacecraft hull and how that risk increases together with its corresponding damage as velocity of spacecraft increases? I mean a dust grain sized particle is probably not very significant if traveling with a few km/sec but if you travel with a significant fraction of c then even a nanogram impact becomes significant. Is there any graph that can illustrate this relationship visually (because my math is a bit poor sad ) ?





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WatsisnameDate: Monday, 18.11.2013, 20:32 | Message # 210
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Certainly! Perhaps the best way to make sense of it is to see how the kinetic energy of an object increases as a function of its speed. This is what most directly determines how much damage it does when it impacts something. We can graph this, too.

The kinetic energy follows a very simple formula,

You could also take the mass out of the equation and consider the specific kinetic energy (how much energy per unit mass). Graphing it, we get this, just a parabola.

This (Newtonian) formula works great for low speeds, but if you consider speeds that are large fractions of the speed of light then it fails badly. The relativistically correct formula for the energy is

where γ (gamma) is just a short hand for the square root stuff, which appears often in relativity.

But don't worry about the formula itself, let's just plot the two together and you can see the difference when the speed gets close to c (in the plot the speed of light equals 1). The Newtonian one incorrectly goes to infinity as the speed goes to infinity, while the relativistic formula correctly goes to infinity as the speed approaches c. At low speeds the relativistic formula simplifies to the Newtonian one and they are almost indistinguishable.

So this gives some insight as to how the damage potential increases with speed, and it's quite tremendous when you get to the relativistic regime. Even impact with dust specks would be a big problem. Fortunately, or perhaps sadly, I don't think we have to worry about us having spacecraft capable of such speeds any time soon. Of course, even at low speeds, say the few to tens of km/s for which we currently operate, impact with particles and debris is still very important. NASA has a whole research branch that studies this stuff, which I think is really cool. smile





 
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