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Solar System in a Can May Reveal Hidden Dimensions
dylanduck writes "A model solar system, made of tungsten and placed in space, could reveal hidden spatial dimensions and test alternative theories of gravity. If the system's 'planets' moved slightly differently to the way predicted by standard gravity, it would signal the presence of new physical phenomena." From the article: "Once at the Lagrange point, the artificial solar system would be set in motion inside the spacecraft. An 8-centimetre-wide sphere of tungsten would act as an artificial sun, while a smaller test sphere would be launched 10 cm away into an oval-shaped orbit. The miniscule planet would orbit its tungsten sun 3,000 times per year."
Read more about tungsten here
There should be a "-1:Groupthink"
If the minature solar system is sent into space, then would it also come under the effect of the gravity of the actual solar system?
Lagrangian Point
FTA:
And the spacecraft components themselves would exert gravitational forces on the spheres. These forces could be minimised by making the spacecraft as symmetrical as possible and putting its heaviest components as far from the artificial solar system as possible.
"Such an experiment would be quite challenging to set up, but I don't think it is technologically impossible," says MOND expert Stacy McGaugh of the University of Maryland, US.
Not impossible can be quite a stretch to feasible, though.
I got black text on a mostly black background. Sheesh! The printable page reads a lot better.
Flyboy 8v)
I might be indecisive, but I'm not really sure. What do you think?
A tungsten sphere 10cm in diameter would have such a tiny gravitational field that I suspect even a hydrogen atom at the ambient temperature of local space would possess escape velocity.
No doubt. The only reason there is any hydrogen on *Earth* is because it binds readily with more massive elements. Helium does not and, as a consequence, any helium released into the atmosphere will ultimately escape. My understanding is that the only reason we have any helium at all is due to radioactive decay from heavier elements
Well yea, no kidding. Seeing as how the average temperature of a hydrogen gas molecule at any reasonable temperature is quite large. That could be why they aren't going to use hydrogen molecules.
Instead they are going to use tungsten, which is quite heavy. And a very slow orbit - it works out to about 8 revs per day or about 0.06 millimeters/second. That's just from their statement of an orbital radius of 10 cm and 3000 rev/year. I don't have the time to work out the orbit at the moment, but it doesn't sound to outlandish to me.
Article states the orbit would be 1/3,000 degree in year.
This is MUCH MUCH less than 3000 times in year
Since it's not explicitly stated in the article or these replies, gravitational effects precisely cancel inside a uniform shell. So if the spacecraft's mass was evenly distributed on a spherical shell there would be zero effect on items inside the shell, even when those items are close to the shell's interior surface.
Of course, the math for that is based on regular-old physics. It might not apply in higher-dimensional physics that these scientist hope to prove.
Of course, the article ignores the difficulty in clearing out L2. Legrange points, as 'stable equilibrium' points in space, are likely littered with debris, even if this debris doesn't directly impact the experiment, it will exert its own gravity that could prove problematic.
Kevin Fox
The proof, involving triple integrals, is left for the reader.
Of course, designing a spacecraft that is as spherically symmetrical and uniform in density as possible will be difficult. TFA refers to this, and before much money is spent on this project, one would hope some number-crunching is done to see how extreme the effect is.
Another problem will be microgravity. Orbital velocity is dependent upon the distance from the center of the object being orbited. In Earth orbit, even a few inches difference can produce a velocity gradient that can result in minute accelerations. At L2, some of these effects might be minimized, although again, number crunching should be done.
The late Robert L. Forward proposed a system of massive spheres that could flatten spacetime in a local region. To further minimize extraneous effects due to microgravity, a system like this might need to be used. One advantage would be that this same system might eliminate some of the problems due to assymetry in the spacecraft. One of the problems with this situation would be mass lofted, which currently tends to be expensive, and additional calculations that might be required to analyze the data.
This sig seemed like a good idea at the time....
Actually, an 8cm tungsten sphere would exert the same gravitational pull on any object 10cm away, regardless of the other object's mass. It would have an escape velocity of 0.013 cm/s or 1.3 microns per second -- which, while very slow, is certainly within the realm of feasability. Your hard drive heads move accurately with tolerances significantly smaller than that.
I calculated the escape velocity using the formula sqrt(2Gm/r):
sqrt((2)(6.6742x10^-11)(5.16)/0.4) = 0.00013m/s or 0.013cm/s
Random and weird software I've written.
Gauss's Law says that the gravitational acceleration of a body anywhere in an enclosed sphere is 0. At L4, L5 Earth and Sun graviational forces are balanced. The only accelerations that don't cancel out are the two body accelerations of interest. It is surprising to me that the bodies orbit as fast as 10 times per day. I wonder why they don't use heavier Uranium as the mass. It is an interesting side note that a body can stably orbit one of these points. They orbit with no body (!) at the focus. The Genesis Probe and WMAP missions have already taken advantage of this.
an ill wind that blows no good
Ok, some orbital mechanics.
./3000 = 10518.912 seconds per orbit .00009506686623103225 orbits per second
.*.14*3.1415926*2 meters per orbit =
.0000836 meters per second
.*1000 = .0836 millimeters per second
.000268 m^3
.*19250 = 5.16 kg .004 m in radius .000000268 m^3
.*19250 = .00516 kg
.00000000006673000000 * 5.16 * .00516 / (.1*.1) .00000000017767262800 Newtons of force, resulting acceleration on the smaller body of
./.00516 = .00000003443267984496 m/s = .00003443267984496 mm/s
.0836 millimeters per second everything should be fine!
Going with a circular orbit because they didn't specify the ellipse:
365.24*24*3600 = 31556736.00 seconds per year
1/. =
Pretty slow orbit. About that tungsten, 19250 kg/m3
3.1415926*(4/3)*.04*.04*.04 =
And let's say the planet is 8 mm in diameter,
3.1415926*(4/3)*.004*.004*.004 =
F = G m1 m2 / r^2 =
gravitational constant = 6.67300 × 10-11 m3 kg-1 s-2
=
Sounds reasonable to me. Assuming they can get a clean launch at exactly
Start Running Better Polls
Dude, quite bogarting that J.
"To hide" is both a transitive and intransitive verb, your selective quoting of dictionary definitions notwithstanding.
Since it can be an intransitive verb, it does not imply an actor. You're not nitpicking, you're being a fucking idiot. Every fucking conversation about science does not have to include ID-bashing just for the sake of it. ID is as big a load of bullshit as your post, but there was no reason to give it a venue by bringing up retarded criticisms disguised as pedantry. Pedantry is bad enough, but what you present is total fucking nonsense.
Calculating the position of the moon throughout the month and deriving the orbit wasn't something I did until I got out of college. It's well within the capability of a Freshman physics student, so in theory we could have confirmed the inverse square law to a decent level of precision.
Tightening the exact value of that exponent (is it really -2?) further is the purpose of the proposed experiment.
If you know that gravity follows an inverse square law, then you know that inside a uniform sphere the gravitational acceleration will be zero.
You are correct. We never demonstrated experimentally for gravity that the net gravitational force inside a sphere was zero. Of course, I never said we did. The term "demonstrate" can, in fact, be used in a mathematical sense. When one of the kids on our dorm floor claimed the Ringworld was unstable, we had no trouble demonstrating that instability -- not that anyone had a Ringworld to work with.
This sig seemed like a good idea at the time....
The way any scientist would. List all known possibilities of your "myriad of possible effects". Then quantitatively estimate and calculate the magnitude of those effects on the orbit's precession. If all effects are less than the gravitional effect by some quantity greater than the experiment's margin of error, then you assume you can measure this and run the experiment.
If the experiment doesn't give you the values you expect, then you find a flaw in either your theory, your experiment, or your assumptions of the possible effects, etc. If it does give you the values you expect, then you have another good data point in the MOND or similar graviational theory. In any case, you either learn something useful, and further solidify our understanding of known physics.
I doubt you can see ultraviolet light, or any other electromagnetic radiation besides that of visible light. However, it still interacts with the physical world. Just because you can't see gravity or another dimension doesn't mean it doesn't exist.
Your application of Gauss's law is flawed- all those formulas only tell you about the total strength of the field generated by objects inside them (Hence enclosed). Otherwise, your argument could be used to show that no electric or gravitational fields exist anywhere except inside the objects that produce them, since you could otherwise always arrange the volume integral to not include any charged objects, making q or g enclosed = 0.
The fact that your comment was modded +5 Informative suggests that more slashdotters need to go back to introductory physics courses. And you call yourself nerds...
Um, I don't think so.
The effects cancel very nicely at the exact center, and nowhere else. As you get off-center, the attraction of the nearest wall exceeds the attraction of the opposite wall.
.
Read my first reply to my comment for more clarification if you want. But as per your comment here, the surface integral of the vector field (dot producted with its infinitesmal area element, of course) is identically zero for any surface enclosing zero net source/sink density (ie, masses or charges). Just because this surface integral (ie, a continuous sum) is zero doesn't imply the local vector field at any point on or within the surface will be zero. As perfectly exemplified by my brick example.
And finally, the gravitational field within a hollow sphere is zero only if the gravitational field is also zero in the absence of that sphere.
Sorry, you're wrong. The gravitational force due to a hollow sphere is exactly zero everywhere inside that sphere.
Proof