Solar System in a Can May Reveal Hidden Dimensions 251
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."
What's tungsten? (Score:1, Informative)
Re:Outside effects? (Score:3, Informative)
Lagrangian Point [wikipedia.org]
Re:Gotchas, we got em (Score:4, Informative)
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.
Black on black SHOULD be a crime (Score:2, Informative)
Flyboy 8v)
Re:Suspect this is rubbish - NS has been had? (Score:5, Informative)
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
Re:Suspect this is rubbish - NS has been had? (Score:1, Informative)
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.
Summary is incorrect. (Score:2, Informative)
This is MUCH MUCH less than 3000 times in year
Re:Gotchas, we got em (Score:3, Informative)
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.
Re:Gotchas, we got em (Score:4, Informative)
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 [aps.org]. 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.
Re:Why L2? (Score:2, Informative)
Re:Suspect this is rubbish - NS has been had? (Score:5, Informative)
I calculated the escape velocity using the formula sqrt(2Gm/r) [yale.edu]:
sqrt((2)(6.6742x10^-11)(5.16)/0.4) = 0.00013m/s or 0.013cm/s
Gauss's Law (Score:5, Informative)
Gauss's Law [wolfram.com] 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.
High School Physics (Score:5, Informative)
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
Re:What if (Score:1, Informative)
Re:Gauss's Law (Score:5, Informative)
What a load of shit. (Score:1, Informative)
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.
Re:Gauss's Law (Score:5, Informative)
No it doesn't, re=read the law you linked to. It says the "surface integral of gravitational acceleration" will be zero over any arbitrarily-shaped closed surface, as long as that surface encloses zero mass. You cannot work backwards from this statement to assume that the local gravitational acceleration will be zero.
Simple example. Imagine a closed surface (say a small sphere) 20 feet above the ground (and also assume there's no air inside) such that the surface is closed. Since it encloses no mass, the net acceleration will be zero as summed over the whole sphere. However, any object placed within this hypothetical spherical surface (eg a brick) will fall to the ground.
Re:Gotchas, we got em (Score:4, Informative)
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.
Re:Too many uncertainties (Score:3, Informative)
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.
Re:High School Physics (Score:2, Informative)
T=3000 rev/year => 9.5066x10^-5 rev/s
take the inverse
P = 10519.007s (some rounding error)
mu is known as the standard gravitational parameter, and can be found by multiplying the mass of the object by the universal gravitational constant:
mu = G*m
to find the mass take the volume of a sphere of diameter 8cm (FTA) and multiply it by the density of tungsten:
m = V*rho
V = (4/3)*pi*r^3 = (4/3)*pi*(4cm)^3
= 268.08 cm^3
rho = 19.25 gm/cm^3
m = (268.08 cm^3)(19.25 gm/cm^3)
= 5160.589gm
= 5.160 kg
mu = G*m
= (6.6742x10^-11 m^3*s^-2*kg^-1)(5.16 kg)
= 3.443 m^3/s^2
Using mu and the period P, we can find the semimajor axis a:
P = 2*pi*sqrt(a^3/mu)
solving for a:
a = (P^(2/3)*(2*mu)^(1/3))*(2*pi^(2/3))^-1
plugging all those fun numbers in:
a = 212.838 m
from there we can find the eccentricity by the formula:
e = 1-(r_p/a)
r_p is the radius at the periapsis of the orbit (the closest point in the orbital path to the mass it is orbiting).
FTA the "planet" will be launched 10cm away into an oval shaped orbit. For an arbitrary assumption, we'll take 10cm to be the periapsis of this orbit.
That would put the eccentricity at e=.99953, which is extremely close to the eccentricity of a parabolic trajectory of e=1, so 10cm is most likely not the periapsis of the orbit.
However since the semimajor axis of the orbit is so large compared to the initial given value of 10cm, it wouldn't make sense for it to be at any other point in the orbit. So essentially, from the numbers given in the article we have an extremely large system (total orbital diameter is 2a = 425.676m) which would be impractical to create and launch to the legrange point.
Unfortunately I believe we have yet another case of not enough actual details in the article to come to a reasonable conclusion about the physics behind the story.
As always, if there's something wrong with the analysis or any of my calculations please point them out. It's been a while since I've done any of this stuff and mistakes do happen.
Re:Forgive me but I have to nitpick (Score:1, Informative)
Gauss's Law does not apply (Score:1, Informative)
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...
Re:Gotchas, we got em (Score:3, Informative)
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.
Re:Confused physical reasoning (Score:3, Informative)
.
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.
Re:Gotchas, we got em (Score:2, Informative)
Proof [nasa.gov]