Slashdot Mirror


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."

52 of 251 comments (clear)

  1. Gotchas, we got em by Ancient_Hacker · · Score: 3, Interesting

    This sounds mighty dubious. The gravitational attaction of the spacecraft is likely to be much larger than the effect looked for.

    1. Re:Gotchas, we got em by d34thm0nk3y · · Score: 3, Informative

      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.

    2. Re:Gotchas, we got em by pilgrim23 · · Score: 5, Funny

      the old L5 Society wanted to place a module they called a High Orbital Mini-Earth there... sort of a H.O.M.E. on LaGrannge.....

      --
      - Minutus cantorum, minutus balorum, minutus carborata descendum pantorum.
    3. Re:Gotchas, we got em by ceoyoyo · · Score: 4, Insightful

      They mentioned that would have to be taken into account. Scientists measure the gravitation attraction between human scaled objects on Earth all the time, yet that's dwarfed by Earth's gravity.

    4. Re:Gotchas, we got em by KFury · · Score: 2, Informative

      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.

    5. Re:Gotchas, we got em by Rob+Carr · · Score: 3, Informative
      In Freshman physics, it's common to demonstrate the net gravitational or electrical attraction inside a uniform sphere is zero. Any force with an inverse-square law will exhibit this peculiarity. If you want the details, there's a Wiki article on the Divergence theorem of vector fields.

      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....
    6. Re:Gotchas, we got em by sharkey · · Score: 2, Funny

      I hear they gotta lotta nice girls.

      --

      --
      "Outlook not so good." That magic 8-ball knows everything! I'll ask about Exchange Server next.
    7. Re:Gotchas, we got em by Ohreally_factor · · Score: 2, Funny

      I think that the experiment will have at least one protocol governing researchers shuffling their feet on the carpet.

      --
      It's not offtopic, dumbass. It's orthogonal.
    8. Re:Gotchas, we got em by Rob+Carr · · Score: 3, Informative
      We demonstrated that forces that follow an inverse square law follow this rule. We demonstrated that a charged sphere followed that rule in a lab by charging the sphere and then measuring the electrical force inside the sphere and out. We demonstrated that electrical forces follow the inverse square law in the lab. I'd argue that stable orbits demonstrate inverse square law for gravity, and we did visit the telescope and look at the moon in Freshman physics. We also calculated G using the old torsion technique.

      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....
    9. Re:Gotchas, we got em by Ancient_Hacker · · Score: 2, Informative
      >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.

      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.

    10. Re:Gotchas, we got em by As_I_Please · · Score: 2, Informative

      Sorry, you're wrong. The gravitational force due to a hollow sphere is exactly zero everywhere inside that sphere.

      Proof

  2. Outside effects? by Clazzy · · Score: 2, Interesting

    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? Granted the effect will be very small (considering one object is very small and is far away anyway) but surely it would cause enough of an effect to make a difference? I'm sure they're trying their best to cancel out these forces, but they'd need absolutely minute amounts of gravity or (impossibly enough) none at all for a good reading.

    --
    If we can hit that bull's-eye, the rest of the dominoes will fall like a house of cards... Checkmate.
    1. Re:Outside effects? by Compholio · · Score: 2, Informative

      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

    2. Re:Outside effects? by HelloKitty · · Score: 2, Insightful

      even the lagrangian point feels miniscule effects from other planets...

      it's lagrangian for the earth/moon system... not for the rest of the planets...

      with that force, and with the gravity from the spacecraft, how can any measurements be useful enough (i.e. free from otside noise) to show anything useful? one ide.... maybe they will model everything (spacecraft, and solarsystem) in a computer and compare to what really happens in the experiment. Even so... wont there be thermal considerations that even a computer can't predict? the point at which you launch the ball, photons hitting the ball, etc...

    3. Re:Outside effects? by GTMoogle · · Score: 2, Funny

      For different reasons, the grandparent post and I would both argue you should attend mass. :)

    4. Re:Outside effects? by Ohreally_factor · · Score: 2, Funny

      Are you implying that I'm dense? =)

      --
      It's not offtopic, dumbass. It's orthogonal.
  3. Yeah, but... by jamestheprogrammer · · Score: 2, Interesting

    You would need to be extremely precise for that to work. The masses of the model planets would have to be PERFECTLY scaled. Do we actually know for a fact the masses of all the other planets, and can we make something that precise?

    Then you have to consider the gravitational effect of the asteroid belt. Do we know the mass of that, too? That might affect the model when put in use.

    Any conclusions made from this experiment would be debated over endlessly because of this...

    --
    "You teach a child to read and he or her will be able to pass a literacy test." - President George W. Bush
  4. What if by Raindance · · Score: 5, Funny

    I wonder if our universe is just a hidden spacial dimension test for a super-advanced alien civilization... still trying to figure out string theory.

    1. Re:What if by stunt_penguin · · Score: 4, Funny

      Tom Cruise seems to think so.

      --
      When the posters fear their moderators, there is tyranny; when the moderators fears the posters, there is liberty.
  5. Suspect this is rubbish - NS has been had? by Toby+The+Economist · · Score: 3, Insightful

    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.

    What exactly are they thinking of putting into orbit around this thing?

    1. Re:Suspect this is rubbish - NS has been had? by erice · · Score: 4, Informative

      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

    2. Re:Suspect this is rubbish - NS has been had? by Cecil · · Score: 4, Informative

      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

    3. Re:Suspect this is rubbish - NS has been had? by njchick · · Score: 2, Interesting

      Fortunately, thermal velocities of macroscopic objects are much lower than those of atoms.

  6. Black on black SHOULD be a crime by flyboyfred · · Score: 2, Informative

    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?
  7. Re:Why L2? by addie · · Score: 4, Interesting
    Well actually the article did say:
    A spacecraft placed there would stay fixed in space, relative to Earth, making it easier to monitor. The Earth would also shield it from the Sun's radiation, which pushes gently on any objects it shines on. Any such push could change the spacecraft's position relative to the tiny "planets" held inside it.

    So they've got that much of it thought out. But in regards to the mass of the spacecraft carrying this jar:
    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.


    So while they're full aware of the problems the mass of the craft can cause, they seem to think it's possible to minimize the effects to a reasonable level.

    My question is, aren't Lagrangian points going to start to get a bit crowded? There are only five to work with in our neighbourhood and who gets to say who uses which and for how long?
  8. Sounds like a fancy version of... by exp(pi*sqrt(163)) · · Score: 2, Interesting

    ...Cavendish's classic experiment. I look forward to seeing the results.

    --
    Doesn't it make you feel good to know that our freedoms are protected by politicans, lawyers and journalists.
  9. I am sure someone has thought of this already. by wickedsteve · · Score: 2, Interesting

    But can anyone explain to me why gravity would be the only force bleeding into other dimensions? Or is it the only one? Also is there any evidence of extra dimensions already? I would think there would already be some evidence since it does not sound very scientific to me to base the very popular string theory on imaginary notions with no basis in reality. If we are just gonna make up dimensions to make the math work isn't that just as bad as making up Thor to explain the thunder and lightning?

    1. Re:I am sure someone has thought of this already. by particle_fizax · · Score: 3, Insightful

      You have to understand that to a theoritician, having a "basis in reality" is a vague phrase. We have these equations, and they work really well for certain things that have been troubling the physics community for quite some time. They happen to require more than 4 dimensions. The theoretician says, "Oh well, find the other dimensions!" It's not a strange concept to make up new physics to "make the math work out". Quarks were hypothesized, then as each quark pair was discovered, we knew how many pairs of quarks there should be and knew roughly where we needed to look for the rest of them. You could say someone "made up" the top quark to "make the math work". But it turns out that it worked pretty well. Also, don't even get me started about the Schrodinger equation... ever seen the derivation? Yeah... it's more of a "here's a couple theoretical ways the equation _could_ be derived... maybe...

  10. Forgive me but I have to nitpick by The+Living+Fractal · · Score: 2, Insightful

    I don't like the word choice "hidden". Hidden is the past participle of hide.

    "hide"
    v. hid, (hd) hidden, (hdn) or hid hiding, hides
    v. tr.
    To prevent the disclosure or recognition of; conceal.


    This fairly clearly implies intelligent action. I.E. something did the hiding. I.E. the dimensions we can't see (if they exist) are purposefully invisible to us because something chose for them to be, something intelligent. Invisible, as another word choice, would've been better.

    Besides, something can't be hidden and yet physically interact with the universe. I believe if a thing interacts with the universe on any degree then we should be able, generally speaking as intelligent beings, to see it. And if we can't see the interaction, despite being able to probe to fundamental scales (planck, anyone?), then, forgive me again but, so fucking what?

    So to me the word is not only implying an intelligent purpose, but is furthermore illogical in choice from the very point of view of physics.

    Maybe I'm full of shit, I probably am. I guess the real reason I write this is that I'm really starting to get tired of people throwing around buzzword catchphrases like 'hidden dimensions' to get attention. You might as well say the UFO's made them. Anyway, explain to me how something that can be measured via our three standard directional dimensions manages to slip into the realm of 'hidden dimensions'? /endrant

    TLF

    --
    I do not respond to cowards. Especially anonymous ones.
    1. Re:Forgive me but I have to nitpick by Tomfrh · · Score: 2, Insightful

      This fairly clearly implies intelligent action.

      "Hidden" doesn't imply intelligent action. E.g. "The sun was hidden behind the clouds"

      Besides, something can't be hidden and yet physically interact with the universe.

      Yes it can. Sub atomic particles were hidden for most of history and yet they had no trouble physically interacting with the universe.

      despite being able to probe to fundamental scales (planck, anyone?)"

      No-one is poking around at those scales.

  11. Simpson's... by grumling · · Score: 2, Funny

    "Well, we're running an experiment to see the effects of gravity on these little screws."

    --
    "Well, good luck finding a judge that doesn't run a bestiality site."
  12. Re:Why L2? by HarveyTheWonderBug · · Score: 2, Informative
    My question is, aren't Lagrangian points going to start to get a bit crowded? There are only five to work with in our neighbourhood and who gets to say who uses which and for how long?
    The lagrangian points are not stable positions (L1 is only the more stable), especially L2. If you put a satellite there, it will eventually drift away. Space agencies are putting satellite in orbits "around" the lagrangian points (only L1 and L2 so far), and proceed regularly to orbit corrections. Here close means a few 10s of kilometers quite enough to avoid collisions.
  13. Re:Risky by ScrewMaster · · Score: 2, Funny

    No big deal, plenty more scientists where they came from. I'd be more concerned about them creating a great big black hole, and us never being heard from again.

    --
    The higher the technology, the sharper that two-edged sword.
  14. Some sanity here by viking2000 · · Score: 3, Insightful

    The key question is: What is the ratio between signal and noise here? The article does not mention this at all except talking about lagrange points, solar wind, etc. I assume placing it at L2 is to get the S/N ratio >1.

    This fails when considering some noise sources:
    1. Accelleration felt by a "grain sized planet" due to a 5kg ball 10cm away is 1m/s/year.
    2. Acceleration felt by same "planet" due to moon 1 million kilometers away: 130 times more
    3. Accelleration felt due to spaceship: ?
    4..? L2 orbit itself, light pressure, magnetic & other fields etc

    This appears unfeasable by orders of magnitude.

    I do not have much faith in statments like "Gravity leaks into other (higher) dimensions." Where does this come from? Efforts to make string theory models fit the real world?

  15. Hmm... by muzammal · · Score: 2, Insightful

    Would the fact that this little universe would be enclosed in a spaceship have any effect on it?

  16. Re:What's tungsten? by JourneyExpertApe · · Score: 4, Funny

    Read more about bots here.

    --
    If you can read this sig, you're too close.
  17. Gauss's Law by amightywind · · Score: 4, Informative

    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
    1. Re:Gauss's Law by bbaskin · · Score: 5, Informative

      If I had a nickel for everytime I heard someone suggest replacing a tungsten weight with uranium, I'd have a buck or so. Uranium (238 anyway) isn't denser than tungsten. Tungsten is the densist material for semi-practical applications. It's more available than iridium or osmium, and far less expensive than platinum, three more dense elements. For a few reasonably obvious reasons, neptunium and plutonium aren't really good alternatives to tungsten if you just want a dense lump of metal.

    2. Re:Gauss's Law by Quantum+Fizz · · Score: 4, Informative
      Gauss's Law says that the gravitational acceleration of a body anywhere in an enclosed sphere is 0.

      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.

  18. interesting but by rucs_hack · · Score: 5, Insightful

    Since we're not able currently even to build a spaceship capable of making it to the moon (having mothballed all the relevent tech and gone for the technical nightmare that is the shuttle, and the hidiously expensive disaster that is the ISS), why bother with these types of experiments?

    Such experiments, while useful, aren't practical when we have a real and current need to figure out how to get construction workers and ordinary people into space, so we can build a realistic presence there.
    Once we're there, we could perform experiments like this at a fraction of the cost.

    Ok, perhaps I'm thinking too fancifully, but it's real concern. Let's face it, every environment we've moved into only becomes liveable when the ordinary people who know how to build stuff and make things arrive. The larger the number of people, the faster things progress.

    So long as it's only scientists and the 'elite' going into space and performing experiments progress will be very slow. That can't be good.

    What we need is people going 'prospecting' for interesting asteroids/orbiting 'junk' that can be exploited, building commercial stations, setting up routine flights into space. In short, we need economic forces active in space.

    1. Re:interesting but by Kopretinka · · Score: 2, Insightful
      Since we're not able currently even to build a spaceship capable of making it to the moon [snip] why bother with these types of experiments?

      Yes, why play with twitching frog legs and your so called "electricity" when we have starving people and battling kingdoms to take care of?

      Funnily enough, fancy abstract "basic research" often has benefits that greatly outweigh the relatively small costs of setting up "these experiments".

      --
      Yesterday was the time to do it right. Are we having a REVOLUTION yet?
  19. Too many uncertainties by mcguiver · · Score: 4, Interesting

    It seems to me, after reading the article, that there are just too many influential factors to be able to conclude anything by such a test. From the article If gravity is leaking into extra dimensions, the slight change in its force should cause the planet's oval-shaped orbit to rotate, or precess, slowly... the orbit would precess by 1/3600 per year - "a reasonable quantity to try and measure," they say.
    I wonder how they could conclude that a change of this magnitude would come from gravity leaking into other dimension and not from any of the other myriad of possible effects. It is a good idea, I just don't see how it could work.

    1. Re:Too many uncertainties by Quantum+Fizz · · Score: 2, Informative
      I wonder how they could conclude that a change of this magnitude would come from gravity leaking into other dimension and not from any of the other myriad of possible effects.

      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.

  20. Wow... by Jugalator · · Score: 2, Funny

    I nominate this for the strangest news article title of 2006.

    --
    Beware: In C++, your friends can see your privates!
  21. High School Physics by Soong · · Score: 4, Informative

    Ok, some orbital mechanics.

    Going with a circular orbit because they didn't specify the ellipse:
    365.24*24*3600 = 31556736.00 seconds per year
    ./3000 = 10518.912 seconds per orbit
    1/. = .00009506686623103225 orbits per second
    .*.14*3.1415926*2 meters per orbit =
    .0000836 meters per second
    .*1000 = .0836 millimeters per second

    Pretty slow orbit. About that tungsten, 19250 kg/m3
    3.1415926*(4/3)*.04*.04*.04 = .000268 m^3
    .*19250 = 5.16 kg
    And let's say the planet is 8 mm in diameter, .004 m in radius
    3.1415926*(4/3)*.004*.004*.004 = .000000268 m^3
    .*19250 = .00516 kg

    F = G m1 m2 / r^2 =
    gravitational constant = 6.67300 × 10-11 m3 kg-1 s-2
    .00000000006673000000 * 5.16 * .00516 / (.1*.1)
    = .00000000017767262800 Newtons of force, resulting acceleration on the smaller body of
    ./.00516 = .00000003443267984496 m/s = .00003443267984496 mm/s

    Sounds reasonable to me. Assuming they can get a clean launch at exactly .0836 millimeters per second everything should be fine!

    --
    Start Running Better Polls
    1. Re:High School Physics by Ctrl-Z · · Score: 3, Funny

      Shhh... he's trying to show off his big brain.

      --
      www.timcoleman.com is a total waste of your time. Never go there.
    2. Re:High School Physics by Napoleon+The+Pig · · Score: 2, Informative

      They don't specify the eccentricity of the ellipse however they do specify the period and the material that makes up their "sun". From a few simple calculations you can figure it out:

      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.

  22. It won't work. by DolomiteZipper · · Score: 2, Insightful

    Wouldn't the space ship exert its own gravity on the system and ruin the whole experiment?

  23. semantics by v1 · · Score: 2, Interesting

    Once at the Lagrange point

    Lets review this. Lagrange point. Last I checked, a point is not a "region". So there's no way to put a titanium anything completely within a Lagrange Point. At the very best they might put the "sun" part of it centered at the LP, but then the "planetoids" would all be outside the LP, and however minorly, would be affected to varying degrees by the gravity of the earth and of the sun.

    This test is invalid. The use of a LP is not going to nullify the effect of gravity of the earth, let alone of the sun. If they are going to do a test that is this sensitive, there is nowhere in the solar system they can hold it and get accurate results.

    --
    I work for the Department of Redundancy Department.
  24. Re:What's tungsten? by Anonymous Coward · · Score: 2, Funny

    Learn more about funny here.

  25. Home on LaGrange by Anonymous Coward · · Score: 3, Funny
    Oh, give me a locus where the gravitons focus
    Where the three-body problem is solved,
    Where the microwaves play down at three degrees K,
    And the cold virus never evolved.

    Home, home on LaGrange,
    Where the space debris always collects,
    We possess, so it seems, two of Man's greatest dreams:
    Solar power and zero-gee sex.

    We eat algea pie, our vacuum is high,
    Our ball bearings are perfectly round.
    Our horizon is curved, our warheads are MIRVed,
    And a kilogram weighs half a pound.

    Home, home on LaGrange,
    Where the space debris always collects,
    We possess, so it seems, two of Man's greatest dreams:
    Solar power and zero-gee sex.

    If we run out of space for our burgeoning race
    No more Lebensraum left for the Mensch
    When we're ready to start, we can take Mars apart,
    If we just find a big enough wrench.

    I'm sick of this place, it's just McDonald's in space,
    And living up here is a bore.
    Tell the shiggies, "Don't cry," they can kiss me goodbye
    'Cause I'm moving next week to L4!

    Home, home on LaGrange,
    Where the space debris always collects,
    We possess, so it seems, two of Man's greatest dreams:
    Solar power and zero-gee sex.
  26. Re:Confused physical reasoning by Quantum+Fizz · · Score: 2, Informative
    Do you even understand Gauss's law? My example did envelop the mass with a "Gaussian surface", the fact you don't understand that and yet resort to namecalling only makes you look both naive and immature.

    .

    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.