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Does Gravity exist ?

  1. Jul 8, 2009 #1
    Another dumb question, sorry.

    I am trying to get my head around if there is some force (Gravity) that acts like an elastic band, pulling everything towards everything else – the more mass a thing has the stronger the elastic band.
    Or is it
    Gravity doesn’t exist. What happens is that everything bends space/time around it causing everything else to follow the bends formed around. The more mass something the stronger the bends.

    Eg. (even though the results are basically the same)
    Gravity exists – A snowflake is pulled down downs the earth and the earth is pull up toward the snowflake. As the earth has more mass than the flake, the flake moves more than the earth does.
    Gravity doesn’t exist – Both the earth and a snowflake bend space/time around themselves. As the earth has more mass than the flake, the flake follows the bends created by the earth’s mass more then the earth follows the bends created by the flake.

    Either way – Both the flake and earth are moving and it is not the movement that matters but the acceleration of the flake from one point (relative to another) that is happening rather than movement.

    "or" is all of the above wrong :confused:

    Please help me to understand this

    Thank you

    Alison
     
  2. jcsd
  3. Jul 8, 2009 #2
    Sounds like semantics. That's kinda like saying air doesn't exist, it's just molecules of gases floating in three dimensional space.
     
  4. Jul 8, 2009 #3
    The rubber band analogy should only be used for the strong force.
    As for gravity, your arguments for gravity existing/not existing (using the snow flake argument) are both the same. Einstein's theory of relativity says that space-time is curved by matter and the path that the snow flake will take in going from point a to point b is the the shortest distance between the two points constrained to the surface of the curved space-time.
    In normal space (i.e. your desk) the shortest distance between two points will be a straight line. Now put your two points on the surface of a trampoline and place a bowling ball between them (but not on the line that connects them!). The trampoline surface is now bent, and the path of the shortest distance between the two points (constrained to the surface of the trampoline!) will look curved. The deviation in the once straight line is what we call gravity.

    Gravity is a very hard concept to grasp. It took einstein 11 years to understand, and he's einstein!
     
  5. Jul 8, 2009 #4
    So - sorry to get this. :-)
    There is no direct force - say a force between 2 masses. Just a curve in space time that is created by the masses. All other objects (mass) will follow a streight line - along the curve. This being governed by the size of the mass - the more mass the more curve (force)

    Is that right ??

    I ask a simler question a while ag ans someone said that (in the snow flake anaolgy) the earth moves outwards in all directions (towards the flake)

    Sorry but I know some people will get this right away but it's taking me a while. How can the surface of the earth move in all directions at once ? and if it does, is there a need for a curve for the flake to follow - it could just wait for the surface to rise up to meet it OMG
     
  6. Jul 8, 2009 #5
    I can't answer all of your questions but I remember seeing a particular illustration for gravity (extremely simplified) that may help.

    Imagine space as a tightly pulled sheet, on which everything in space lies. The greater the mass of an object, the greater the "dip" it created in the sheet, which causes objects near it to fall in towards it. However, this doesn't account for orbital inertia, among other things. Don't know if that helps, but it helped me.
     
  7. Jul 8, 2009 #6
    You had best not worry about spacetime "bending" until you get a little further along. General relativity is a very math-intensive subject. No analogy involving snowflakes will ever come close to doing it justice.
     
  8. Jul 9, 2009 #7

    A.T.

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    The math is complicated, but the general idea isn't and can be understood by everyone The rubber sheet analogy is misleading, because it omits the time dimension of space time. Try the visualizations linked here:
    https://www.physicsforums.com/showthread.php?p=2244927#post2244927
     
  9. Jul 9, 2009 #8

    A.T.

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    Not all other objects. Only the free falling ones follow a straight line in curved space-time. And they don't need to have mass.
    That is nonsense. People are confusing movement and acceleration. The surface of the earth is accelerated away from the center in the sense that it doesn't follow a straight line in curved space-time, This doesn't mean that it moves away from the center.
     
  10. Jul 9, 2009 #9
    OMG - this is so much fun. I am learn all the time - 1st to everyone I want to thank you. I am really interested and people like you all make it so easy to understand. I say thank you again to every one, you are so kind to take the time to share your understanding and skill.

    2nd - sorry about my english not being good enough to explain I all want. I hope you understand.
     
  11. Jul 9, 2009 #10
    Hi Alison,
    yes, that small piece of an analogy is ok: just as you stated, but it is NOT good as a broader analogy....
    Which has a stronger gravitational attraction, two masses close together, or the same two masses further apart?
    They have a STRONGER attraction close together, right?; but that's when an elastic band would be LESS stretched reflecting LESS attraction ....you need an analogy that is strong for short distances and weak for long distances....

    But your rubber band analogy does work well for hadron interactions, like quark attraction in a nucleus, where the force of attraction is stronger with greater distance. That's why individual quarks are not found floating around one by one.
     
  12. Jul 9, 2009 #11
    The Earth had bulged outward because there is more centrifugal force near the equator so it has an orange-like shape.
     
  13. Jul 9, 2009 #12

    A.T.

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    If attraction increases with distance (let's say proportionally like in Hooke's law) you still have elliptical orbits. But the source of attraction is in center of the ellipse (instead in one of the focal points, like in Newtons inverse square law).

    Ironically on the one pound note you see Newton next to a diagram of Hookesian gravity:
    http://www.avemaria.edu/uploads/pagesfiles/734.jpg [Broken]
     
    Last edited by a moderator: May 4, 2017
  14. Jul 9, 2009 #13
    I was not decribing anything with elliptical orbits...but now that you mention it:
    I'm reminded the force between quarks does not diminish as they are separated...stays steady rather than increase, I think....
    a little different than an "elastic" analogy. I think maybe the self repairing string and glueball is a better analogy for nuclear particles covered by QCD: when quarks are separated, say by a large force, the "gluons" forming the string repair to elongate the string and eventually the quarks are pulled back together...
     
  15. Jul 9, 2009 #14
    Gravity does not exist as a force, it is the effect of space-time, or dark matter warping.

    I must admit I also find the 2 dimentional rubber sheet a really bad example. I also find this pulling term for gravity also a bad description, as it actually now being understood as a pushing force. I find the over simplistic concept of a ball immersed in a pool of water easier to get my head around, where gravity can be interpreted as the force of the water acting on the outside of the ball. If we then swap say Earth for the ball and dark matter for the water then it becomes clearer what gravity possibly is, and it also goes some way in explaining why we will never find what we could call the "Graviton" particle. This hypothsis also explains why we become lighter the deeper underground we travel, and why we become lighter the further we travel away from Earth. Also don't think of the Moon pulling on the oceans, consider it as sheilding the mass of water from some of the effects of the dark matter and releasing some of the dark matter pressure allowing the water to rise.

    Of couse it is far more complex than this as the dark matter, whatever it is is 25% of the missing 96% of the missing universe, the rest currently being assumed as dark energy.

    My current studies are in the area of dark matter and why we can only see the effects of it as it holds our galaxy together. Is it simply particles that are out of sync with our galactic time plane? And manifest themselves as neutral particles that we cannot dierectly detect other than the indirect pressure effect of this dark matter quasi-neutral soup that the galactic halo excerts or us? And is it something at the quantum level where particles flit in and out of existence from our galactic plane into other dimentions at various angular planes around our galactic halos sphere?
     
    Last edited: Jul 9, 2009
  16. Jul 9, 2009 #15
    Suppose I made a tunnel right through the earthcore to the other side of the earth, Australia in my case, unfortunately I fall in my tunnel, would this be a correct description of my journey?:

    Initially I'd accelerate faster and faster, but as I approach the center, I will slow down and at one point I will stop accelerating since the earth as a almost spheric shape (a side question, how does a charge in a charged sphere behave, would it move, since there is no potentialdifference?) and the gravitational forces cancel eachother out. So the formula with 1/r² only counts as long as I don't fall into the object. Because according to the formula, I would have infinite gravitational force working on me in the core.
    At the end of my trip I will be shot in the air, any ways of calculating how high I'd fly?
     
    Last edited: Jul 9, 2009
  17. Jul 9, 2009 #16

    A.T.

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    Oh really? You must be living in the http://en.wikipedia.org/wiki/File:Antipodes_LAEA.png" [Broken] then? :wink: In fact your tunnel would work for free fall only along the rotation axis. But lets forget about earths rotation and the lava.
    No, you would accelerate less and less, but still get faster and faster. The force inside a uniform sphere is proportional to r. It is like you were connected with a spring to the center.
    You would slow down after passing the center with max speed.
    In the center, for a brief moment you are not accelerated by Newtons force. In GR terms you are never accelerated in free fall.
    It would have zero net force.
    Yes, inside a unifom sphere it is a formula with r.
    The same height you started at the other side (any loses by friction omitted).
     
    Last edited by a moderator: May 4, 2017
  18. Jul 9, 2009 #17
    Ah the middle of the Atlantic, that's funny :biggrin:
    Do you have any derivation for the force inside a sphere? I'm wondering right now, if the earth were an empty sphere, I wouldn't accelarate at all? (just like the potential in an empty, charged sphere is the same in the whole sphere) So the formula you have is for a massive sphere?
     
    Last edited by a moderator: May 4, 2017
  19. Jul 9, 2009 #18

    diazona

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    The r dependence is for a "full" sphere, yes, not a shell. As for a derivation: start with the fact that the gravitational force exerted by a spherical shell on an object outside the shell is
    [tex]\mathrm{d}F = G \frac{\rho m 4\pi R^2 \mathrm{d}R}{r^2}[/tex]
    Here R is the radius of the shell, [itex]\rho[/itex] is the volume mass density of the shell, and [itex]\mathrm{d}R[/itex] is the shell's thickness. r is the distance from the center of the shell to the object outside the shell and m is the object's mass.

    The gravitational force exerted by the shell on anything inside the sphere is 0. You can do an integral to figure this out, or just use a clever argument that patches of the sphere in opposite directions exert forces that cancel each other out. (see http://hyperphysics.phy-astr.gsu.edu/hbase/mechanics/sphshell2.html)

    Now to compute the force due to an object partway inside a solid sphere, you divide it up into shells and integrate over only the shells which the object is not inside:
    [tex]F = \int_0^r G \frac{\rho m 4\pi R^2 \mathrm{d}R}{r^2} = 4\pi \rho m G \frac{r^3 - 0^3}{r^2} = 4\pi \rho m G r[/tex]
    You can see that it's proportional to r.
     
  20. Jul 10, 2009 #19
    shell.. sphere wouldn't it be the same..?

    dil
     
  21. Jul 10, 2009 #20

    DaveC426913

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    A sphere is solid; a shell is hollow.

    A hollow shell could have a mass approaching zero, depending on how thin the shell it is.
     
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