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Clarification about gravity and floating in space

  1. Feb 12, 2010 #1
    I'm really confused about how we float in space. What is it that keeps us from falling through it?

    I've heard that astronauts in space are actually always falling, but move so fast that the Earth curves away from them.

    If that's the case, what would happen if we were in space but extremely far from any object with it's own gravity?

    Also, how do the Sun and our planets and other stars stay where they are?

    Thanks!
     
  2. jcsd
  3. Feb 13, 2010 #2
    Gravity has no range, so it will affect you no matter how far you are from an object with mass. However, its effect does drop off quickly with distance. If we imagine that you were very very far from the Earth, and that the Earth was the only object with mass then while you would fall towards it the rate would be so slow you would probably not notice it.

    The astronauts are in orbit around the Earth, same as the Moon. The Earth is in orbit around the Sun. The Sun is in orbit around the core of our galaxy. The reason they don't hit whatever they are in orbit around is because they are moving so fast to the side that the curved surface moves downward as fast as they move downward.

    http://en.wikipedia.org/wiki/File:Newton_Cannon.svg

    In that picture five separate canon balls are fired with different speeds. Cannon balls A and B are slow enough where they eventually fall to Earth. Cannon ball E is so fast that it flies away from the Earth. Cannon balls C and D are in the range where they will forever circle the Earth, forever falling towards it but never actually hitting it.
     
    Last edited: Feb 13, 2010
  4. Feb 13, 2010 #3

    D H

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    An astronaut floating in space is falling, and falling all the time. An astronaut floating around in the International Space Station is still very much subject to Earth's gravitational pull. The gravitational force on that astronaut is only reduced by about 10% compared to the gravitational force the astronaut experiences on the surface of the Earth. The reason we call that phenomenon "weightlessness" is because the astronaut does not feel that gravitational force. In fact, an astronaut standing on the surface of the Earth doesn't feel that force, either.

    What we feel as "weight" is all of the other forces acting on us. Suppose you are standing still on the surface of the Earth. The upward normal force exerted by the ground pretty much cancels the downward pull of gravity. You feel that upward force in your feet as your weight. That force propagates through your body. You feel the tension in your guts, your joints, and in your inner ear. Take away that tension (e.g., skydive out of an airplane) and you too will feel "weightless" even though the gravitational force hasn't changed a bit.
     
  5. Feb 13, 2010 #4

    russ_watters

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    Nothing at all. But you'd have to travel a very long way to get to such a place.
    All objects we can see with our eyes. The planets orbit the sun, the sun and other stars orbit their common center of gravity (the center of the galaxy). Looking on a slightly larger scale, our local group of galaxies is gravitationally bound, but it is a bit of a pinball game up there: we're apparently going to collide with the Andromeda Galaxy relatively soon.
     
  6. Feb 13, 2010 #5
    Yep! The Sky really is falling! But collision isn't due for ~3 billion years and stars are too far apart, mostly, for even one collision to occur - though watch out for a big surge in massive star formation as the gas clouds collide.
     
  7. Feb 14, 2010 #6
    I don't think this or the answer above really answer his question fully. He is wondering why any object that was in space a good billion light years from anything else....wouldn't just fall down and why it is just "floating" (given no force ever has or will ever act on it).

    Unfortunately, we do not yet know what it is that the "space-time" fabric is made out of. Therefore we do not understand how or why gravity works the way it does when anything with mass is rested upon the "space-time" fabric.

    Perhaps once we figure out what "dark matter" and or "dark energy" really is and what it is made out of....then we can get a better picture of what it is galaxies, stars and planets all really rest upon both at a grand level and at a molecular one.
     
  8. Mar 27, 2010 #7
    So could you say that "weight" is a function of resistance to free fall?

    So, for example, a sky-diver falling through air has weight equal to the amount of wind-friction pushing up on her?

    Related question: is weight reduced by the rotation of the Earth? I.e. if Earth was rotating half as fast, would the weight of objects be greater?

    Before answering consider that if the Earth was rotating fast enough that its matter would break apart into a cloud of free-falling/orbiting fragments, then presumably all the fragments would be weightless in that they would be in free-fall.

    So maybe the ground is only pushing up against our feet to the extent that it's not free-falling away from them due to the planet's rotation.
     
  9. Mar 27, 2010 #8

    DaveC426913

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    Yes. By about half a pound per hundred pounds.
     
  10. Mar 27, 2010 #9
    I don't know who else's answers could have sufficed,however, it is in short that they follow geodesics. In technical forms, it can be a curve in time and space, but since recent confirmation we have a near-flat universe, even straight lines remain even in relation to curvature, the quickst route in an Einstein Universe.
     
  11. Mar 27, 2010 #10
    ''If that's the case, what would happen if we were in space but extremely far from any object with it's own gravity?''

    They won't They will break away because of the absence of gravtational strength - perhaps collide towards a stronger source, like Andromeda, whic if memory serves, is already colliding towards us at 300,000 m/p/s.
     
  12. Mar 27, 2010 #11
    DH, i well-respect you! a xouple of questions:

    I had two professors:

    1st said: In free fall, a classical limitation is given as [itex]F=mg[/itex]. If [itex]g=0[/itex] since it is not near a surface of a gravitational body.. does it experience zero-gravity? As far as i am aware, in free fall, every particle is dragged equally within the gravitational field - so, how can there be an absolute vanishing value for [itex]g[/itex] in [itex]F=Mg[/itex].

    Thanks in advance!
     
  13. Mar 28, 2010 #12

    Chronos

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    NYS, you are falling into a black hole of ignorance. Spacetime is a coordinate system, it has no 'properties'. You missed the essence of Einstein's explanation.
     
  14. Mar 28, 2010 #13
    Which "recent confirmation" are you referring to?

    How are you defining "straight lines" and "curvature?"

    How do you define "near flat" for that matter?

    I think the whole issue of curved space-time arises from the fact that light appears to travel in a straight line even when it follows a curved path. You can't see light bending, only the image-patterns that it generates at the point of observation.

    The only clue you have that light is bending is that objects and physical laws appear different in what a given light image seems to represent. Galaxies appearing to rotate faster than the speed of light are a good candidate for considering the possibility of gravitational lens effects, for example, because it is not logical that matter-energy would behave as it appears with the assumption that the light is traveling in straight lines over a given distance.
     
  15. Mar 28, 2010 #14

    DaveC426913

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    There's no reason in principle why you should not be able to see it bending though.

    Just like in a smoky concert hall, you would be able to see a curved path through a large dust cloud surrounding the star if you used a laser.

    It would just be fabulously difficult to set up.
     
  16. Mar 28, 2010 #15
    Correct me if I'm wrong, but I believe the only reason you can see light beams or lasers through a cloud of dust, smoke, etc. is because the light reflects or refracts off the particles toward you.

    So what you're seeing is not the beam of light or laser itself as you perceive it, but rather a large number of refracted beams of light that travel from the particles to your retinas.

    Without any reflective/refractive medium to redirect light, how could you see it except by having it reach your retinas? If the light is traveling through a vacuum, it could be doing loop-d-loops and all an observer would see is the image carried by the light.

    I think the key to understanding light-bending lies in interpreting the images carried by the light and whether those images themselves contain evidence that the light has travelled in a non-straight path and why/how.
     
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