Hypothetical Gravity Question

Discussion in 'Astronomy, Exobiology, & Cosmology' started by c20H25N3o, Sep 22, 2005.

  1. c20H25N3o Shiny Heart of a Shiny Child Registered Senior Member

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    Hi,

    If it were possible to bore a 2 meter hole through the earth, shoring up the sides as you go and a person jumped in from one side of the hole and another person jumped from the other side, what would happen? Would they get stuck together in the middle spinning about?

    Just mildly curious

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    c20
     
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  3. Tyltyl Registered Member

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    If the hole is large enough for them not to crash into each other, they would probably oscillate back and forth I think.
     
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  5. c20H25N3o Shiny Heart of a Shiny Child Registered Senior Member

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    I imagine such a hole would completely mess up the stability of the entire solar system but maybe I am underestimating natures ability to compensate?

    c20
     
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  7. Tyltyl Registered Member

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    I thought it was a hypothetical question. If anything I'd like to know how would you dig such a deep hole through the Earth's core.
     
  8. Pete It's not rocket surgery Registered Senior Member

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    Why?
    You could blow up the whole Earth without significantly changing the stability of the Solar System.
     
  9. c20H25N3o Shiny Heart of a Shiny Child Registered Senior Member

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    I'm no scientist although very interested in all this stuff. How could removing a whole planetary body 'not' affect the way the other bodies in the solar system intereact with one another?

    peace

    c20
     
  10. Pete It's not rocket surgery Registered Senior Member

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    I didn't say it wouldn't affect it, I said it wouldn't significantly change the stability.

    Really, the gravitational affect of Earth on any other planet is so tiny it makes almost no difference.
     
  11. Blindman Valued Senior Member

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    1,425
    Let open the hole up a bit..

    20 meters wide.. We will fill it with air, insulate it from the heat, and use it very quickly before the various movements of the earth twist it into oblivion.


    So being a sky diver I jump into the hole, as my twin does on the other side of the world. In about ten seconds we reach terminal velocity. From the arch, at 200kmh to head first in a suit, at 400km, you start racing, well slowly racing towards the centre of the earth.. Head first at 400kmh it would take 15 hours to see your twin..

    But it gets worse.. As we go down the pressure increases thus increasing the drag slowing you down.. Not to far down the air becomes a liquid, and you're going nowhere. So let just assume it’s a constant pressure...


    So we speed down at 400km and hour. At about 1000km down you will notice that your air speed is reducing.. This is because as you move into the earth the gravity of the earth above you starts to pull you up reducing your downward acceleration. As your acceleration decreases drag will slow you down. At about 5000km (1000km from the centre) you will be moving slower then walking speed. In the end you will never meet your twin. Friction will stop you and you will die before your bones gently settle together

    At the centre of the earth you are weightless.

    So lets take the air out.. Put on a space suit, and jump.. In about 3hours you will be passing your twin at about 2km per second. 3hours after that your head will pop out at the other end and for a moment you will stop moving.. Then back again. So on and so on.. If you hit each other, then you will make a good sized bomb… KaaaBooomm. (numbers are head calcs and may be out by about 200%. You would be in orbit but you can’t use a two body solution. )


    Luck would have it that there are objects that don’t need a hole.. Where solid rock is like air to it.. Take a small chunk of a neutron star the size of the ball in you pen. Drop it and it will fall through rock. Eventually it would get to the centre. But as in the sky dive example, it will slow as it approaches the centre due to friction..
     
  12. Dinosaur Rational Skeptic Valued Senior Member

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    Blindman: You might be correct in your statements about the pressure and drag as you approach the center, but the following is erroneous.
    The Earth is a fair approximation to a sphere whose density is a function of radial distance from the center.

    The gravitational force inside a hollow spherical shell is zero everywhere. This leads to the conclusion that there is no gravitational effect due to the Earth farther from the center than the person falling down the hole. Of course, there is less mass below, which would decrease the acceleration as an object fell toward the center.

    The deviation from a sphere due to the small hole would not have much effect.

    BTW: I am not sure of the effect of Earth's rotation on the dynamics of this situation. I suspect that an object would not fall directly toward the enter. I think such an object would hit the sides as it fell toward the center.
     
  13. Pete It's not rocket surgery Registered Senior Member

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    I think it's valid.

    The gravitation force inside a hollow sphere is zero everywhere because the part of the sphere "above" any pont pulls up with the same strength as the part of the sphere "below" that point.
     
  14. cato less hate, more science Registered Senior Member

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    to Dinosaur:
    I don't think you would hit the sides, when you jump in, you would have the same angular velocity as the earth, and when you fell in, you would stay stationary WRT the XY-plane and only fall down the Z (where the XY-plane is the plane that makes a circular cross section with the hole, and Z is orthogonal to it). I could be wrong, I am just going on what I think would happen. there is no physical basis, other than a gedanken experiment, for it.

    to pete:
    I think what happens is that any gravity from a shell "above" you would be canceled by the part of the shell still "below" you. these ideas used to be clearer in my head (when I was in physics class for example) but I think I am right.
     
  15. Blindman Valued Senior Member

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    Dinosaur: I think you misunderstood me. Sorry I should be calling the Earth a ball not a sphere. The term sphere only refers to the surface of the Earth.


    Take the mass of the portion of the sperical cap above you cutting along a plane perpendicular to the hole, calculate the centre of mass, and then plug that into the F= Gm/r^2

    At the centre you will have equal and opposite forces exerted on you effectively zero gravity. 100km form the centre it will be almost negligible. With a in my head calculation, at around 1200km down from the surface you will have an acceleration of about 8.5g


    As for the hollow sphere, the gravity is not zero everywhere. Gravitational force is inversely proportional to the square of the distance. The gravity in the centre would be zero but the closer you get to one wall the greater the attraction from that part of the wall assuming the shell has mass.

    Dropping in from the equator would send you into the wall. Just as a dancer spins up by pulling in her arms, the person falling would increase their rotational speed around the centre of the earth. Thus if they where falling in a frictionless environment, down, not a straight hole but a curved hole that at some point would be in the directed of the tangent of the sphere. At this point the centrifugal force will balance the gravitational force and you would be in orbit inside the earth. This problem can be solved by dropping in from the poles.

    Looks like there is some interesting figures to discover. The curve representing the decreasing acceleration when falling in from the poles, and the diameter and orbital period of the person falling in the curved hole.
     
  16. c20H25N3o Shiny Heart of a Shiny Child Registered Senior Member

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    2,017
    Interesting replies. Thanks for taking it seriously

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    I would love to see some graphical representation of the ideas presented here if anyone can be assed?

    peace

    c20
     
  17. Dinosaur Rational Skeptic Valued Senior Member

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    blindman: The following cannot be used to calculate the gravitational force due to any object other than a sphere whose density is radially symmetric.
    • Take the mass of the portion of the spherical cap above you cutting along a plane perpendicular to the hole, calculate the centre of mass, and then plug that into the F= Gm/r^2
      • For most objects, you must integrate (G/d<sup>2</sup>)dM over the object.
      The above assumes uniform density, and would not be correct for a cap sliced from the Earth. Density would have to be taken into account, making Dm a function of position.

      Inside a hollow spherical shell of uniform density, there is no net force in any direction. If the shell has zero thickness, there is a discontinuity. For a shell with finite thickness, there is zero force at a point on the inside surface of the shell and at a point on the outside of the shell, the shell acts like a point mass at the center. There is a continuous gradient between the two surfaces.

      The above statement is valid for a shell of finite thickness if the density is a function of distance from the center. This is why most stars, planets, et cetera can be treated as point masses. In most cases, they very closely approximate a sphere with density a function of distance from the center.

      Pete: You are correct, the part of the shell above cancels the part below. The point I tried to make is that this is not intuitively obvious. Without doing the mathematics, I would not expect anybody to correctly guess that there is no gravitational force inside a hollow spherical shell of uniform density. Note blindman's opinion expressed in a previous post.
     
  18. Flunch Registered Senior Member

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    67
    Dinosaur and Cato are right with respect to the fact that any point inside a shell of uniform density feels no net gravitational force from that shell. And you're still right that this isn't intuitively obvious.

    Interestingly, a tunnel between any two points on the earth’s surface would allow a body to free fall (e.g., slide along a perfect, frictionless tunnel surface, with no aerodynamic drag) with a period that is the same as a body in free-fall (like a satellite) orbiting near the Earth’s surface. The period is merely a function of the radius of the Earth and gravity. It is independent of tunnel length.

    Assuming the radius and density of the earth is constant and that gravity is constant at the typical intensity at the Earth’s surface (which it is not… but humour me) the period T = 2*Pi(20.9x10^6 ft / 32.2 ft/sec)^(1/2) = 84.4 minutes.

    This means that in an ideal world (no friction, aerodynamic drag, constant gravitational force, etc.) you could move from any point on the Earth’s surface to any other in 1/2 this time (42.2 minutes) through this tunnel.

    Interestingly, this apparently has been semi-seriously proposed for making tunnels between cities like New York and Washington D.C. Way too expensive obviously.
     
  19. Gudgeon Registered Member

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    12
    if you dig into the core of the earth, you will disable the gravitational pull, and the magnetic field
     
  20. Flunch Registered Senior Member

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    seriously doubt that.
     

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