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Another strange physical event: This time regarding gravity.

  1. Jun 8, 2009 #1
    Imagine,
    You have dug a hole clear through the Earth. From one side to the other. Don't ask me how, you just did.
    Let's pretend that the hole is totally perfect: Same diameter thoughtout, no cave-ins, and it somehow doesn't get hot or filled with magma in the center. It's just a clear shot of a hole.

    Now, picture someone jumping into it. I have two discussion topics:

    1. Would the person fall through the entire hole and come out on the other side?
    - I mean, wouldn't gravity stop them in the center, as it comes at the person from all sides? Would this cause them to sort of float in the center of the Earth (remember, we're pretending that there's no magma in the center, so our guy is safe).
    - Or, would the great pressure in the center of the Earth crush our jumper to death?

    2. What if he made it to the other side?
    - What would the people watching him leave the hole see? Would he come out the ground, followed by gravity forcing him down back to Earth?
    - Or, would his velocity keep him falling (or in this case, flying) out the hole, into the atomsphere? (This is obviously not possible, but whatever)

    What do you think?
     
  2. jcsd
  3. Jun 8, 2009 #2

    sylas

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    Ignoring issues with temperature and so on, just worrying about gravity, the force downwards will be proportional to your distance from the center. A particle in a hole moving under gravity alone will therefore trace out simple harmonic motion, from one side of the Earth to the other.

    The net gravitational at a point inside the Earth is proportional to all the mass that is closer to the center than you are. The forces from everything further from the center than you are cancels... it corresponds to a spherical shell of mass, for which there is no net gravitational force inside the shell.

    The mass close to the center than you is proportional to volume, or R3. The force also varies as the inverse square of your distance from the center, or R-2. Hence gravitational force is proportional R, and you have simple harmonic motion.

    Cheers -- sylas
     
  4. Jun 8, 2009 #3
    As you fall through the hole, at distance R from the center, only mass inside this radius will attract you, but you will be a distance R from the center, so the net gravitational force F =R M g/6378,000 (where 6378,000 is radius of earth in meters, g = 9.81 m/sec^2, and M is your body mass).
    If the hole is drilled from the North Pole to the South pole, you will fall straight through, and just reach the other pole in about 42 minutes. If somebody doesn't catch you, you will fall back through in another 42 minutes. If the hole is drilled on the equator (like Brazil to India maybe), you will have to drill a curved hole to account for the fact that the Earth is rotating, and the rotational velocity is 1000 miles per hour at the equator. If you drill a straight hole on the equator, you will bounce around and get stuck at the center of the Earth.
     
  5. Jun 8, 2009 #4
    You're going to want to ignore air friction too, or your intrepid explorer will find his ride wimpering to a stop in the middle.
     
  6. Jun 9, 2009 #5

    QuantumPion

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    You are neglecting though that the force of gravity will decrease as you approach the center. So you will continue to accelerate to a very high speed as you fall towards the center, but the acceleration will decrease to zero at the center, and then gradually increase in the opposite direction, slowing you down. If there are no resistances then you would stop once you reach the other end of the hole, and then fall back through.
     
  7. Jun 9, 2009 #6

    Doc Al

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    He doesn't neglect that fact, he states it in the part that you quoted.
     
  8. Jun 10, 2009 #7

    QuantumPion

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    But he said he assumed a constant acceleration of 9.8 m/s^2? Wouldn't the acceleration due to gravity would decrease linearly to zero at the center (assuming uniform density for simplicity)?

    eta: Here's a nice picture from wiki showing the variation by density: http://en.wikipedia.org/wiki/File:Earth-G-force.png" [Broken]
     
    Last edited by a moderator: May 4, 2017
  9. Jun 10, 2009 #8

    Doc Al

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    No, he said g = 9.8 m/s^2. (g is just a constant.)
    Yes, exactly as he described. See his formula for net gravitational force, which is linearly proportional to the distance from the center.

    (I think you just misread what he meant.)
     
  10. Jun 10, 2009 #9
    Please re-read the following from my previus post:
    "As you fall through the hole, at distance R from the center, only mass inside this radius will attract you, but you will be a distance R from the center, so the net gravitational force F =R M g/6378,000 (where 6378,000 is radius of earth in meters, g = 9.81 m/sec^2, and M is your body mass)."

    Force at radius R is = R M g /6378,000, where 6378,000 is radius of Earth, so it is linearly proportional to radius.
     
  11. Jun 10, 2009 #10

    QuantumPion

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    Oh I see what you mean now. Nevermind :p
     
  12. Jun 10, 2009 #11
    I have actually been wondering this question as long as I can remember. Thank you for posting it! (sorry I didn't contribute to the topic)
     
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