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Tunnel through the centre

  1. Dec 11, 2004 #1
    this is supposed to be a very easy one, but i am still putting it up for fun.

    suppose a tunnel is dug right through the centre of the earth, from one end to another,( say from north pole to south pole), and then from one end of the tunnel a ball is dropped in it.

    what will finally happen to the ball????

    keep in mind this is a hypothetical situation, so please post sensibaly.
  2. jcsd
  3. Dec 11, 2004 #2


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    it will combust or melt.
    Assuming we ignore the role of heat, it will hover almost stationary half way down/up/however you care to phrase it.
  4. Dec 11, 2004 #3
    Doesn't it just oscillate forever? (with no air resistance)
  5. Dec 11, 2004 #4
    IF there is no air resistance it will ossilate forever. It is intresting because it is proven that its period is the same as a satilite orbiting the earth!
  6. Dec 11, 2004 #5


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    Yes, it would.

    First extra question : How long would it take to travel from one end to the other ?
  7. Dec 11, 2004 #6
    I get around 42 minutes :biggrin:
  8. Dec 12, 2004 #7


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    59.75 minutes.

    The radius of the Earth isn't the semi-major axis.

    Look at it this way.

    You start wtih a circular orbit with a semi-major axis that matches the center of the Earth.

    Start making the orbit more elliptical, but keep apogee at the surface of the Earth. For example. With the center of the Earth still the focus, and an eccentricity of .1, an object would be at 5218.48 at its closest point to the center of the Earth - 6378.14 at its furthest (still at the surface of the Earth). The semi-major axis is reduced to 5798.

    At .2 - perigee at 4252, apogee at 6378, semi-major axis at 4252

    Etc.etc. to eccentricity of .9999 - apogee 6378, perigee 0.03, semi-major axis 3189. The satellite reaches perigee and heads right back to the same point on the surface of the Earth that it started on.

    Continue increasing eccentricity until, finally, eccentricity is undefined (object headed straight at the center of the Earth). You can use the the extremely high eccentricity of .9999 something and the semi-major axis of 3189 as a pretty good esitmate (the limit).

    Apogee is the surface of the Earth, perigee is the center of the Earth. The big difference is that with no tangential velocity, the flip side of the orbit (perigee to apogee) is in the same direction as apogee to perigee side of the orbit.

    In other words, reaching the other side of the Earth (apogee) is the equivalent of completing one 'orbit'. It takes two 'orbits' to reach apogee on the same side of the Earth.

    There is no way you can have two apogees and no perigee!

    That also means its period is not the same as an orbiting satellite.
    Last edited: Dec 12, 2004
  9. Dec 12, 2004 #8


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    Ooh Bob-gee, you lost me there (actually I didn't read the post carefully, after finding your anwer disagree with mine) ! :grumpy:

    I first got about 85 minutes, then realized that was twice the time...so my number is pretty close to what ascky gets.

    Now I guess I'll really have to read your post
  10. Dec 12, 2004 #9


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    BoBG.. :uhh:
  11. Dec 12, 2004 #10
    If you assume the earth to be of the same density throughout, the gravity _decreases_, linearly, as you go from the surface to the center. I do not know the proof of this result but I read it somewhere... so you can't treat it as just an ordinary orbit. I'm not sure how to solve it, though, without just using a computer program for an approximation.
  12. Dec 12, 2004 #11
    Solving d2x/dt2 = -k*x , with g=9.81m/s^2 at R_earth=6366km , I've got

    42m 10s
  13. Dec 12, 2004 #12
    Bartholomew is absolutely right!

    Considering g(x) the gravity acceleration at distance 'x' from the center:
    g(x) = -G * M_earth/x^2

    But , at x < r_earth , the mass outside (x>r) doesn't affect the object, so the "new" mass to be considered is:
    M_earth * x^3/r_earth^3

    So, the "new" g , if x<r_earth, is:
    g(x) = -G * M_earth * x / r_earth^3 , or

    finally g(x) = -K*x ,

    where K = G*M_earth/r_earth^3 = 9.81/(6366000*s^2)
  14. Dec 12, 2004 #13
    Easiest solution is to use the formula for the period of a simple pendulum:

    [tex] T = 2\pi\sqrt{\frac{l}{g}}[/tex]

    And put l (pendulum length) equal to Earth radius:

    [tex] T = 2\pi\sqrt{\frac{6378100}{9.80665}} \approx 5067 seconds = 84 minutes, 27 seconds[/tex]

    This is the time taken for the return trip.
  15. Dec 12, 2004 #14


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    I don't see any similarity between this problem and that of a pendulum...

    OTOH, it is like a spring (compare the force of gravity to Hooke's Law)
  16. Dec 12, 2004 #15
    The pendulum analogy is perfect given the way this question is normally phrased:

    "A straight tunnel is driven between two points on the Earth's surface. How long would a train powered by gravity alone, take to traverse the tunnel? Ignore friction and air resistance."

    Now the reasoning goes like this:

    1. The question wouldn't have been asked like that unless the time taken is independent of tunnel length, so if we solve it for one case, we have automatically solved it for all.

    2. So that we don't have to worry about decreasing gravity with depth we will consider a very short tunnel, so that it only penetrates a negligible distance below the Earth's surface.

    3. So we draw a 'tunnel' one metre long, and work out the accelerating force on the 'train'. This involves a bit of pythagoras, and we see that the force pushing the train towards the centre of the tunnel is proportional to the distance from the tunnel centre (for the very small angles involved).

    5. Now, we recognise that this is exactly the same situation as a pendulum bob, with the length of the pendulum equal to the Earth's radius. Bingo! we already know the formula for that!

    6. 30 seconds on a calculator later, and we're done!
  17. Dec 12, 2004 #16
    Most such tunnels would actually be affected by all sorts of coriolis effects, but I notice that the O.P. specifically used an example that isn't - the tunnel that connects the two poles. :smile:
  18. Dec 12, 2004 #17


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    Aside from the fact that Bartholomew's point about the acceleration due to gravity decreasing due to the distribution of mass both above and below, there's a problem with your solution even in the simplified example (all the Earth's mass exists at the center).

    A pendulum assumes 'l' stays constant. If the two points are more than 12,756 km apart, the pendulum is so long that it swings past the center of the Earth. That definitely causes a problem. At the 'bottom' of the pendulum swing, the pendulum is 3640 km past the center of Earth, meaning your motion is going to be a little more complicated than a simple pendulum swing.

    Edit: Oh, I read your original solution. Except your original solution is for a curved tunnel with the openings 12,756 km apart, not a tunnel straight through the Earth. And I don't see how a pendulum applies to a straight tunnel in any event. The pendulum length will be constantly changing, reaching a minimum length at the 'bottom' of its swing and then expanding back to its original length.
    Last edited: Dec 12, 2004
  19. Dec 12, 2004 #18
    I looked up the formula for the period of a spring and if you plug in it works out to the same answer ceptimus got.

    Edit: How would you do that analytically, though, going from acceleration as a function of distance to acceleration/speed/distance as a function of time?
    Last edited: Dec 12, 2004
  20. Dec 12, 2004 #19
    Take a look at my 2 first posts, and see how to get the differential equation d2x/dt2 = -K*x , where ' x' is the distance to the center of the earth, ' t' is time, and 'K' is a positive constant.

    The (trivial) solution is x= A*sin(sqrt(K)*t+B) , where A and B are constants to be determined from the contour conditions.

    Here, at t=0 we have x=R_earth and dx/dt = 0 , so B=0 and A=R_earth.

    Using the known values, you'll get something like 42m 10s.
    Last edited: Dec 13, 2004
  21. Dec 13, 2004 #20
    No it isn't. Try reading it again.

    Why do you think the pendulum method comes up with exactly the same answer that you did (except you have slight errors in your constants)? I used the officially accepted values for Earth radius and little g.

    I admit that the pendulum method does take advantage of the assumption that the time taken to traverse a straight tunnel is independent of the length of the tunnel. It is by no means obvious that this should be the case. But if we accept that this has already been proved, then the pendulum method is sound.

    Remember the formula for the period of a simple pendulum also makes assumptions, (small angles of swing, sin(angle in radians) = angle for small angles, etc.

    Also, the Earth isn't actually made of uniformly dense material. The core is denser than the mantle. If you take this into account, you'll find that the time taken to traverse a deep straight tunnel (in a vacuum, in the absence of planet spin) is actually less than that taken to traverse a short tunnel.

    Yeah - I've studied this problem before. :smile:
  22. Dec 13, 2004 #21


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    There's still a problem with this example. All three, Hooke's Law, the formula for a pendulum, and the formula for the period of an orbiting object should give the same answer if the problem's set up right.

    The acceleration due to gravity is:


    Substituting this into the formula for a pendulum and you get the formula for computing the period of an orbit:


    As long as 'r' stays constant, everything works out. If 'r' varies, it falls apart. In an elliptical orbit, it takes less time to go from one covertex to perigee to the opposite covertex than it does to go from one covertex to apogee to the opposite covertex.

    In fact, to account for elliptical orbits, the average radius (the semi-major axis) is substituted for r.

    For it to work for the tunnels (with a varying radius), you have to assume the highest point is directly above the fulcrum and the lowest point directly below the fulcrum. In other words, your pendulum length is one half the radius of the Earth with a period of 59.75 minutes.

    Actually, I just rounded off the WGS-84 (1996) value for the equatorial radius of the Earth (6378.137 km), since WGS-84 is the datum most commonly used on maps (thanks to GPS).
  23. Dec 13, 2004 #22
    just wondering, i know nothing about pendulums and orbits or epogee and apogees, but im just wondering, if your calculations involved the acceleration of an object due to gravity, (which is essentially due to mass/weight, whatever you want to call it) wouldnt you need to compensate for the weight loss in the making of the tunnel, and as such change your values, so that the world in theory is a little lighter?
  24. Dec 13, 2004 #23
    Of course not!

    For a 100cm diameter tunnel, the relative difference is less than 1/200,000,000,000,000 .
    Last edited: Dec 13, 2004
  25. Dec 13, 2004 #24
    Okay, Rogerio, I think I get it--thanks.
  26. Dec 13, 2004 #25

    Doesn't the fact that gravity acts on the pendulum in the direction of its motion only in two places on the arc and is diminished elsewhere, in contrast with the tunnel, where gravity always points in the direction of motion, suggest that the pendulum model will overstate the time taken ?
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