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Easy differential equation

  1. Dec 27, 2013 #1
    1. The problem statement, all variables and given/known data

    Inside the earth, the force of gravity is proportional to the distance from the center. If a hole is drilled through the earth from pole to pole and a rock is dropped in the hole, with what velocity will it reach the center?

    3. The attempt at a solution

    I think that the proper way to set it up is

    [tex] \frac{ d^2 x}{dt^2} = C (R - x), [/tex]

    where R is the radius of the earth. Then, since the acceleration is ## g ## when the rock is at R, aka when x = 0, C must equal g / R. Thus, the equation is

    [tex] \frac{d^2 x}{dt^2} = \frac{ g}{R} (R-x) [/tex]

    Is this the correct way to proceed?

    [tex] \int \frac{d^2 x}{g/R (R - x)} = \int dt^2 [/tex]
  2. jcsd
  3. Dec 27, 2013 #2


    Staff: Mentor

    I would be inclined to let x be the distance from the center of the earth, with x = 0 being at the center and x = R being at the earth's surface.

    For a differential equation I would start with m x'' = Cx, which is a 2nd-order DE with constant coefficients. Presumably you have already worked some problems like this.
    No. Your integral is not correct.
  4. Dec 27, 2013 #3
    Ok, so I have

    [tex] \frac{ d^2 x}{dt^2} = \frac{g}{R} x [/tex]

    Can I do this?

    [tex] \frac{ R}{g} \int \frac{ d^2 x}{x} = \int dt^2 [/tex]

    Does that even make sense?

    By the way, I haven't learned anything about second order d.e.'s yet
  5. Dec 27, 2013 #4


    Staff: Mentor

    No, you can't do that, and it doesn't make sense.
    Maybe I'm looking at this the wrong way, but I don't see any way around writing the equation as a 2nd order DE. I can't think of any trick to get it into the form of a 1st order DE.
  6. Dec 27, 2013 #5
    Well, I don't need to solve it, I only need to find an expression for ## dx / dt ## since the problem is asking for the velocity at x = 0. Any ideas?
  7. Dec 27, 2013 #6


    Staff: Mentor

    The DE could be written this way:
    m dv/dt = kx, x(0) = R

    or dv/dt = (k/m)x
    If you integrate the left side, you get v, but the right side would have to be left as an integral, since you're integrating a function of t (i.e., x(t)) with respect to t.
  8. Dec 27, 2013 #7


    User Avatar
    Homework Helper

    You have a sign error. x is the distance from the centre of Earth, which decreases if the rock is falling down. The correct equation is [tex]\frac{ d^2 x}{dt^2} = -\frac{g}{R} x[/tex]

    The velocity v is the first time derivative of x, so d2x/dt2=dv/dt. You can consider v as function of x: v(x(t)). Whith the Chain Rule, dv/dt=dv/dx dx/dt = dv/dx v= 0.5 d(v2/dx). Your equation can be written as

    [tex]\frac{ d(v^2)}{dx} = -\frac{g}{R} x[/tex].

    that is a first order equation. Can you solve it for v2(x)?

  9. Dec 27, 2013 #8


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    2017 Award

    Staff: Mentor

    Do you have to use a differential equation?
    It is easier with energy conservation - ehild's last equation goes in the same direction.
  10. Dec 27, 2013 #9
    Nice catch, ehild. (By the way, I was defining up as positive, so the g i was using had a self-contained negative, but you're right, using using -g works better here)

    [tex] \frac{dx^2}{dt^2} = v \frac{ dv}{dx} = - \frac{g}{R} x [/tex]

    [tex] \int v \, dv = -\frac{ g}{R} \int x \, dx [/tex]

    [tex] v^2 /2 = - \frac{g}{R} x^2 /2 + C [/tex]

    Since v= 0 when x = R:

    [tex] 0 = - gR + 2C \implies C = gR /2 [/tex]

    Plugging in x = 0:

    [tex] v^2 = gR \implies v = \sqrt{gR} [/tex]

    Interesting, even though I'm pretty sure hollowing out the earth would take away its gravitational properties.
  11. Dec 27, 2013 #10

    D H

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    Staff Emeritus
    Science Advisor


    This is an immediate consequence of conservation of energy -- assuming "the force of gravity is proportional to the distance from the center", that is. This isn't the case. That assumption requires a constant density throughout. The Earth's interior has anything but a constant density. The Earth's core is mostly iron and is highly compressed. Gravitational acceleration at the core/mantle boundary (about halfway down) is greater than acceleration at the surface of the Earth! That the Earth does not have a constant density has been known since 1774. Even though it's rather unrealistic, this constant density assumption makes for lots of interesting and fairly simple math/physics problems.

    We're not hollowing out the Earth here. We're just making a hole through the Earth, presumably one with an insignificantly small diameter.
  12. Dec 27, 2013 #11
    Good point, but I think that's also impossible
  13. Dec 27, 2013 #12

    D H

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    Science Advisor

    Of course it's impossible. That's not the point of the exercise.
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