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Math Methods in Physics, find FODE

  1. Feb 16, 2007 #1
    1. The problem statement, all variables and given/known data
    Suppose that a hole has been drilled through the center of the Earth, and that an object is dropped into this hole. Write a first-order differential equation for the object's velocity, v as a function of the distance r from the Earth's center (i.e., an equation involving dv/dr), and solve it to determine the speed the object achieves as it reaches the center of the Earth. Check this speed with the result you get from simple conservation of energy considerations. Consider the Earth's mass density to be uniform throughout. [Hint: recall Gauss' Law as it applies to the gravitational field of a spherically symmetric mass distribution.]


    2. Relevant equations
    I think the Gauss' Law for gravitation is simply to point out that the force is only effected by the mass enclosed, not the outer mass.


    3. The attempt at a solution
    Well, here is what I know...

    The mass enclosed is a function of radius so [tex] a = \frac{dv}{dt} = \frac{Fm(r)}{r^2} [/tex]

    I thought, to get [tex] \frac{dv}{dr} [/tex] I multiply both sides by [tex] \frac{dt}{dr} [/tex]. This gives me [tex] \frac {dt}{dr} \frac {Gm(r)}{r^2} = \frac {dv}{dr} [/tex]. Now I have a function in terms of dv/dr, but Im not sure if this is right or what to do next.

    Thx for any tips or help!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
    Last edited: Feb 16, 2007
  2. jcsd
  3. Feb 16, 2007 #2

    Kurdt

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    What you have here:

    [tex] a = \frac{dv}{dt} = \frac{Gm(r)}{r^2} [/tex]

    looks fine. What you have to do is find how the mass of the Earth varies with radius (i.e. look at the hint given in the question). once you have this it becomes a separable differential equation and you can solve that quite easily through integration.
     
  4. Feb 17, 2007 #3
    I'm sorry I won't be able to help, but I have to ask...do you to go ASU?
     
  5. Feb 17, 2007 #4
    Still not there...

    So...

    [tex] a = \frac{dv}{dt} = \frac{Gm(r)}{r^2} = \frac{4}{3} G \rho \pi r [/tex]

    ... there doesnt seem to be any variable to separate, is there? I can move the dt over to the right hand side..

    [tex] dv = \frac {4}{3} G \rho \pi r dt [/tex]

    and integrate both sides to get...

    [tex] v = \frac {4}{3} F \rho \pi r t + C [/tex]

    I dont think this is right... With this function as r approaches zero, the velocity minimizes, but the opposite would be the case.
     
  6. Feb 18, 2007 #5

    Kurdt

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    Ok let me start again. The question hints at using Gauss' law for gravitation. Do you know what this is? This may give you a clue as to how to proceed.

    The differential version of Gauss' law for a spherically symmetric distribution just has one variable of radius.
     
  7. Feb 18, 2007 #6
    I am a little familiar with the integral form of Gauss' law, but I dont understand the differential version at all.

    I thought that the reason for the Gauss' law hint was that the only mass affecting the velocity/acceleration is the mass enclosed.
     
  8. Feb 20, 2007 #7
    so i'm sorry for the text version but i can help you (Pi = 3.1415blah blah blah) and (P(row) = p) so Pi and p might be confusing but bare with me. (a = acceleration) (G = gravitational constant) i think you can figure out the rest.

    so (4/3)(Pi)(G)(p)(r) = a

    and a = (dv/dt) = (dv/dr)(dr/dt)

    so (4/3)(Pi)(G)(p)(r)(dr) = v(dv)
    (*1st side*) ----------- (*2nd side*)

    integrate both sides,

    first side from radius of earth, to zero.

    Second side between velocity zero and V(velocity at center of earth) and goto algebra city! :smile:
     
  9. Feb 20, 2007 #8
    Hey I know you! Thx, I did get it already. The part we were missing was (dv/dt) = (dv/dr)(dr/dt)
     
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