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Classic Hole through Earth Problem. Period of Oscillation with Varying Density.

  1. Apr 26, 2012 #1
    1. The problem statement, all variables and given/known data

    This is the exact problem: http://hyperphysics.phy-astr.gsu.edu/hbase/mechanics/earthole.html

    However, they assume that the Earth has a uniform density. I know how the density of the Earth varies with the distance from the center of the Earth. I also know the acceleration of gravity at each location within the Earth. However, how would I go about finding the period of oscillation as the mass falls through the Earth?
     
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  3. Apr 26, 2012 #2
    You could just use Gauss's law for gravitation. Then to find the period of oscillation, and if you still get a SHM form for the differential equation, you do the same thing as hyperphysics.
     
  4. Apr 26, 2012 #3

    D H

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    You don't get SHM with a non-uniform density.

    A simple model of gravitation acceleration inside the Earth is that it remains constant at 10 m/s^2 from the surface down to halfway to the center of the Earth and then drops linearly that point inward, reaching zero at the center. This yields a period of 76.41 minutes.

    To do even better you'll have to do numerical integration. The Preliminary Earth Reference Model gives gravitational acceleration as a function of distance from the center of the Earth broken down by different regions inside the Earth. I did this a long time ago here, but can't find the post. The period I calculated with the PREM and a simple numerical integration scheme was 76.38 minutes.
     
  5. Apr 27, 2012 #4
    This is exactly what I need to do. I'm still confused on how you would find the period using either of these methods. I know that Period=2*pi*sqrt(R/g) but how would you apply this equation to the 2 methods described? I would appreciate any help very much. Thank you.
     
  6. Apr 27, 2012 #5

    D H

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    That makes this sound like homework.

    We don't do student homework for them at this site. Our goal is to help you do your own homework, helping you through some trouble spots. If this is homework, you need to show some work. Read our rules.

    Assuming this is homework, what exactly is the problem you have to solve and what have you done to solve it?
     
  7. Apr 27, 2012 #6
    There is no explicit homework question. In class we found the period of oscillation for a mass falling through a hole in the Earth when we assumed the Earth has uniform density. Our teacher asked, how would we go about finding the period of oscillation if the mass is not uniform density?

    So, I then found the density of the Earth with respect to the distance from the core. I then found the acceleration due to gravity at each location within the earth with respect to the distance from the core and the Earth's varying density.

    I know the period equation. How can I relate the information that I found with the period equation? That is where I am stumped and would like tips/guidance/suggestions.

    I wasn't asking for someone to "do my homework" for me. I was asking for help on *how* to do it. Thank you for your time and help.
     
    Last edited: Apr 27, 2012
  8. Apr 27, 2012 #7

    D H

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    You mentioned that you found acceleration inside the Earth, so you should see that my simple, ad hoc model is a (somewhat) reasonable fit to that acceleration. It is also very simple. Calculating the period with this simple model is easy. The time it takes to reach the center of the Earth is 1/4 of the period by symmetry. Find this time and multiply by four.

    To find this time you split the fall into parts. The first part is the fall to the halfway point. This part is uniform acceleration: High school physics. Find the time to that point and the velocity at point. The second part is the fall from the halfway point to the center. This is fairly simple calculus. The differential equation is just [itex]d^2r/dt^2 = -kr[/itex], where that constant k makes gravitational fall linearly from 10 m/s^2 at the halfway point to 0 at the center.


    If you want to use the exact values from the PREM the only option is numerical integration. Do you know what numerical integration is?
     
  9. Apr 27, 2012 #8
    I really don't understand why I just can't figure this out. This is insanely frustrating. Its been about 2 months since we first dealt with this problem so maybe that is why.

    I have taken Calc I, II, III, and Differential Equations. I know what numerical integration is.

    For the first part I got 798.6 seconds and the velocity halfway to the center is 7986m/s.

    I just don't know what to do with that second equation. The solution is r(t)=sin(sqrt(k)*t) correct? I found k to be .000003136. r is equal to 3189000 m (half the radius of the earth). Obviously plugging these values in gives me the incorrect answer.

    Something just isn't clicking right now. I'm sure it's right in front of me but I'm just not seeing it.
     
  10. Apr 27, 2012 #9

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    Good so far, but you should have the velocity as negative (radial distance is decreasing) rather than positive.


    That's one of an infinite number of solutions, and you should have an arbitrary multiplicative constant there. You could also use cosine in lieu of sine. Or a sum of a sine and a cosine terms, each with their own arbitrary multiplicative constants.

    Or you could use [itex]r(t) = r_0 \cos(\omega t + \phi)[/itex], where [itex]r_0[/itex] and [itex]\phi[/itex] are arbitrary constants. You know radial distance and velocity at the halfway point. These values, with some simple calc and trig, should let you solve for those two constants. All that's left is solving for r(t)=0.
     
  11. Apr 27, 2012 #10
    Could you also integrate so that r=(-k/2)*r*t^2+Vo*t+ro?

    Plugging in .000003136 for k, 0 for r, -7986m/s for vo and 3189000m for ro gives t=399.3 seconds. (399.3+798.6)=1197.9*4=4791.6/60= 79.86min

    This is different than 76.41min but not by much.
     
  12. Apr 27, 2012 #11

    D H

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    No. What is that r on the right hand side: Is it some constant, or is it r(t)?

    Either way, you don't have a solution to [itex]\ddot r(t) = -kr(t)[/itex]. It's always a good idea to double check your solution.
     
  13. Apr 29, 2012 #12
    [itex]r(t) = r_0 \cos(\omega t + \phi)[/itex]

    Okay. Ro=3,189,000m (half the radius of the earth)

    I tried to solve for phi and got inverse sine of the radial distance (3189000m) over the velocity at the center (-7986m/s). Plugging this into the calculator I get an error.

    What could I possibly be doing wrong?
     
    Last edited: Apr 29, 2012
  14. Apr 29, 2012 #13

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    One of the things you should always do when solving a physical problem is check your units. Here you have distance over velocity, which has units of time. The argument of inverse sine should be unitless. So obviously something is wrong here. Your ratio has the wrong units.

    We're using [itex]r(t)=r_0\cos(\omega t + \phi)[/itex] as the general solution to [itex]\ddot r = -kr(t)[/itex]. Differentiating yields [itex]v(t) = -r_0\omega\sin(\omega t + \phi)[/itex]. Dividing distance by velocity yields [itex]r(t)/v(t) = - 1/(\omega \tan(\omega t + \phi))[/itex].

    This suggests using [itex]v(t)/r(t) = -\omega\tan(\omega t + \phi)[/itex]. Setting t=0 to the time at which the object reaches the halfway point yields [itex]v(0)/r(0) = -\omega\tan\phi[/itex]. You know [itex]\omega[/itex], you know [itex]v(0)[/itex], and you know [itex]r(0)[/itex]: You know everything that you need to know to be able to find [itex]\tan\phi[/itex] and hence [itex]\phi[/itex].


    Another good practice to follow is to keep everything symbolic as long as possible.
     
  15. Apr 29, 2012 #14
    That makes sense.

    I was looking at this problem the wrong way at each step. I was trying to directly apply certain formulas and pages of my book when they didn't exactly work. I should have stepped back and just thought about what I already knew.

    Thank you so much for helping and your patience with me. It really is a pretty easy problem. haha.
     
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