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Simple Harmonic Motion Inside earth

  • Thread starter Mindscrape
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  • #1
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Homework Statement


A particle is dropped into a hole drilled straight through the center of the Earth. Neglecting rotational effects, show that the particle's motion is simple harmonic if you assume Earth has uniform density. Show that the period of the oscillation is about 84 min.


Homework Equations


[tex]F = -G m \int_V \frac{\rho(r') e_r}{r^2}dv'[/tex]


The Attempt at a Solution



I was going to use Newton's second law to show that
[tex] m \frac{d^2 r}{dt^2} = -G m \int_V \frac{\rho(r') e_r}{r^2}dv'[/tex]

Where the volume integral should produce some function of r. So, I started off the integration by choosing an arbitrary point in the sphere a distance r' away, and using R as the distance from the origin to the point mass, r as the distance between the arbitrary distance and the distance of the point mass, theta as the asimuthal angle, and phi as the rotational angle I got.

[tex]m \frac{d^2 r}{dt^2} = -G m \int_0^{R} \int_0^\pi \int_0^{2\pi} \frac{\rho r'^2 sin\theta}{r^2}dr' d\theta d\phi[/tex]

integrating with respect to phi brings in a factor of 2π, and now I used law of cosines

[tex]r^2 = r'^2 + R^2 - 2r'Rcos\theta[/tex]

[tex] 2r dr = 2r'Rsin\theta d\theta[/tex]

[tex]\frac{sin\theta}{r}d\theta = \frac{dr}{r'R}[/tex]

so the substitute the law of cosines stuff into the force equation

[tex]F = \frac{-2 \pi G m \rho}{R} \int_0^R r'^2 dr \int_{r'-R}^{r'+R} \frac{1}{r}dr[/tex]

doing the r' integral I can see that this ultimately won't give a linear function of R

[tex]\frac{2 \pi}{3} \frac{G m \rho}{R} R^3 \int_{r'-R}^{r'+R} \frac{1}{r}dr [/tex]

Can someone help me out, point out what I did wrong, put me on the right track?

I think that once I find the right equation for force, which I actually know from experience should be [tex]F(r) = -\frac{4 \pi}{3}G m r \rho[/tex], that I can do the differential equation stuff.
 

Answers and Replies

  • #2
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You're making it too complicated. Assume that the density of the earth is a constant function, and realize that because of the character of the 1/r^2 force law, you can turn this into a surface integral. Once you do that, you're golden.
 
  • #3
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At any point in the fall, the only mass exerting a nett force on the test body is that contained inside the current radius because the shell of matter outside the radius has no effect. So the force at point x is

[tex]F(x) = -\frac{4}{3}\pi Gm\rho x^3/x^2[/tex]

which gives

[tex]F(x) = -Kx[/tex]

QED I think. Assuming uniform density.
 
Last edited:

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