S.h.m. through a hole in the earth

  • Thread starter Mechdude
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  • #1
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Homework Statement



this is the problem:
assume that the earth is spherical and that the density is uniform. show that particle dropped into a straight hole drilled through the center of the earth , will execute harmonic motion. find the period of oscillation and show that it depends on the density not the size of the earth.

Homework Equations



the force on a mass m outside a spherical shell is given by:
[tex] F = \frac{Gm M_{sh}}{r^2} [/tex]
and the potential inside a spherical shell (and the force by consequence) is zero.


The Attempt at a Solution



the fact that potential inside a spherical shell is zero led me to think i should consider the earth as being composed of a series of concentric shells, and only the shells with a radius smaller than the displacement of the mass from the center of the earth will contribute to the force, now to find the total force on the mass from these considerations is causing me some trouble, since if i integrate the equation of the force due to a shell from the center of the earth to the distance the mass is at will give me the expression for potential and taking that integral as the expression for that force (and this is a stretch since the assumption that im right is far fetched) taking the limits between 0 and x, x being the displacement from the earth gives me an indeterminate expression since there's a zero ending up in the denominator,
[tex] F_{sph} = - \int_{0}^{x} \frac{ G m M_{sh}}{r^2} dr [/tex]

[tex] = -GmM_{sh} \left( - \frac{1}{r} \right)_{0}^{R} [/tex]



i doubt im even approaching the problem properly, what would be the correct way of doing this?

but anyway given the expression for the force the period would follow easily. that i have no trouble with.
 

Answers and Replies

  • #2
Doc Al
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Find an expression for the force on a test mass as a function of distance from the center of the earth. Hint: What part of the Earth's mass contributes to that force?

How does your expression compare to the force from a spring?
 
  • #3
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thanks for the reply, here i go , only the part of the earth the contributes to the force is that within a sphere of a radius smaller than the distance between the falling mass and the center of the earth .
so i assume if all the density of the earth is [itex] \rho [/itex] then
[tex] F = \frac{ G m_1 m_2 }{r^2} [/tex]
becomes:
[tex] F = \frac {G m_1 \rho 4 \pi r^3}{r^2 3} [/tex]

or
[tex] F = \frac{G m_1 \rho 4 \pi r}{3} [/tex]
is that it?
if it is then using newtons 2nd law:
[tex] m \frac {d^2 r}{d t^2} + m \frac {G \rho 4 \pi }{3} r = 0 [/tex]

or
[tex] \frac {d^2 r}{d t^2} + \frac {G \rho 4 \pi }{3} r = 0 [/tex]

by inspection then :
[tex] \omega = \sqrt{ \frac{G \rho 4 \pi }{3} } [/tex]
 
  • #4
Doc Al
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Perfecto!
 

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