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SHM and Gravitation

by authgeek
Tags: gravitation
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authgeek
#1
Sep29-09, 07:18 PM
P: 7
I'm working on a problem on relating gravitation and simple harmonic motion. The idea is that a mass dropped in a hole drilled through the earth will oscillate (no friction, etc).

The question asks this:
"Show that the motion of the mass is simple harmonic motion and find a formula for r(t)"

So, I'm starting with the basic r(t) equation for simple harmonic motion, r(t) = Acos(wt) where A is the total radius of the earth (the amplitude), t is the time and w is the angular frequency. My problem is that any attempt to find an angular frequency seems to involve dictating the distance from the center, r, which is what I'm trying to find as a function of time.

The best I've been able to come up with is w = sqrt(k/m) where k is the "gravitational spring constant" involving (4/3)pi * p * R * G, R being the radius that I'm trying to find.

How should I find w in this case?
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Quiablo
#2
Sep29-09, 09:25 PM
P: 9
First, considerer Newtons gravitation law:

[tex]F = \frac{-GmM}{r^{2}}[/tex];

where the minus sign indicates that the force is attractive. Now consider a body moving through the center of the Earth, at a distance r(t) from the center. Suppose that the Earth has constant mass density [tex]\rho[/tex]. Tha total mass "below" (e.g, that will effectively exerct a resultant force uppon) the particle located at position r(t) (supposing that r(t) is always smaller than the outter radius of the Earth) will be clearly a function of r of the form:

[tex]M = \frac{4 \rho \pi r^{3} }{3}[/tex];

Plugging that formulae on Newton's gravitation law and writing F = ma we have:

[tex]a = \frac{-4 G \rho \pi r^{3}}{3 r^{2}} = \frac{-4 G \rho \pi r}{3} [/tex];

Then, writing [tex] a = \frac{d^{2}r}{dt^{2}}[/tex], we get:

[tex] \frac{d^{2}r}{dt^{2}} = \frac{-4 G \rho \pi r}{3} [/tex];

Solving for r (which should be easy enough) you get the harmonic solution and, furthermore, the angular frequency of oscilation.
authgeek
#3
Sep29-09, 09:59 PM
P: 7
Quiablo -

Thanks for your reply, it's very helpful. One question: When you say "the harmonic solution", what does that mean conceptually? I follow the math, but I'm a little fuzzy on the terminology.

Quiablo
#4
Sep30-09, 08:22 AM
P: 9
SHM and Gravitation

I just wanted to mean that you will get, as a solution to the differential equation, a harmonic function of the form A cos(Bx), as you said it should be. You can right solutions of the exponential form also, complex exponentials, but that won't help you much in this case.

The general solution to the the second order diff equation:

[tex]\frac{d^{2}r}{dt^{2}} = -C r[/tex]

where C substitutes the group of multiplying constants, can be written in the form:

[tex]r(t) = A_{1} cos (Bt) + A_{2} sin (Bt)[/tex]

or equivalently:

[tex]r(t) = A_{3} cos (Bt + \phi )[/tex];

Where A1, A2, A3, B and [tex]\phi[/tex] are all constats, to be determined by the initial conditions of the problem (except for B, which will be determined directly by C, which is known by looking at the orignal equation).


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