# SHM and Gravitation

by authgeek
Tags: gravitation
 P: 9 First, considerer Newtons gravitation law: $$F = \frac{-GmM}{r^{2}}$$; 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 $$\rho$$. 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: $$M = \frac{4 \rho \pi r^{3} }{3}$$; Plugging that formulae on Newton's gravitation law and writing F = ma we have: $$a = \frac{-4 G \rho \pi r^{3}}{3 r^{2}} = \frac{-4 G \rho \pi r}{3}$$; Then, writing $$a = \frac{d^{2}r}{dt^{2}}$$, we get: $$\frac{d^{2}r}{dt^{2}} = \frac{-4 G \rho \pi r}{3}$$; Solving for r (which should be easy enough) you get the harmonic solution and, furthermore, the angular frequency of oscilation.
 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: $$\frac{d^{2}r}{dt^{2}} = -C r$$ where C substitutes the group of multiplying constants, can be written in the form: $$r(t) = A_{1} cos (Bt) + A_{2} sin (Bt)$$ or equivalently: $$r(t) = A_{3} cos (Bt + \phi )$$; Where A1, A2, A3, B and $$\phi$$ 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).