Proportional acceleration due to changing density of the Earth My question has to to do with a recent video Minutephysics posted about the time it takes for a person to fall through the earth, found here
http://www.youtube.com/watch?v=urQCmMiHKQk
At around 4:05, he mentions some "mathemagical dust" that nets him the answer of 348 seconds:

Wanting the total time it takes to fall through the earth, he splits the earth into two portions: the first where the density is relatively constant and acceleration equal to gravity, where the time calculation is a simple application of kinematics knowing the radius of the earth and acceleration.
The second portion is somewhat more involved. Here, the density increases more quickly, increasing the mass that causes acceleration due to gravity at a rate different than the first portion. As a result, acceleration changes from the constant 10 m/s to a variable one, apparently proportional to the radius. He calculates this new acceleration by doing the following.
$$\ddot{R}(t) = a(t)=-36.36\frac{R}{ R_\oplus}$$ 
then noting the initial velocity $v_0=\dot{R}(t=0)=-7580\,{\rm m}\,{\rm s}^{-1}$.
He then goes on to conclude 
$$R(t) = 3.5 \times 10^6 \cos\left( \sqrt{\frac{ R_\oplus}{36.36}}t\right) - 3.2 \times 10^6 \sin\left(\sqrt{\frac{R_\oplus}{36.36}}t\right)$$
$$t = \tan^{-1}(1.1)\sqrt{\frac{R_\oplus}{36.36}} = 348\,{\rm s}$$ 
My question has to do with the constant $-36.36$ he notes and the source of the two trig equations (somewhat limited physics knowledge, so please excuse me if this is something basic).
I was attempting to figure out the logic behind the second part of his calculations, but couldn't get very far. I attempted to do what he did earlier in the video, plugging in the Earth's mass, in terms of radius and density, into the universal gravity equation with the different density of the core, but don't understand either of the trig equations he has or how the second derivative denoted by the two dots relates to time. Any help would be appreciated; I simply couldn't get to bed without knowing the answer. 
 A: I'm not quite sure how he gets a value of $36.36$ for that constant, because it depends a bit on exactly how he does his approximations and what values he assumes for different constants. But basically it's this:
For a spherically symmetric mass distribution like the one considered here, the gravitational acceleration is simply, from Newton's law for the gravitational force:
$$a=\frac{-GM(<R)}{R^2}$$
For the mass enclosed within radius $R$, which I write $M(<R)$, he assumes a constant density, so the mass is just the volume of the sphere (of radius $R$, which depends on where you are) times the density:
$$M(<R) = \left(\frac{4}{3}\pi R^3\right)\sigma$$
Putting the two equations together,
$$a = -\frac{4}{3}G \pi R\sigma$$
Or making it look like his:
$$a = -\left(\frac{4G\pi\sigma R_\oplus}{3}\right)\frac{R}{R_\oplus}$$
The quantity in brackets is what he evaluates to 36.36 (that should be ${\rm m}\,{\rm s}^{-2}$). I'm not sure what he uses for $R_\oplus$; it's not clear whether it's the diameter of the Earth (which is what it looks like from the equation) or the diameter of the Earth less $\Delta x=2.87\times10^6\,{\rm m}$, which is what it looks like from the diagram. He also doesn't mention what he uses for the density, but from one diagram it looks like it's about $10^4\,{\rm kg}\,{\rm m}^{-3}$. The value of the constant is suspicious; since at $R=R_\oplus$ the falling person is just leaving the "constant acceleration" portion of the trip, I'd expect the acceleration here to be $10\,{\rm m}\,{\rm s}^{-2}$, not $36.36$...
As to where the trig equations come from, this is related to the dot notation. A dot typically indicates a time derivative, so:
$$v=\dot{R}=\frac{{\rm d}}{{\rm d}t}R(t) \\ a = \ddot{R} = \frac{{\rm d}^2}{{\rm d}t^2}R(t)$$
The equation: 
$$\frac{{\rm d}^2}{{\rm d}t^2}R(t) = -k^2R(t)$$
where $k$ is a constant is a differential equation. The solution to this equation is a function $R(t)$ such that the equality holds for all values of $t$. This is probably the best known differential equation in existence, the simple harmonic oscillator equation. You can check that the general solution:
$$R(t)=A\sin(kt)+B\cos(kt)$$
solves the equation by substituting it into the differential equation and evaluating the derivative. This is where his equation comes from, once he works out what the constants $A$, $B$ and $k$ should be. I won't go through that since solving the simple harmonic oscillator equation is such a ubiquitous exercise that you can look it up in basically any book or on the Internet.
