# 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

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.

• Hello, and welcome to Physics Stack Exchange. There's a really good question in here, but as the post is set up right now, it's needlessly difficult for a reader to figure out exactly what you're asking. The problem is that we're required to click on a video link and watch it! Please take all of the information relevant to your question and put it right there in the post. This is required by site standards. Note that you can type math using mathjax. I look forward to seeing the updated post :) – DanielSank May 24 '16 at 5:03
• physics.stackexchange.com/q/2481 See my answer (user82794), equation (B-01), and SECTION 2, Proposition C, (c2). – Frobenius May 24 '16 at 5:08
• Other related: physics.stackexchange.com/q/7346/2451 , physics.stackexchange.com/q/18446/2451 and links therein. – Qmechanic May 24 '16 at 5:45
• There are some rocks in that mathemagical dust. His value of 36.36 is wrong. – David Hammen May 24 '16 at 19:32
• @DavidHammen He never specifies units for $R_{\oplus}$, so it's right... for some choice of units ;) – Kyle Oman May 25 '16 at 6:08

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.

I am not sure about the derivation in this video although it might be correct?

My doubt is with the equation for the acceleration $a$ for the inner phase of the motion where $R$ is the distance from the centre of the Earth and the radius of the Earth is $R_\oplus = 6.37 \times 10^6$ m.

$$\ddot{R}(t) = a(t)=-36.36\dfrac{R}{ R_\oplus}$$

During the constant acceleration (outer) phase of the motion the distance covered is $2.87 \times 10^6$ m and so the inner phase motion is within the following bounds $3.5 \times 10^6 \le R \le 3.5 \times 10^6$ m

Putting $R = 3.5 \times 10^6$ m into the equation for the acceleration during the inner motion phase gives $-20$ m s$^{-2}$ not the expected $-10$ m s$^{-2}$.

Also the expression given in the video is not dimensionally correct.

$$R(t) = 3.5 \times 10^6 \cos\left( \sqrt{\dfrac{ R_\oplus}{36.36}}t\right) - 3.2 \times 10^6 \sin\left(\sqrt{\dfrac{R_\oplus}{36.36}}t\right)$$

The brackets for the cosine and sine functions should be $\left( \sqrt{\dfrac {36.36}{ R_\oplus }}t\right)$ not $\left( \sqrt{\dfrac{ R_\oplus}{36.36}}t\right)$

The expression $t = \tan^{-1}(1.1)\sqrt{\dfrac{R_\oplus}{36.36}} = 348\,{\rm s}$ can be used to find the value of the radius of the Earth which is used in the video.

When this is done this gives $R_\oplus = 6.35 \times 16^6$ m which is in agreement with the accepted value.

Going back to the beginning for a uniform density Earth the expression for the acceleration $a$ as a function of distance from the centre can be written as $a = - \dfrac {9.8}{R_\oplus} R$ which is of the form of the equation of motion for shm $a = - \omega^2 R$ where $\omega$ is a constant of the motion.

Using $R_\oplus = 6.37 \times 10^6$ m gives $\omega = 1.24\times 10^{-3}$ s$^{-1}$.

The period of this motion is $T = \dfrac {2 \pi}{\omega}$ gives a half period of $2532$ s $\approx 42$ minutes.

For this motion $R(t) = 6.37 \times 10^6 \cos \omega t = 6.37 \times 10^6 \cos \left( \sqrt{\dfrac {10}{ R_\oplus }}\; t\right)$.

If the expression for $R(t)$ is differentiated once

$$v(t) = \dot R(t) = -6.37 \times 10^6 \omega\sin \omega t$$

then at time $t =0$ the velocity is zero and if differentiated again

$$a(t) = \ddot R(t) = -6.37 \times 10^6 \omega^2\cos\omega t$$

which shows that $R(t) = 6.37 \times 10^6 \cos \omega t$ is a solution the equationto $a = - \omega^2 R$.

The procedure for the mixed composition Earth is in two parts and it is the second part that you have queried.

The magnitude of the acceleration at a distance from the centre $R_i = 3.5 \times 10^6$ m is given as $10$ m s $^{-2}$ and changes linearly with distance from the centre of the Earth.

The equation of motion for that inner phase is $a = -\dfrac {10}{R_i} R = -\dfrac {10}{R_\oplus}\dfrac{R_\oplus}{R_i} R = -\dfrac{18.2}{R_\oplus}R$ whereas it is given as $a = -\dfrac {36.36}{R_\oplus} R$ in the video.

Is there a reason for this?

This gives $\omega = \sqrt{\left(-\dfrac{18.2}{R_\oplus} \right )} = 1.69 \times 10^{-3}$ s$^{-1}$.

$R(t) = 3.5 \times 10^6 \cos \omega t$ is a solution to this equation but the velocity at $t =0$ is zero which is not correct.

To have a velocity of $-7580$ m s$^{-1}$ at the start an extra term $-4.06\times 10^6 \sin\omega t$ must be added to give the expression

$$R(t) = 3.5 \times 10^6 \cos \left (\sqrt{\dfrac{18.2}{R_\oplus} }\; t \right) - 4.49\times 10^6 \sin \left (\sqrt{-\dfrac{18.2}{R_\oplus} }\; t \right)$$

Note that the additional sine term is negative at the start of the motion which will make the distance from the centre of the Earth decrease faster than if there was just the cosine term present.
Just as you might expect as the mass has an initial inward velocity when $t=0$.

To find the time to reach the centre make $R(t) = 0$ and this gives

$$t = \tan^{-1}\left (\dfrac{3.5}{4.49}\right )\sqrt{\dfrac{R_\oplus}{18.2}} = 392\,{\rm s}$$