# Math Help - Mechanics: 4 HH2G Fans

1. ## Mechanics: 4 HH2G Fans

In one of the episodes of The Universe I saw on Netflix, a scientist mentioned that by digging tunnels in the Earth—straight point to point tunnels—one could travel from place to place on the Earth. The striking revelation given in this episode was that each traverse through the Earth would take the same amount of time, i.e., 42 minutes. Hitchhiker's Guide to the Galaxy fans will note that this is why the mice built the Earth.

G = the gravational constant
g = standard gravity = 9.80665 $\frac{m}{s^2}$
R = radius of the Earth = 6371 km
r = radius variable (r R)
M = mass of the Earth
$M_r$
= mass of the portion of the Earth enclosed in a sphere of radius r
δ = density of the Earth
$a_r$
= acceleration due to gravity at distance r from the Earth's center
t = time of transit through the tunnel, with t = 0 corresponding to transit's start
a(t) = acceleration within the tunnel at time t
x = distance along tunnel with maximum negative value at entrance, x = 0 at the center, and maximum positive value at exit
l = least distance of tunnel from Earth's center (middle of tunnel to Earth's center)
a(t) = acceleration after time t

$g=G\frac{M}{R^2}\hspace{4mm}M=\delta\frac{4\pi}{3} R^3\hspace{4mm}M_r=\delta\frac{4\pi}{3}r^3\hspace{ 4mm}\frac{M_r}{M}=\frac{r^3}{R^3}\hspace{4mm}M_r=M \frac{r^3}{R^3}$

$a_r=G\frac{M_r}{r^2}=G\frac{M}{R^2}\frac{r}{R}=g\f rac{r}{R}$

$x''(t) = a(t) = -\frac{x}{r}a_r = -g\frac{x}{R}\hspace{4mm}x''(t) = -kx\hspace{4mm}(k = \frac{g}{R})$

Solution: $x(t) = c_1 sin(t\sqrt{k}) + c_2 cos(t\sqrt{k})$

By symmetry $c_1 = 0$. Thus $c_2 = -\sqrt{R^2 - l^2}$.

The time it takes to traverse the tunnel is therefore $\frac{\pi}{\sqrt{k}} = \pi\sqrt{\frac{R}{g}}$ or 42 minutes, 12 seconds.

Of course, the Earth does not have a
constant density, nor is the radius a constant 6371 km, so the actual times may not be constant and independent of the tunnel length and location. However, 42 minutes is likely the closest number of minutes for the traverse.

Are there any comments, flaws, or improvements to the calculation?

2. Originally Posted by NowIsForever
In one of the episodes of The Universe I saw on Netflix, a scientist mentioned that by digging tunnels in the Earth—straight point to point tunnels—one could travel from place to place on the Earth. The striking revelation given in this episode was that each traverse through the Earth would take the same amount of time, i.e., 42 minutes. Hitchhiker's Guide to the Galaxy fans will note that this is why the mice built the Earth.

G = the gravational constant
g = standard gravity = 9.80665 $\frac{m}{s^2}$
R = radius of the Earth = 6371 km
r = radius variable (r R)
M = mass of the Earth
$M_r$
= mass of the portion of the Earth enclosed in a sphere of radius r
δ = density of the Earth
$a_r$
= acceleration due to gravity at distance r from the Earth's center
t = time of transit through the tunnel, with t = 0 corresponding to transit's start
a(t) = acceleration within the tunnel at time t
x = distance along tunnel with maximum negative value at entrance, x = 0 at the center, and maximum positive value at exit
l = least distance of tunnel from Earth's center (middle of tunnel to Earth's center)
a(t) = acceleration after time t

$g=G\frac{M}{R^2}\hspace{4mm}M=\delta\frac{4\pi}{3} R^3\hspace{4mm}M_r=\delta\frac{4\pi}{3}r^3\hspace{ 4mm}\frac{M_r}{M}=\frac{r^3}{R^3}\hspace{4mm}M_r=M \frac{r^3}{R^3}$

$a_r=G\frac{M_r}{r^2}=G\frac{M}{R^2}\frac{r}{R}=g\f rac{r}{R}$

$x''(t) = a(t) = -\frac{x}{r}a_r = -g\frac{x}{R}\hspace{4mm}x''(t) = -kx\hspace{4mm}(k = \frac{g}{R})$

Solution: $x(t) = c_1 sin(t\sqrt{k}) + c_2 cos(t\sqrt{k})$

By symmetry $c_1 = 0$. Thus $c_2 = -\sqrt{R^2 - l^2}$.

The time it takes to traverse the tunnel is therefore $\frac{\pi}{\sqrt{k}} = \pi\sqrt{\frac{R}{g}}$ or 42 minutes, 12 seconds.

Of course, the Earth does not have a
constant density, nor is the radius a constant 6371 km, so the actual times may not be constant and independent of the tunnel length and location. However, 42 minutes is likely the closest number of minutes for the traverse.

Are there any comments, flaws, or improvements to the calculation?
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