I do something difference today, ha.

The problem is simple, How long does it take to tunneling through the center of the earth to the other side?

Assuming uniform density of the earth ($\rho$), the acceleration inside the earth at radius $r$ is

$\displaystyle - G \frac{4\pi}{3} \rho r = -k r$

$k = 1.5413\times 10^{-6}$ in 1/sec^2.

because only the mass within the radius matter. Thus, the equation of motion is

$\displaystyle \frac{d^2 r}{dt^2} = - k r$

the solution is

$\displaystyle r(t) = R \cos( \sqrt{k} t )$

Thus, the time for a trip is $T = \pi / \sqrt{k}$  = 2530.5 sec or 42 min and 10.5 sec.

The maximum speed when passing the core. The speed is $R \sqrt{k} = 7910 m/s$

How about we use a realistic earth density?

The density, and acceleration can be found in the web, for example, here.

The travel time is 2291 sec, or 38 min 11 sec.

It is interesting that, in the uniform density calculation, the travel time is independent of the radius, but density. The peak velocity is

$\displaystyle R \sqrt{k} < c \rightarrow \rho < \frac{c^2}{R^2}\frac{3}{4\pi} \frac{1}{G}$

$\displaystyle R_S = \frac{2GM}{c^2} \rightarrow \rho_S = \frac{1}{2}\frac{c^2}{R^2}\frac{3}{4\pi} \frac{1}{G}$

The Schwarzchild density is half of the maximum density by classical argument.

In case of point mass.

The equation of motion is

$\displaystyle \frac{d^2r}{dt^2} = -\frac{GM}{r^2}$

change of variable $v(r) = \frac{dr}{dt}$

$\displaystyle \frac{d(v^2/2)}{dr} = -\frac{GM}{r^2}$

$\displaystyle v^2(r) = 2 \frac{GM}{r} - 2 \frac{GM}{R}$

The above solution assume $v(R) = 0$. We can see that, at $r = 0$, the speed go to infinity. Something wrong…..

Update: that is not wrong at all. imagine two neutrino with head-on collision, the released gravitational energy will be infinity! But, because of uncertainly principle, their separation distance can never to be zero.

Taking account of the earth rotation with the centripetal force, the equation of motion becomes,

$\displaystyle \frac{d^2 r}{dt^2} = - k r + \omega^2\cos(\theta) r$

where $\omega$ is the angular velocity of earth, and $\theta$ is the polar angle from the north pole. The travel time would be

$\displaystyle T = \frac{\pi} {\sqrt{k - \omega^2 \cos(\theta)}}$

$\displaystyle \omega = \frac{2\pi}{24 \times 60\times 60} = 7.3 \times 10^{-5}$ rad/s

Thus,

$\omega^2 = 5.3 \time 10^{-9}$ 1/sec^2.

which is a very small correction.