Falling to the centre of the Earth

Problem

Falling to the centre of the Earth. Estimate the time it would take for an object to fall to the centre of the Earth through a straight tunnel, released from rest at the surface. Neglect friction and rotation. Hint: inside a homogeneous Earth, $g(r) = g_0\,r/R$.

Solution

Inside a homogeneous Earth, gravity grows linearly with distance from the centre: $g(r) = g_0\,\dfrac{r}{R}$, directed towards the centre. The equation of motion along the tunnel is $m\ddot r = -m\,g_0\,\frac{r}{R} \quad\Longrightarrow\quad \ddot r = -\frac{g_0}{R}\,r$

This is the equation of simple harmonic motion $\ddot r = -\omega^2 r$ with $\omega = \sqrt{\frac{g_0}{R}}, \qquad T = 2\pi\sqrt{\frac{R}{g_0}}$

With $R = 6{,}37\cdot 10^6$ m and $g_0 = 9{,}8$ m/s$^2$: $T = 2\pi\sqrt{\frac{6{,}37\cdot 10^6}{9{,}8}} = 2\pi\sqrt{6{,}5\cdot 10^5} \approx 2\pi \cdot 806 \approx 5{,}06\cdot 10^3\ \text{s} \approx 84\ \text{min}$

The object released at the surface starts with an amplitude equal to $R$ and reaches the centre after a quarter of an oscillation: $t = \frac{T}{4} \approx \frac{84}{4} \approx 21\ \text{minutes}$

$\ev{t = T/4 \approx 21\ \text{min}\quad(\text{harmonic oscillation with } T = 2\pi\sqrt{R/g_0} \approx 84\ \text{min})}$

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Topics: Gravitation Concepts: Simple harmonic motion · Weight force Skills: Fermi estimates