We have a hole that leads straight through the center of the Earth. We neglect air resistance and assume that the Earth has uniform density ρ. How long would it take to fall through the center of the Earth and reach the other side of the planet?

###### Solution.

According to the Gauss’s law for gravity, the gravitational force on an object of mass m at a distance r from the center of a spherically symmetric mass M is -Gm/r^{2} times only the total mass M within a smaller distance than r. All the mass at a greater distance than r from the center can be ignored. In other words, we have

F = -GmM/r^{2} (1). (Similar to Newton’s law of universal gravitation.)

We know that M = ρ*V = ρ*(4/3)*π*r^{3} (2).

We substitute (2) into the formula (1) and we get

F = -Gm*ρ*(4/3)*π*r (3).

We can see that the gravitational force decreases proportionally to r as we approach the center of the Earth.

Also, we know that the gravitational acceleration g on the surface of the Earth is

g = G*M_{Earth}/R_{Earth}^{2} (4).

We use formula (2) and rewrite formula (4) as

g = G*ρ*(4/3)*π*R_{Earth}

or as

g/R_{Earth} = G*ρ*(4/3)*π (5).

Now we substitute (5) into the formula (3) and we get

F = – m*g*r/R_{Earth} (6).

m, g, R_{Earth} are constant, so we can write

F = – k*r (7),

where k = m*g/R_{Earth} (8).

The formula (7) has the same form as Hooke’s Law for a mass on a spring.

It means that we can use the formula for the period of the oscillations

T = 2π*√(m/k) (9).

We substitute (8) into (9) and we get

T = 2π*√(R_{Earth}/g) (10).

We know that R_{Earth} = 6378*10^{3} m and g = 9.8 m/s^{2}.

We substitute the values into the last formula and we get

T = 2π*√(6378*10^{3}/9.8) ≈ 5068.85 sec ≈ 84 min .

A traveler oscillate back and forth through the center of the Earth with the period of 84 min.

Hence, time required for fall through the center of the Earth and reach the other side of the planet is

84 min / 2 = 42 min.

Image By Luther1968 (Own work) [Public domain], via Wikimedia Commons