The classic 1864 science fiction novel by Jules Verne, titled "Journey to the Center of the Earth" has mesmerized many generations. The desire to discover the secrets inside the earth is perhaps as old as the desire to discover the secrets of the stars.

In reality, the mysteries of the earth are not as fantastic as envisioned by Jules Verne in his novel, and in this article we are going to take a very simple mathematical journey inside the earth, following a set of basic rules.

The Fantasy Ride

Our plan is to discuss a rather nice (and old) result from Newton's theory of Gravitation, using a hypothetical situation.

Let us imagine that we fall into a long tunnel dug into the earth, passing through the center, and coming out of the opposite end on the surface of the earth.

Obviously, this is a technologically impossible (at the moment) and downright stupid thing to do, but it presents a nice picture.

Now, we ask a question - Assuming that we are going to survive the fall, how long will it take for us to come out at the other end?

For this, we would have to study two very important (and simple) ideas in physics, and the connection between the two, which leads to the required answer.

The Physics of Springs

The first important idea that we need is the mathematical description of a body attached to a spring. We know from experience that, more we stretch the spring, greater is its pull.

Robert Hooke in 1676, gave the formal statement of this law as

The force needed to extend or compress a spring by some distance scales linearly with respect to that distance

In mathematical terms, it means

$$F \propto x$$

where $F$ is the force applied, and $x$ is the extension or compression in the spring.

The proportionality is fixed by choosing a constant $k$, which depends only on the material properties of the spring. Therefore, the constant is called the spring constant. The final form of the law is

$$F = k x$$

Let us first understand how a spring works - If we stretch it by some amount and release it, we observe the following -

• The ball begins to move towards the wall where the spring is mounted.
• It starts from rest, and gradually picks up the speed.
• The speed of the ball will keep on increasing till the point where the spring begins to compress.
• The ball will stop at the point when the spring is fully compressed.
• The compressed spring, pushes the ball , and the motion of the ball is repeated in the opposite direction.
• The ball comes back to the initial point, and stops.

In the situation above, we have assumed ideal conditions, with no energy losses (like due to friction). This would imply that the ball will keep on oscillating forever. If needed, we can develop a more realistic description of the motion.

The simple equation, given by Hooke's law, can be used to explain the entire motion of the ball attached to the spring, in a mathematical setup. We could obtain many physical parameters associated with the ball, like its position and speed, but here we are only interested in one such property - the time taken by the ball from the point of release to the point where is bounces back. The formula for this interval of time is given as

$$T = \pi \sqrt{\frac{m}{k}}$$

The steps to obtain this scary-looking formula involve a bit of calculus, and is beyond the scope of the article.

The Physics of Gravity

The second important idea that we need, is the description of how gravity works. The discovery of the law of universal gravitation by Issac Newton, published in 1687, is considered to be one of the greatest achievements of all time. It is mostly due to the fact that while it is incredibly powerful and astonishingly accurate, its mathematical description is really simple and compact. Although, it has been superseded by another version of theory (given by Einstein), the simplicity and practical utility of Newton's approach is hard to dismiss.

The law can be stated as

Every particle attracts every other particle in the universe with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.

This can be mathematically written as

$$F \propto \frac{m_1m_2}{r^2},$$
where $F$ is the force of gravity, $m_1$ and $m_2$ are the masses of the two bodies and $r$ is the distance between their centers.

The proportionality can be fixed by choosing a constant, which is independent of anything related to the bodies or the surroundings. This constant is denoted by $G$, and is called Newton's constant of gravitation. Then, the final form of the law becomes

$$F = G \cdot \frac{m_1m_2}{r^2}.$$

This little formula is capable of unlocking the secrets of all the bodies in motion (under gravity), ranging from a tiny stone in our backyard to the motion of giant stars, planets and other heavenly bodies.

Let's see how to use this formula for our case. We have a diagram below that depicts the situation of a body falling towards the center of the earth, due to gravity.

The quantities that we know -

• Mass of the falling body is $m$
• Radius of the earth is $r$.
• The falling body is at a distance of $x$ from the center of the earth.
• We assume that the earth is a perfect sphere and has a uniform density of matter.

In order to use Newton's formula, we need to find the mass of the earth, and the distance between the two bodies. The first body is of course the one which is falling into the earth, and the second body is the earth itself.

Now, we have a problem - Since the earth is not a point, from where on earth, should we measure the distance from? Well, there is a cool mathematical idea, that tells us that we can effectively choose the center of the earth as a point where all its mass is assumed to be concentrated. This sounds crazy, right? But, it's true!

So, in the diagram above, the distance between the "earth" at $O$ and the falling body at $P$ can be taken as $x$.

Next, we evaluate the mass of the earth. We know from basic mathematics that,

$$Mass = Density \times Volume,$$

if the matter is distributed uniformly throughout the volume.

The same cool idea, also tells us that the effect of gravity caused by the (non-shaded) region of the earth "above" point $P$ is zero! (A discussion of this "cool" idea is reserved for a future article)

So, we are allowed to work with only the shaded region of the earth. If we denote the density of earth by $d$, then the mass of the region of the earth that exerts force, becomes

$$M = d \times V,$$

where $V$ is the volume of the shaded region. Its value can easily be obtained by using a formula that most of us learned in high school mathematics - the volume of a sphere, which is given by

$$\text{Volume} = \frac{4}{3} \pi \times \text{ radius }^3$$

Therefore, we can write the mass of shaded region as

$$M = d \times \frac{4}{3} \pi \times x^3$$

That's it. We have collected all the pieces required to use the Newton's formula and we can now obtain the force exerted by the earth on the falling body when it's at a distance of $x$ from the center -

\begin{aligned}F &= G \frac{m \times M}{x^2}\\ &= G \times \frac{m}{x^2} \times d \times \frac{4}{3} \pi \times x^3\\ &= \frac{4\pi Gdm}{3} \times x\end{aligned}

This is a very special formula. It can be seen that the factor

$$\frac{4\pi Gdm}{3},$$

is a constant. It only depends on the density of the earth, the mass of the falling body, the Newton's constant and some pure numbers. All of these quantities are fixed for a given situation, and we can collectively pack this into another constant called $k$. Therefore, we can write this formula in a simple form as

$$F = k \times x$$

But, this is exactly what we had for the body connected to a spring!

So, we have discovered a weird connection between a body falling into earth and a body attached to a spring, and it means we can make a really wild statement -

A body falling freely into the earth, oscillates like a body attached to a spring!

Therefore, we can actually calculate the time taken by the falling body from the initial point on earth, to a point on the opposite side of the earth, through the tunnel. We use the formula for the time duration discussed earlier, and get

\begin{aligned}T &= \pi \sqrt{\frac{m}{k}}\\ &= \pi \sqrt{\frac{3m}{4\pi Gdm}} \\ &= \pi \sqrt{\frac{3}{4\pi Gd}}\end{aligned}.

The values of these quantities are known from experiments -

• $G = 6.67 \times 10^ {-11}$ in standard units.
• $d = 5.51 \times 10^3$ in standard units.

At last, we can plug these values into the formula and get approximately 42 minutes as the time taken by the body falling into the tunnel, to come out at the other end!

If you haven't realized it yet, this is an incredibly small duration to cover a journey of 12,800 km (diameter of earth). The body must have been traveling with an unbelievable average speed of more than 18000 km/hr !

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