Last updated on 13/02/2025

“That was when I saw the Pendulum.
The sphere, hanging from a long wire set into the ceiling of the choir, swayed back and forth with isochronal majesty.
I knew – but anyone could have sensed it in the magic of that serene breathing – that the period was governed by the square root of the length of the wire and by $\pi$ , that number which, however irrational to sublunar minds, through a higher rationality binds the circumference and diameter of all possible circles. The time it took the sphere to swing from end to end was determined by an arcane conspiracy between the most timeless of measures: the singularity of the point of suspension, the duality of the plane’s dimensions, the triadic beginning of $\pi$ , the secret quadratic nature of the root, and the unnumbered perfection of the circle itself.” Incipit of “Foucault’s Pendulum“, Umberto Eco – 1988

The solemn opening of Umberto Eco’s masterpiece Foucault’s Pendulum evocatively refers to the formula that gives the period of the ‘small’ oscillations of a pendulum placed near the Earth’s surface. The formula is:

(1) $\begin{equation*} T = 2 \pi \sqrt{\dfrac{L}{g}} \end{equation*}$

where $g$ is gravity near Earth’s surface.

The period is said to be isochronous because the pendulum will always swing with the same period regardless of the amplitude of its oscillations, as long as they are small, i.e. $\alpha \ll 1$ , or the distance between the extreme points touched by the oscillating mass very small compared to $L$ .

According to (1), the period increases as the length of the wire ( $L$ ) increases. For a very long pendulum, $L$ and $T$ tend to infinity. Since the period is nothing more than the time it takes the mass to return to the same point after a complete oscillation, in the case of an infinite pendulum the suspended mass should be stationary.

But let’s have a look at the following picture. If we imagine lengthening the string until it becomes comparable in length to the Earth’s radius, we see that the suspended mass cannot remain stationary for any non-zero angle $\alpha$ , because it would be attracted towards the center of the Earth.

Indeed, (1) is valid under the assumption that the pendulum is in a uniform gravitational field, so that the weight has a constant magnitude $mg$ and is always directed downwards. The second figure, however, shows that the direction of the attractive force $\vec{F}$ is not aligned with the string except when $\alpha=0$ . The component of $\vec{F}$ perpendicular to the string, namely $\vec{f}$ , tends to cause the mass to oscillate, thereby producing a periodic motion.

We deduce that it is not true that the period of the oscillations tends to infinity as the string lengthens indefinitely. So, what is the limiting value of the period?

Suppose that $L \rightarrow \infty$ . The pivot point $C$ moves away from Earth and the pendulum becomes gigantic. When $C$ is at infinity, we have an astronomical pendulum.

In such a limit case $\alpha \rightarrow 0$ , as shown in the following figure. The suspended mass, assumed to be in $P$ , will then oscillate along the line $\gamma$ , which is perpendicular to the string and at a distance $H$ from the center $O$ of the Earth.

When the extension of the string does not pass through the center of the Earth, the attractive force $\vec{F}$ will have a component $\vec{f}$ , along $\gamma$ , responsible for the oscillatory motion.

Let us denote by $x$ the position of $P$ along $\gamma$ , and let’s choose the reference frame so that $x=0$ coincides with the position of $P$ when the extension of the string passes through the center $O$ of the Earth.

Clearly, $\cos \vartheta = \dfrac{x}{|OP|}$ . Since $f=F\cos\vartheta$ , we have $f= -F \dfrac{x}{|OP|}$ , that is

$f=-G M m\dfrac{x}{|OP|^3}$

Newton’s law of dynamics ( $f = m \ddot{x}$ ) becomes:

$m \ddot{x}=-G M m\dfrac{x}{|OP|^3}$

(2) $\begin{equation*} \ddot{x}+ \dfrac{G M}{|OP|^3}x = 0 \end{equation*}$

The distance $|OP|$ is a function of $x$ . However, when the line $\gamma$ is close to the Earth’s surface and $x$ is small relative to the radius of the Earth ( $R_t$ ), we can accept the following approximation:

$|OP| \approx R_t$

Equation (2) then approximates an harmonic motion:

(3) $\begin{equation*} \ddot{x}+ \dfrac{G M}{R_t^3}x = 0 \end{equation*}$

The angular frequency of this harmonic motion is

(4) $\begin{equation*} \omega = \sqrt{\dfrac{GM}{R_t^3}}\end{equation*}$

which can also be written by introducing the average density $\delta = \dfrac{M}{\frac{4}{3}\pi R_t^3}$ , as:

(5) $\begin{equation*} \omega = \sqrt{\dfrac{4}{3}\pi G \, \delta}\end{equation*}$

We see that the period of small oscillations of the astronomical pendulum depends only on the universal gravitation constant ( $G$ ) and the average density of the body generating the gravitational field (e.g. the Earth).

Using (4), we can also express the period as:

(6) $\begin{equation*} T_a = 2 \pi \sqrt{\dfrac{R_t^3}{GM}} \end{equation*}$

Using equation (5), we have a second equivalent formulation:

(7) $\begin{equation*} T_a = \sqrt{\dfrac{3\pi}{G \delta}} \end{equation*}$

Let’s move on to evaluate the numerical value of this period. Given that:

$R_t = 6378 \, km = 6'378'000 \, m$
$G = 6.6726\, 10^{-11} \, \dfrac{N m^2}{kg}$
$M = 5.972 \, 10^{24} \, kg$

we get:

$T_a = 5'060.6\, s$ that is $T_a = 84 \, min \, 21 \, s$ .

Observations

1. There exists a maximum value for the period of the pendulum

The first observation to make is that as the string lengthens, the period does not tend to infinity and the suspended mass does not tend to stop. Instead, the period tends to a finite value, given by (6): this is the period of the astronomical pendulum when the oscillating mass is near the Earth’s surface.

2. The period of small oscillations of the astronomical pendulum is exactly equal to the period of satellites in low Earth orbit

What terrestrial phenomenon has a period of about $84 \,min$ ?

The International Space Station, which orbits just over $400 \,km$ from the Earth’s surface, that is approximately $6\%$ of the Earth’s radius, takes about $92 \, min$ to complete an orbit. This is a typical orbital period for satellites in low Earth orbit, placed at a distance between $200 \, km$ and $2000 \, km$ from Earth.

The lower limit ( $200 \,km$ ) is imposed by the need to avoid the thicker atmosphere, which would slow down the satellite too quickly. In a scenario where there was no friction and satellites could be closer to Earth, the period of their orbit would be shorter. Hypothetical satellites skimming the Earth’s surface, assuming no air friction, would take $84 \,min$ and $21 \,s$ to complete an orbit.

Indeed, ignoring air friction, the orbital period of a satellite in low Earth orbit is given by $T_s = 2 \pi \sqrt{\dfrac{R_t^3}{GM}}$ , which is exactly the same formula for the period of the astronomical pendulum.

3. There is a simple pendulum, of finite length, that has the same period as the astronomical pendulum

The weight force near the surface of Earth is given by $m \, g = G \frac{Mm}{R_t^2}$ , from which we have $g= G \frac{M}{R_t^2}$ . Equation (6) can thus also be written as $T_a = 2 \pi \sqrt{\frac{R_t}{g}}$ .

In other words, the period of small oscillations of the astronomical pendulum coincides with the period of small oscillations of a simple pendulum suspended at a pivot point that is as far from the Earth’s surface as the radius of the Earth.

4. The period of the astronomical pendulum coincides with the period of the gravitational trains

Let’s suppose we dig a diametral tunnel across the Earth. A body that fell into such tunnel would move with harmonic motion, with period $T_a$ (if friction is negligible). That is, such a body would take about $42 \, min$ to move from a point on the Earth’s surface to the point located at its antipodes.

Let’s now suppose that we dig a straight tunnel, which does not pass through the Earth’s center. Such tunnel would connect two points which are not at their antipodes. Again, assuming zero friction, a body that was free to move inside this tunnel would oscillate back and forth because of gravity with a period exactly equal to $T_a$ .

(For more information: https://en.wikipedia.org/wiki/Gravity_train)

5. Gravitational transport along horizontal bridges

The oscillating mass of the astronomical pendulum moves along the straight line $\gamma$ . Therefore, any mass free to move without friction along a horizontal line is equivalent to the suspended mass of the astronomical pendulum. As such, it would oscillate with a period equal to $T_a$ .

Let’s further develop the consequences of this idea. Let’s imagine we build a horizontal bridge connecting two points $A$ and $B$ , as shown in the figure. In the absence of friction, a mass placed on the bridge would move from $A$ to $B$ in a time equal to $T_a/2$ , i.e. in about $42 \, min$ , regardless of the distance ( $2x$ ) between $A$ and $B$ . Starting from $A$ with zero velocity, the mass would accelerate for half the distance and decelerate for the second half, arriving at $B$ with zero velocity. A convenient way to travel!

Since this is a harmonic motion, it is easy to calculate the maximum acceleration and the maximum speed as a function of $x$ . Let’s rewrite (4) in a more compact form, using $g$ , as follows:

(8) $\begin{equation*} \omega = \sqrt{\dfrac{g}{R_t}}\end{equation*}$

The maximum acceleration (deceleration) will be in $A$ ( $B$ ) and will be:

(9) $\begin{equation*} a_{max}= g\dfrac{x}{R_t}\end{equation*}$

The maximum acceleration is a fraction of $g$ proportional to the ratio between $x$ and $R_t$ . For example, for a trip of $2x = 1000 \,km$ (which, remember, would take only $42 \, min$ ) the maximum acceleration would be equal to about $8 \%$ of $g$ .

The maximum speed is given by

(10) $\begin{equation*} v_{max}= x\sqrt{\dfrac{g}{R_t}}\end{equation*}$

In this example ( $2x = 1000 \, km$ ), we have $v_{max} = 2230 \,km/h$

A fast and eco-friendly way to travel – if only there was an eco-friendly and efficient way to eliminate friction!

6. All bodies resting on a horizontal surface tend to oscillate harmonically with a period $T_a$

The oscillation of period $T_a$ would characterize any body free from friction and resting on a horizontal surface, even of small dimensions, such as a table in the laboratory or at home. However, I am not aware of this oscillation having ever been observed. It would be an excellent demonstration of the sphericity of the Earth.

7. The period of the astronomical pendulum is of the same order of magnitude as the sidereal day of the giant planets of the Solar System.

Let us imagine that there is an observer sitting at the pivot point of the astronomical pendulum. Ancients would say that this observer is attached to the fixed stars. If this observer were to measure the rotation time of a planet around its axis, he would measure an interval of time called a sidereal day.

The sidereal day is different from the usual day that a hypothetical inhabitant of that planet would measure as -say- the interval between two successive sunrises or sunsets, because the apparent motion of the Sun, which determines sunrises and sunsets, depends not only on the rotation of the planet around its axis, but also on the simultaneous rotation of the planet around the Sun during its orbit.

For example, a terrestrial sidereal day is about $4 \, min$ shorter than a solar day. To be precise, a sidereal day is $0.99726968$ times a solar day.

The ratio of the duration of terrestrial sidereal day to the period of the astronomical pendulum (assuming $R_t=6378000$ , $G = 6.6726 \, 10^{11}$ and $M=5.9720 \, 10^{24}$ , and using SI units, is $16.995$ , which can be approximated to $17$ with a tiny error of about $0.03\%$ .

If we look at the other rocky planets, we find that this ratio can take on very different values, ranging from $14.8$ for Mars to $4045.3$ for Venus.

However, the sidereal day of the gas planets (Jupiter, Saturn, Uranus and Neptune) is of the same order of magnitude as the period of the astronomical pendulum that would oscillate near their surfaces, the ratio being, respectively: $3.46$ , $2.65$ , $5.87$ , and $6.25$ , with mean $m = 4.56$ (let’s keep in mind the number $m$ , we will use it soon).

The same can be said for the major bodies in the asteroid belt, namely Ceres and Vesta, which orbit between Mars and Jupiter.

The following figure shows the ratio: sidereal day to period of the astronomical pendulum for the planets and some other major bodies in the Solar System.

For all bodies orbiting between Mars and Pluto, the length of the sidereal day is of the same order of magnitude as the period of the astronomical pendulum. Also, it is interesting to note that this ratio is always greater than $1$ .

This suggests that the rotation speed around the axis was favored by the tangential impact of smaller bodies that were rotating too slowly to become satellites.

For the sake of completeness, it should be noted that the great relative length of the sidereal days of bodies closest to the Sun is presumably to be attributed to tidal effects, which slowed down their rotation. The same could also be true for Pluto, assuming it formed closed to the Sun to be then pushed away by gravitational perturbations toward the periphery of the Solar System, but only after the tidal effects had time enough to increase the length of its sidereal day.

8. Length of day on Planet X

If there were indeed a planet X (https://science.nasa.gov/solar-system/planet-x/), which would be a gas giant with a mass between $5$ and $10$ times the mass of the Earth, and if it had a density close to the density of the outer gas planets (we will take here the average between the density of Uranus and that of Neptune) and finally if its sidereal day was $m$ times the period of its astronomical pendulum, with $m$ already defined above, then a day of planet X would last about $12.5$ Earth hours.

This is of course a bold and rather unjustified hypothesis.

But it is certainly also an intriguing one, because it creates an ideal connection between simple and familiar objects, such as a pendulum, and nothing less then the utterly mysterious planet X: nothing better to close this bizarre post than showing how far you can go by simply stretching a pendulum…

The period of small oscillations of the astronomical pendulum and related curiosities