A pendulum, no matter the gravitational strength (g) or the length of the string (l), finds its north star in the dimensionless oscillation period: $T/l/g$ is always $2π$. In understanding this statement, students seem to pass through similar stages as I did. First, I didn't even understand the claim. Second, I understood it only symbolically, as I would on hearing that my child was running a fever of 40 °C: “Yes, and?” Third and finally, I felt its force, as I would on hearing that the fever was 104 °F: “Oh, damn, that's dangerous!” I became amazed that I could swing my key chain from a string, measure the period, compute $T/l/g$, and get a number valid for pendula everywhere and of all lengths.

This lengthy process of understanding happened for a dimensionless constant, a zero-parameter universal or dimensionless function. Still harder to grasp is the next-simplest kind of universal function, a one-parameter dimensionless function. A famous example, featured in many treatments of dimensional analysis, is Taylor's analysis of the Trinity atomic-bomb test.1

A less warlike example hides among the motion of the planets. Take the orbital data in Table I that include perihelion ($rmin$) and aphelion ($rmax$) distances. The data could serve for a course that I've long wanted to teach: “Patterns in nature as seen in the CRC Handbook of Chemistry and Physics.” If the mask …2 … lifts, I might even teach such a course in person. Each lecture would begin with CRC data on one topic—perhaps the boiling points of long-chain hydrocarbons, the temperature versus height in the atmosphere, or the thermal conductivity of several materials—and explain the trends qualitatively and quantitatively, approximating as needed.

Table I.

Orbital data for the planets. V is the effective potential: the effective potential energy, $Ueff$, divided by the planet's mass. $rmin$ and $rmax$ are the perihelion and aphelion distances, respectively.

$rmin$ (m)$rmax$ (m)V ($m2/s2$)
Mercury $4.6001×1010$ $6.9818×1010$ $−1.1462×109$
Venus $1.0748×1011$ $1.0894×1011$ $−6.1339×108$
Earth $1.4710×1011$ $1.5210×1011$ $−4.4369×108$
Mars $2.0666×1011$ $2.4923×1011$ $−2.9119×108$
Jupiter $7.4067×1011$ $8.1601×1011$ $−8.5277×107$
Saturn $1.3498×1012$ $1.5036×1012$ $−4.6523×107$
Uranus $2.7350×1012$ $3.0063×1012$ $−2.3122×107$
Neptune $4.4598×1012$ $4.5370×1012$ $−1.4755×107$
$rmin$ (m)$rmax$ (m)V ($m2/s2$)
Mercury $4.6001×1010$ $6.9818×1010$ $−1.1462×109$
Venus $1.0748×1011$ $1.0894×1011$ $−6.1339×108$
Earth $1.4710×1011$ $1.5210×1011$ $−4.4369×108$
Mars $2.0666×1011$ $2.4923×1011$ $−2.9119×108$
Jupiter $7.4067×1011$ $8.1601×1011$ $−8.5277×107$
Saturn $1.3498×1012$ $1.5036×1012$ $−4.6523×107$
Uranus $2.7350×1012$ $3.0063×1012$ $−2.3122×107$
Neptune $4.4598×1012$ $4.5370×1012$ $−1.4755×107$

In these orbital data, V is the planet's effective potential: its effective potential energy, $Ueff$, divided by its mass, m. $Ueff$ itself is most evident in the total energy written in polar coordinates:

$E=12mṙ2+12mr2θ̇2−GMmr︸Ueff,$
(1)

where M is the solar mass. Then,

$V≡Ueffm=12r2θ̇2−GMr.$
(2)

If the orbital data are plotted with V on the vertical axis and distances ($rmin$ and $rmax$) on the horizontal axis, the result is a mess with no interesting pattern (Fig. 1). Sure, the outer planets are nearer the horizontal axis than are the inner planets, and all the points lie near a curve that might be $V∝−1/r$. But this pattern arises from the form of the gravitational potential energy, $−GMm/r$, so it's not news.

Fig. 1.

Orbital data from Table I. Each planet gets one point for $rmin$ and another for $rmax$. Both points lie on a horizontal line due to conservation of energy. For at the extremes of the orbit, $ṙ=0$. So, from Eq. (1), $E=Ueff$, meaning that $Ueff$, and thus V, cannot vary from one point to the other. (For the Earth and Venus, the inner planets with nearly circular orbits, the two dots almost blend because of the huge distance range on the r axis.)

Fig. 1.

Orbital data from Table I. Each planet gets one point for $rmin$ and another for $rmax$. Both points lie on a horizontal line due to conservation of energy. For at the extremes of the orbit, $ṙ=0$. So, from Eq. (1), $E=Ueff$, meaning that $Ueff$, and thus V, cannot vary from one point to the other. (For the Earth and Venus, the inner planets with nearly circular orbits, the two dots almost blend because of the huge distance range on the r axis.)

Close modal

But let's replot the data in dimensionless form. To rescale the horizontal axis, we need a length scale. The best scale is

$r0≡2rminrmaxrmin+rmax,$
(3)

the harmonic mean of $rmin$ and $rmax$. (We'll return to why we should form that particular length.) The rescaled horizontal axis is then $r¯≡r/r0$.

For rescaling the vertical axis, we need a specific energy (energy per mass). A quantity with these dimensions and based on the chosen length scale is $GM/r0$. Thus, the rescaled vertical axis will be $V¯≡V/(GM/r0)$.

The resulting dimensionless plot (Fig. 2) is cleaner than the preceding unscaled graph. The data now lie on the decent curve

$V¯=12r¯2−1r¯.$
(4)
Fig. 2.

Dimensionless version of Fig. 1, plotting $V¯≡V/(GM/r0)$ versus $r¯≡r/r0$. From the top downward, the planets are Mercury, Mars, Saturn, Jupiter, Earth, Uranus, Neptune, and Venus. All points lie on a decent curve with a simple expression, $V¯=0.5/r¯2−1/r¯$.

Fig. 2.

Dimensionless version of Fig. 1, plotting $V¯≡V/(GM/r0)$ versus $r¯≡r/r0$. From the top downward, the planets are Mercury, Mars, Saturn, Jupiter, Earth, Uranus, Neptune, and Venus. All points lie on a decent curve with a simple expression, $V¯=0.5/r¯2−1/r¯$.

Close modal

As promised, let's return to the choice of length scale: Why is the harmonic mean, r0, the correct length scale? Other natural length scales include the semi-major axis, a, and the semi-minor axis, b; see Fig. 3, taken with minor changes from my column on zeroth powers.3

Fig. 3.

The three means in a Kepler orbit. The semi-major axis (a) is the arithmetic mean (AM) of the perihelion ($rmin$) and the aphelion ($rmax$). The semi-minor axis (b) is their geometric mean (GM), and the semi-latus rectum (our r0, usually denoted l) is their harmonic mean (HM).

Fig. 3.

The three means in a Kepler orbit. The semi-major axis (a) is the arithmetic mean (AM) of the perihelion ($rmin$) and the aphelion ($rmax$). The semi-minor axis (b) is their geometric mean (GM), and the semi-latus rectum (our r0, usually denoted l) is their harmonic mean (HM).

Close modal

One reason not to choose the semi-major axis as the length scale is that the resulting plot is too regular. The total energy in a Newtonian-gravitational or Kepler orbit turns out to be

$E=−GMm2a.$
(5)

At the orbital extremes (where $ṙ=0$), Eq. (1) shows us that $E=Ueff$. Then,

$V≡Ueffm=Em=−GM2a.$
(6)

Thus, at the extremes, any planet has $V/(GM/a)=−1/2$, placing all the data on one horizontal line (with this choice of length scale).

Meanwhile, the semi-minor axis and indeed all length scales except r0 are disfavored for a physical reason. The orbital energy, Eq. (1), depends on r and $θ̇$. In using energy conservation to find the orbital shape, we would get a two-variable differential equation (from $dE/dt=0$), far more than twice as difficult as a single-variable differential equation. Fortunately, the energy's joker term, $mr2θ̇2/2$, which ropes in the second variable, can be rewritten using the other conserved quantity in a Kepler orbit, the angular momentum

$L≡mr2θ̇.$
(7)

Thus,

$12mr2θ̇2=L22mr2$
(8)

and voilà: The $θ̇$ has gone, making the effective potential a function now only of r (and ostensibly of L, but it's a constant).

$V=(L/m)22r2−GMr.$
(9)

This rewriting helps us because r0, unlike other length scales, is a function just of L (and physical constants). In a Kepler orbit, it turns out that

$r0=(L/m)2GM.$
(10)

If we wisely choose r0 as our length scale, the dimensionless plot in Fig. 2 becomes an analog computer that implicitly incorporates the conservation of angular momentum.

Let's put the plot and the curve Eq. (4) to a moderate test by using an orbital extreme case—Halley's comet.4 Here are its orbital data:

$rmin=8.7874×1010 m;rmax=5.2378×1012 m;V=−2.4926×107 m2 s−2.$
(11)

Do its two points lie on the dimensionless curve of Eq. (4)? Numerically,

$rminr0=0.508 388; rmaxr0=30.303 030.$
(12)

For $r¯=rmax/r0$, Eq. (4) predicts

$V¯=12·(30.303 0302)−130.303 030=−0.032 454$
(13)

and the $V¯$ calculated from Eq. (11) is almost identical, −0.032 456. (The reader is invited to confirm that the closest approach, where $r¯=0.508 388$, also works.) In Newtonian gravity, any planet orbiting around any sun is described by one universal function.

Having passed through the stages of understanding a zero-parameter universal function and having thought for a while about the orbital one-parameter function, I feel myself nearing the final, third stage of understanding and my amazement growing.

The author has no conflicts to disclose.

Sanjoy Mahajan is interested in the art of approximation and physics education and has taught varying subsets of physics, mathematics, electrical engineering, and mechanical engineering at MIT, the African Institute for Mathematical Sciences, and the University of Cambridge. He is the author of Street-Fighting Mathematics (MIT Press, 2010), The Art of Insight in Science and Engineering (MIT Press, 2014), and A Student's Guide to Newton's Laws of Motion (Cambridge University Press, 2020).

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2.
The editor censored the two words around the footnote to avoid my offending readers who cannot abide sarcasm about mask science.
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Despite widespread contrary claims in the philosophy of science, evidence available when you made your theory (“old evidence”) can help confirm a theory, and such reasoning is Bayesian. For the claim, see
Clark N.
Glymour
,
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(
Princeton U. P
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63
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For why the claim is incorrect, see
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Bayesians can learn from old data
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