The moon test

By 1670 there were two separate bodies of exact knowledge about motion, one for the heavens and one for the ground, and nothing whatever connecting them.

Kepler's three laws, from the earlier lesson, describe how planets move and offer no cause. Galileo's law of fall describes how bodies drop and offers no cause either. Both were solid, both were quantitative, and each was silent about the other. Isaac Newton's achievement is to show that they are two consequences of one rule, and the argument that persuaded him is a single arithmetical comparison that this lesson works through in full.

Where the inverse square comes from

Start with the one thing everyone agreed on by 1670: Kepler's third law, that the square of a planet's period goes as the cube of its distance, $T^{2} = k r^{3}$ for some constant $k$ shared by all the planets.

Then take the result from the lesson on the mechanical philosophy. A body moving in a circle of radius $r$ at constant speed is continually changing direction, and Huygens had published in 1673 the size of the acceleration required to keep it there:

a = \frac{v^{2}}{r}

The speed of a body going once round a circle in time $T$ is $v = 2 π r / T$ , so

a = \frac{(2 π r / T)^{2}}{r} = \frac{4 π^{2} r}{T^{2}}

Example. Substitute Kepler's third law into that expression and see what the acceleration towards the sun depends on.

Putting $T^{2} = k r^{3}$ into the denominator:

a = \frac{4 π^{2} r}{k r^{3}} = \frac{4 π^{2}}{k} \cdot \frac{1}{r^{2}}

The $r$ in the numerator cancels two of the three in the denominator and an inverse square law falls out, with a constant that is the same for every planet because $k$ is. So if planets are held in their orbits by an acceleration directed at the sun, that acceleration must weaken as the square of the distance. Nothing in this derivation is difficult, which is why several people had it: Newton by about 1666, Robert Hooke by 1670, and Edmond Halley and Christopher Wren independently. In January 1684 the three Londoners discussed it over coffee and none of them could do the hard part, which is to show that an inverse square force produces an ellipse rather than merely being consistent with a circle. In August, Halley went to Cambridge and asked Newton, who replied that it would be an ellipse and that he had calculated it.

Now you. Suppose observation had given $T^{2} \propto r^{2}$ instead. What force law would follow, and what does that tell you about the status of the inverse square?

Answer

With $T^{2} = k r^{2}$ the same substitution gives $a = 4 π^{2} r / (k r^{2}) = (4 π^{2} / k) (1 / r)$ , an inverse first power law. In general $T^{2} \propto r^{n}$ gives an acceleration proportional to $r^{1 - n}$ . The inverse square is not a deduction from first principles, and it is not a guess: it is read off a measured relation, and a different measurement would have given a different exponent. That matters for how the achievement should be described. Newton did not invent the law of gravitation and then check it. The exponent came from Kepler's data, which came from Tycho's arcminute, and the derivation above turns one empirical regularity into another. What Newton added is the claim that the same rule reaches down to the ground.

Why the circle is the easy case

Everything above assumes a circular orbit, and no planetary orbit is circular. That restriction is not a detail, and it is the reason four capable men could all have the inverse square law and still have nothing.

The derivation runs in the wrong direction. It starts from an observed regularity across the planets, Kepler's third law, and extracts the exponent of a force. What is needed is the reverse: assume a force falling off as the inverse square of the distance from a fixed point, let a body move under it with whatever speed and direction it happens to have, and prove that the path traced out is a conic section with that point at a focus. Only then are Kepler's first two laws consequences of the force rather than facts sitting alongside it, and only then does the theory apply to a body whose distance is changing, which is every real planet and every comet.

That problem needs a mathematics of continuously changing quantities, and it is genuinely hard. Hooke wrote to Newton in 1679 with the physical idea, that orbital motion is straight-line motion continuously bent by a central attraction, and Hooke could not do the mathematics; he said as much, and later claimed the result anyway. Halley could not do it. Wren offered a prize of a book worth forty shillings to whoever could produce a proof within two months, and nobody collected it.

Newton could, and the nine-page tract he sent Halley in November 1684 is the proof. It also settles the converse, that an inverse square attraction admits ellipses, parabolas and hyperbolas depending on the speed at a given distance, which is what makes a comet on an enormous elongated orbit the same kind of object as a planet. Kepler's first two laws stop being descriptions of what planets happen to do and become theorems.

Comparing an apple with the moon

The bold step is to suppose that the acceleration holding the moon in orbit is the very same one that makes a stone fall in an orchard, weakened by distance according to the law just derived. That is testable with numbers available in the 1660s.

The moon's distance had been known reasonably well since antiquity, from the size of the earth's shadow in a lunar eclipse and from parallax, at about 60 earth radii. The modern figures are an earth radius of $6371$ km and a mean lunar distance of $384 400$ km, so

\frac{r}{R} = \frac{384 400}{6371} = 60.3

The moon's sidereal period is $27.3217$ days, which is $2.3606 \times 10^{6}$ seconds.

Example. Compute the moon's acceleration towards the earth, and compare it with $g$ divided by the square of $60.3$ .

The centripetal acceleration is

a = \frac{4 π^{2} r}{T^{2}} = \frac{4 \times 9.8696 \times 3.844 \times 10^{8}}{(2.3606 \times 10^{6})^{2}} = \frac{1.5177 \times 10^{10}}{5.5724 \times 10^{12}}

which is $2.723 \times 10^{- 3}$ metres per second squared. The prediction from an inverse square law anchored at the earth's surface is

\frac{g}{{60.3}^{2}} = \frac{9.81}{3636} = 2.698 \times 10^{- 3}

in the same units. The two agree to $1.1$ per cent. The moon is falling, continuously, at the rate a stone would fall if a stone could be carried out to sixty earth radii, and the two numbers were obtained from completely independent measurements: one from an orchard and a pendulum, the other from an eclipse and a calendar.

Now you. Newton's early attempt used the common estimate that a degree of latitude is 60 miles, giving an earth radius of $3438$ miles or $5530$ km. Redo the comparison with that figure and see what happens.

Answer

The moon's distance was known in earth radii, about 60 of them, so a wrong earth radius makes the moon's distance wrong in the same proportion: $60 \times 5530 = 3.32 \times 10^{5}$ km. The acceleration becomes $4 π^{2} \times 3.32 \times 10^{8} / (2.3606 \times 10^{6})^{2} = 2.35 \times 10^{- 3}$ , while $g / 3600 = 2.73 \times 10^{- 3}$ is unaffected, since it depends only on $g$ and the ratio of the distances. The comparison now fails by $13.7$ per cent, which is far too much to shrug off and far too little to look like a different law. Newton later said this was why he set the calculation aside, and that he took it up again after Jean Picard's survey of 1669 to 1670 gave a degree of latitude as $69.1$ miles. Historians treat that story with suspicion, since it comes from Newton himself decades later and serves his interest in an early priority date, and there was a much more serious obstacle in the way.

The obstacle that actually mattered

The comparison above quietly assumes something enormous. It treats the distance from the earth to the moon as the distance between two points, and the distance from the earth to the apple as the earth's radius. But the earth is not a point. It is a ball of matter, and every part of it is a different distance from the apple, with the near parts a few metres away and the far parts twelve thousand kilometres away, each pulling with a different strength and in a different direction.

For the moon, sixty radii out, treating the earth as a point is obviously a fair approximation. For an apple three metres above the ground it is not obviously anything, and until you know what the total pull of a whole sphere amounts to, the number $g$ cannot be compared with anything.

The result Newton needed is the shell theorem: a uniform spherical shell attracts an external body exactly as though its entire mass were concentrated at its centre, provided the attraction of each piece falls off as the inverse square. It is a special property of that exponent and it fails for others. He proved it in the spring of 1685, twenty years after the first version of the moon test, and it is Proposition LXXI of the Principia. Only with that theorem in hand does the apple's distance from the earth become the earth's radius, and only then does the calculation above mean anything.

There was a second obstacle, conceptual rather than mathematical. Newton's early work, following Huygens and Descartes, was framed in terms of a body's outward endeavour in a circle, balanced by something. The modern framing, in which a body would travel in a straight line and is continuously deflected inwards by a force, arrived clearly in a letter Hooke wrote him in November 1679, proposing exactly that decomposition. Newton found the suggestion useful and spent the next thirty years refusing to admit it, which is one reason the two men detested each other.

Predicting rather than checking

A comparison of two numbers is one thing; a prediction is stronger. Turn the relation round and use the surface value of $g$ to predict how long the moon should take to go round.

Setting the required acceleration equal to the inverse square law gives $4 π^{2} r / T^{2} = g R^{2} / r^{2}$ , so

T = 2 π \sqrt{\frac{r^{3}}{g R^{2}}}

Example. Evaluate that with $r = 3.844 \times 10^{8}$ m, $R = 6.371 \times 10^{6}$ m and $g = 9.81$ .

The numerator inside the root is $r^{3} = 5.680 \times 10^{25}$ and the denominator is $9.81 \times 4.059 \times 10^{13} = 3.982 \times 10^{14}$ , so the ratio is $1.4265 \times 10^{11}$ and its square root is $3.777 \times 10^{5}$ . Multiplying by $2 π$ gives $2.373 \times 10^{6}$ seconds, which is $27.47$ days. The observed sidereal period is $27.32$ days, so the prediction is high by $0.53$ per cent. One measurement made in a laboratory, plus two distances, predicts the length of the month.

Now you. Where does that remaining half a per cent come from?

Answer

Chiefly from the moon's own mass. The earth does not stand still while the moon goes round it; both circle their common centre of mass, and the correct relation involves the sum of the two masses rather than the earth's alone. The moon is $1 / 81.3$ of the earth's mass, and including it shortens the predicted period by a factor of $\sqrt{1 + 1 / 81.3} = 1.0061$ , giving $27.47 / 1.0061 = 27.30$ days against the observed $27.32$ . The remainder is accounted for by smaller effects: the measured $g$ at the surface is reduced by about $0.35$ per cent at the equator by the earth's rotation, the earth is not a perfect sphere, and the sun perturbs the lunar orbit substantially. The important point is that the residual is not noise. Each part of it is a physical effect that can be identified and calculated, and a century of celestial mechanics after Newton consists largely of doing exactly that, which is how a theory earns trust: not by fitting perfectly, but by having its misfits turn into further results.

What the test does and does not establish

Be careful about the size of the conclusion. The moon test establishes that one law of attraction towards the earth, weakening as the inverse square, fits both a falling stone and the moon's orbit. That is already a repudiation of the division that opened this subject, since a single rule now spans the sublunary and the celestial.

It does not establish that gravitation is universal. It says nothing about whether the moon pulls back on the earth, whether Jupiter pulls on Saturn, or whether two stones on a bench attract each other. Those claims need the third law of motion, that action and reaction are equal and opposite, and a great deal more work, and they bring consequences the moon test does not: the tides, the precession of the equinoxes, the perturbations of one planet by another, and the shape of the earth.

Nor does it explain anything. The whole apparatus says how strongly bodies are attracted, and nothing whatever about why or by what means, across a quarter of a million miles of empty space, with no contact of any kind. Everything the mechanical philosophy stands for forbids it. That is the bill for the Principia, and the next lesson is about what it delivered and what it cost.