The Theory of Gravitation

The bench starts with two pins 10 cm apart and a 30 cm loop of string around them. Pulled taut by the tip of a pencil, the loop forms a triangle whose side between the pins never changes, so the other two sides, the distances \(r_1\) and \(r_2\) from the pencil to each pin, always add up to 30 − 10 = 20 cm. The curve the pencil draws under that constraint is an ellipse, and the pins mark its two foci.

Half of that sum is the semi-major axis, \(a = 10\) cm, the distance from the centre to the far end of the curve. The distance \(c\) from each focus to the centre, measured in units of \(a\), is the eccentricity \(e = c/a\). It is 0 when the pins coincide and the pencil draws a circle, and it approaches 1 as the gap \(2c\) between them approaches \(2a\) and the ellipse flattens into a segment. At the top of the curve the pencil is the same distance \(a\) from both pins, and the right-angled triangle formed there gives the semi-minor axis, \(b = a\sqrt{1-e^2}\), or 8.66 cm with \(e = 0.5\).

Kepler reached the ellipse early in the seventeenth century, while trying to fit the orbit of Mars to Tycho Brahe's observations. What tends to surprise people most is that the Sun does not sit at the centre of the curve but at one of the foci, and that the other focus holds nothing at all. The pins-and-string construction is old and appears in Feynman too; its merit is that it turns a geometric definition into a gesture anyone can repeat on a table.

The Areas scenario places the Sun at the right-hand focus and splits each of the planet's turns into eight equal time intervals. Each interval leaves a coloured sector between the Sun and the arc travelled: near the Sun the sectors are short and fanned out, on the far side they are long and thin, and all of them measure 34.0 cm². This is Kepler's second law, by which the line joining the Sun to the planet covers the same area in any two intervals of the same length, and it forces the planet to move faster when it is close to the Sun.

The ratio between the speed at perihelion, the point nearest the Sun, and the speed at aphelion, the farthest, is \((1+e)/(1-e)\), or 3.00 with \(e = 0.5\). The Earth's orbit, with \(e = 0.0167\), has a semi-minor axis only 0.014% shorter than the major one, a difference that on the scale of the bench would be 14 µm, finer than the pencil line. The Sun, however, sits 1.7 mm off centre, and the Earth's speed varies by about 3% over the year. Mercury, at 0.2056, already has a speed ratio of 1.52, and Halley's comet, at 0.967, passes perihelion nearly 60 times faster than aphelion.

In Feynman: §7-2 Kepler’s laws ↗

The bench plots the eight planets on a graph with both axes on a logarithmic scale, the semi-major axis \(a\) in astronomical units and the period \(T\) in years. On such a scale a power law \(T \propto a^p\) becomes a straight line of slope \(p\), and the eight points, from Mercury, at 0.387 AU and 88 days, to Neptune, at 30 AU and 165 years, fall on a line of slope 1.500. Put another way, \(T^2/a^3\) is the same for all of them to within 0.2%, and that is Kepler's third law.

The red line shows what a force with a different dependence on distance would produce. Suppose the force per unit mass at a distance \(r\) from the Sun is \(k/r^n\) and the orbit is a circle. The centripetal acceleration \(v^2/r\) has to equal that force, so \(v^2 = k/r^{n-1}\), and the period \(T = 2\pi r/v\) grows as \(r^{(n+1)/2}\). With \(n = 3\), the bench's starting value, the line passes through the Earth but predicts 904 years for Neptune, more than five times the measured period.

Only one exponent puts the line on the planets: the slope \((n+1)/2 = 3/2\) requires \(n = 2\), and with it the error in Neptune's period drops to 0.06%. A force falling as \(1/r\) would give the outer planets turns that are too short, and one falling as \(1/r^3\), turns that are too long. This argument, with the orbits taken as circles, is probably the shortest route from Kepler's third law to the inverse-square force, and it is where Newton started; the proof that the same force produces exact ellipses, with the law holding for the semi-major axis, came later and takes a good deal more work. Feynman mentions the result without doing the sum, which here fits in two lines.

The calculation says nothing about the value of \(k\), which only sets the height of the line. Jupiter's moons follow the same law, with the same slope of 3/2, but on a line shifted upwards, because Jupiter's \(k\) is smaller and the same distance takes longer to cover. That shift measures the ratio of the masses of Jupiter and the Sun, close to one thousandth, and the secondary-school lesson on gravitation does that weighing with the moons themselves.

In Feynman: §7-4 Newton’s law of gravitation ↗

The bench puts a point on a circle of radius 1 m, going round once per second, which corresponds to a speed \(v = 2\pi\) m/s, or 6.28 m/s. The dashed line is the tangent to the circle at the starting point. After \(\Delta t = 0.05\) s, a twentieth of a turn, the point has moved 18° round the circle and sits 4.89 cm below the tangent, measured towards the centre. That gap \(x\) is what the bench uses to measure the acceleration.

Measuring the gap from the tangent, rather than from the starting point, is the choice inertia imposes. With no force at all, the point would carry on along the dashed line at the same speed, and the force shows up only in the difference between that line and the actual path. On the bench that difference always points to the centre of the circle, and the motion along the tangent looks after itself. The principle of inertia comes from Galileo; reading every departure from the straight line as the mark of a force is Newton's.

The geometry of the gap comes from Pythagoras' theorem. Calling \(S\) the distance the point advances along the tangent, the centre, the final position and the foot of the perpendicular form a right-angled triangle with legs \(S\) and \(R - x\) and hypotenuse \(R\), so that \(x(2R - x) = S^2\) exactly. Over a short interval \(x\) is much smaller than \(2R\), which leaves \(x \approx S^2/2R\), a gap that grows with the square of the advance. On the bench, \(S = 30.9\) cm and \(S^2/2R\) gives 4.77 cm, about 2% below the exact value.

A gap that grows with the square of time looks like a fall. Since \(S \approx v\,\Delta t\), we have \(x \approx v^2\Delta t^2/2R\), and comparing it with a fall from rest, \(x = \tfrac12 a\,\Delta t^2\), gives an acceleration towards the centre \(a = v^2/R\), 39.478 m/s² on the bench. With \(\Delta t = 0.05\) s the estimate \(2x/\Delta t^2\) comes to 39.155 m/s², 0.8% low; with 0.001 s the two agree to the five figures shown, because the error falls as \(\Delta t^2\).

The calculation assumes that the advance along the tangent and the fall towards the centre add up without interfering with each other. It is the same independence that, in the secondary-school lesson on projectiles, makes a ball thrown horizontally reach the ground together with another one simply dropped. The difference is that on the circle the direction of the fall turns with the point, and at each instant it starts afresh from a new tangent. The result is the familiar centripetal acceleration, obtained here only from how far the body strays from the straight line, and this is the calculation the next step applies to the Moon.

In Feynman: §7-3 Development of dynamics ↗

The bench releases a planet one unit of distance from the Sun, 15% faster than a circular orbit would need, and computes the motion in short steps, with a force that always points to the Sun and falls with the square of the distance. The orbit comes out as an ellipse of eccentricity 0.32, and every eighth of a turn the bench paints a sector between the Sun and the arc travelled. The rate at which the Sun–planet line covers area, measured at every step, varies by 0.000%, turn after turn.

The reason lies in a triangle. Over a short interval \(\Delta t\) the planet leaves the position \(\vec r\) and moves by \(\vec v\,\Delta t\), and the area covered is half of \(|\vec r \times \vec v|\,\Delta t\). Only the part of the displacement perpendicular to the radius counts in that product. A tug directed at the Sun adds to the velocity a piece parallel to \(\vec r\), which vanishes in the cross product, and the next triangle covers the same area; a component of the force perpendicular to the radius, on the other hand, makes that area grow or shrink.

The Switch on tangential push button adds that component, along the motion and equal to a fraction of the attraction. At 2%, the areal rate rises by about 11% per turn, the sectors swell and the orbit opens into a spiral. Read backwards, the calculation is Newton's argument: if the area covered grows in step with time, as Kepler measured, the force has no tangential component and points, at every instant, at the Sun. The law of areas says nothing about how that force varies with distance, and that part fell to step 2.

The Moon scenario puts the result of step 3 to work. The Moon travels a circle of \(3.844\cdot10^8\) m in 27.32 days, which gives it an acceleration towards the Earth of \(4\pi^2 r_M/T_M^2 = 2.72\cdot10^{-3}\) m/s². That distance is 60.3 Earth radii, and the 9.81 m/s² at the surface divided by \(60.3^2\) gives \(2.69\cdot10^{-3}\) m/s². The secondary-school lesson on gravitation makes this comparison with round numbers; here it can answer one more question.

The ratio between the two measured accelerations, the one at the surface and the Moon's, is 3,602, and the ratio between the distances to the Earth's centre is 60.34. A force falling as \(1/r^n\) would give \(60.34^n = 3{,}602\), that is, \(n = 1.997\). Because the exponent comes out of a logarithm, the 1% difference between the accelerations only touches the third decimal place, and the result is 2 with a margin of about three thousandths. The Moon test measures the exponent along a route that does not pass through the planets, with a different central body, and lands on the same 2 as step 2. With \(n = 1\) the Moon would have to fall at 0.163 m/s², 60 times more, and with \(n = 3\), at \(4.5\cdot10^{-5}\) m/s², 60 times less.

The 1.1% gap between the two bars probably has an explanation. The Earth does not stay put: it and the Moon turn about their common centre of mass, and the correct calculation replaces the Earth's mass with the sum of the two, 1.2% larger, since the Moon has 1/81 of the Earth's mass. What remains, a few tenths of a per cent, may come from effects such as the eccentricity of the orbit and the pull of the Sun.

In Feynman: §7-4 Newton’s law of gravitation ↗

The bench sets up two stars bound to each other by gravity, with 2 solar masses in total and a semi-major axis of 10 AU for the orbit of one relative to the other. Neither of them stays still in the middle. Both turn about a point between them, the centre of mass, marked with a cross, and each traces an ellipse with one focus at that point. The two ellipses have the same eccentricity and the same period but different sizes, and the stars are always on opposite sides.

The size of each ellipse depends on the mass. Since the centre of mass does not move, the heavier star has to stay closer to it, and the semi-major axes \(a_1\) and \(a_2\) of the two ellipses are in the inverse ratio of the masses. With star 2 holding 0.6 of the mass of star 1, star 1's ellipse comes out at 3.75 AU and star 2's at 6.25 AU. The two add up to the semi-major axis \(a = 10\) AU of the relative orbit, the one you would see looking at one star from the other.

The total mass comes from Kepler's third law. In step 2 the period of an orbit was \(T = 2\pi a^{3/2}/\sqrt{k}\), and in a double star \(k = G(M_1 + M_2)\), because the relative motion is that of a body pulled by the two masses combined. With \(a\) in AU, \(T\) in years and the masses in solar masses, the Earth going round the Sun becomes the yardstick, with \(a = T = M = 1\), and the constants drop out. On the bench, 10 AU and 22.4 years give \(10^3/22.4^2 \approx 2.0\) solar masses.

The Sirius A and B scenario brings the brightest star in the night sky to the bench. The pair completes a turn in 50.1 years, with \(a \approx 19.8\) AU, and the third law gives \(19.8^3/50.1^2 = 3.09\) solar masses. Sirius A's ellipse is about half the size of Sirius B's, so A takes some two thirds of the total, 2.06 solar masses, and B the remaining third, 1.03. The companion was inferred before it was seen. In 1844 Bessel noticed that Sirius A wobbled against the background stars, and in 1862 Alvan Graham Clark found it while testing a new lens.

Sirius B packs a mass similar to the Sun's into a body the size of the Earth; it is a white dwarf, what is left of a star that has used up its nuclear fuel. None of this enters the calculation, which uses only the period and the two ellipses. What matters here is that the law fitted to the planets, around a single star, also describes two stars going round each other. Since the period and the size of the two ellipses are all it takes, double stars are probably the most direct way of weighing a star. In the sky the orbit appears tilted, and Feynman shows, with four decades of measurements of Sirius, why the projection moves Sirius A off the apparent focus. The bench looks at the orbit face on.

In Feynman: §7-5 Universal gravitation ↗

The bench shows from above the Sun, the Earth at 1 AU and Jupiter at 5.2 AU, with the orbits taken as circles in the same plane, which is the model's simplification. At time zero the Earth lies between the Sun and Jupiter, in the position called opposition, and the two planets are 4.20 AU apart, the smallest possible distance. The Earth goes round once a year and Jupiter takes 11.86 years. A little over half a year later the Earth passes to the far side of the Sun, and the distance reaches 6.20 AU at conjunction.

Io, the innermost of Jupiter's four large moons, goes round in 1.769 days and on every turn enters the planet's shadow. Seen from the Earth, the eclipse is a sudden switching off, and the sequence of them works as a clock. With the interval measured near opposition one can draw up a table of the eclipses to come, but the table soon starts to fail. While the Earth moves away from Jupiter, each eclipse arrives up to about 14 s later than the previous one would lead you to expect; while it approaches, each arrives as many seconds early.

The bench's graph shows the accumulated effect. The light of the eclipse has to cross the distance between Jupiter and the Earth, and every extra AU costs 499 s of travel. The delay relative to the table is therefore the extra distance divided by \(c\). It grows to 16.6 min at conjunction, when the extra distance is the diameter of the Earth's orbit, 2 AU. It only returns to zero at the next opposition, after 1.09 years, because Jupiter has also moved in the meantime.

Ole Rømer, who worked at the Paris Observatory, put forward this explanation in 1676, with eclipses of Io followed over several years. By his estimate, light took about 22 minutes to cross the diameter of the Earth's orbit. Christiaan Huygens combined that figure with the size of the orbit as estimated at the time and arrived at a speed of around 220,000 km/s. The gap to today's 16.6 min probably comes from the difficulty of timing an eclipse with seventeenth-century telescopes and clocks. Near conjunction, moreover, Jupiter is lost in the Sun's glare and cannot be observed.

Rømer did not need gravitation to reach this reading. In 1676 Newton's Principia was still eleven years away, and what he had was his trust in Io's rhythm. Nor did the explanation convince at once: Cassini, who ran the observatory, did not accept it, and the finite speed of light only won wide acceptance after Bradley discovered the aberration of light, in 1729. Feynman looks at the episode in reverse, with gravitation already in hand. Io's orbit depends on Jupiter's pull and not on where the Earth happens to be, so a discrepancy like this one would have put the law in serious trouble unless something else accounted for it; here, that something else is light.

In Feynman: §7-5 Universal gravitation ↗

The bench sets up the torsion balance with Henry Cavendish's numbers: a 1.86 m rod hung by its middle from a thin wire, with a 0.73 kg lead ball at each end, and two 158 kg balls beside the small ones, with centres 0.23 m apart. Each pair attracts with \(GMm/r^2 = 1.5\cdot10^{-7}\) N, the weight of about 15 micrograms, roughly a quarter of a grain of table salt. The two forces pull the ends of the rod in opposite directions and add up to a torque \(GMmL/r^2\), which twists the wire until its resistance, \(\kappa\,\theta\), balances the pull.

With these numbers the equilibrium angle is \(9.6\cdot10^{-4}\) rad, less than a tenth of a degree, too small to read on the rod itself. A mirror fixed to the wire solves the problem: when it turns by \(\theta\), the reflected beam turns by \(2\theta\), and on a scale 5 m away the spot of light moves \(2\theta D = 9.6\) mm. Swapping the large balls to the other side reverses the torque and takes the spot across, a travel of 19.2 mm, so the measurement does not depend on knowing where the untwisted wire would sit.

Cavendish, in 1798, did not use a mirror. He shut the balance inside a room, to shield it from draughts and temperature differences, and followed the ends of the rod from outside, with telescopes aimed at fine scales lit by lamps. The mirror with the light beam came later, in nineteenth-century versions, and that is the arrangement the bench draws, with the rod's rotation not to scale.

What is still missing is the wire's constant \(\kappa\), which Cavendish did not need to measure separately. Left free, the rod oscillates about equilibrium with period \(T = 2\pi\sqrt{I/\kappa}\), where \(I = mL^2/2\) is the moment of inertia of the small balls, and in his measurements one oscillation took about 7 min. Taking \(\kappa\) from the period and carrying it into the balance of torques, the mass \(m\) cancels, and what remains is an expression for \(G\) made of two distances, an angle, a time and the mass of the large balls.

With \(G\) in hand, the mass of the Earth follows. The acceleration at the surface is \(g = GM_E/R_E^2\), and with \(g = 9.81\) m/s² and \(R_E = 6.37\cdot10^6\) m the mass comes out as \(M_E = 5.97\cdot10^{24}\) kg. Cavendish presented his result as the Earth's mean density, 5.48 times that of water, and the experiment is remembered as the one that gave the planet its mass; the \(G\) that follows from his numbers is within about 1% of today's value. The secondary-school lesson on gravitation works out the same attraction between two spheres, without the balance.

In Feynman: §7-6 Cavendish’s experiment ↗

The Weakness scenario places two electrons \(10^{-10}\) m apart, about the size of an atom, and compares the two forces between them. The electric repulsion is \(2.3\cdot10^{-8}\) N and the gravitational attraction \(5.5\cdot10^{-51}\) N, and the bars, on a logarithmic scale, end up more than 42 powers of 10 apart. Since both forces fall with the square of the distance, the ratio \(4.17\cdot10^{42}\) stays the same at any separation and depends only on the charges and masses. Between two protons, with the same charge and 1,836 times the mass, it drops to \(1.23\cdot10^{36}\), a number that is still enormous.

If gravity loses by so much, it is fair to ask why it is gravity that rules planets and stars. Charges come in two signs and cancel, and a large body has almost exactly as many protons as electrons; masses have a single sign and always add up. In the lead balls of step 7, one spare charge for every \(4\cdot10^{17}\) protons would already be enough to produce an electric force as large as the attraction the balance measures, and it is the near-perfect neutrality of matter that leaves gravity on its own at large scales.

Newton's law also says nothing about how the attraction crosses empty space. In a 1693 letter to Richard Bentley, Newton called absurd the idea of one body acting on another at a distance with nothing in between, and even so he did not venture a mechanism. Proposals have not been lacking since the eighteenth century, and none has survived the observations; Feynman discusses one of them in §7-7. The bench, like the law, only computes how strongly the masses attract, without saying how.

The Light scenario sends a ray past the Sun at a distance \(b\) from its centre. Treating light as a particle travelling at \(c\) under Newton's law, the calculation gives a deflection of 0.875″ for the grazing ray, close to what Johann von Soldner computed in 1801. Einstein's general relativity, from 1915, gives twice that, 1.75″, and in both theories the deflection falls as \(1/b\). Light bends because, in Einstein's theory, energy gravitates just as mass does, and the half that Newton misses comes from the curvature of space near the Sun.

The measurement needs a total eclipse, which darkens the sky around the Sun and reveals the stars near its edge. In 1919, expeditions organised by Frank Dyson and Arthur Eddington photographed an eclipse at Sobral, in the Brazilian state of Ceará, and on the island of Príncipe, and the measured deflections, with stated uncertainties of roughly 6 to 20%, favoured Einstein's value; the quality of those plates was debated for decades. Today, the positions of quasars observed with radio telescopes when the Sun passes close to them confirm the relativistic value to better than 0.1%.

What led Einstein to tamper with Newton's law was not light. In Newton's formula the force follows the current position of the masses, so shifting the Sun would alter the force on the Earth at that very instant, a signal faster than light, which relativity forbids. The correction is small in the Solar System, and besides light it shows up in the perihelion of Mercury, which advances 43″ per century more than the pull of the other planets explains.

In Feynman: §7-7 What is gravity? ↗ · §7-8 Gravity and relativity ↗