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John Flamsteed

John Flamsteed, the King’s Astronomical Observator, first Astronomer Royal

A course in two strands: the life of a self-taught, often ill clergyman who founded the Royal Observatory at Greenwich and spent forty-three years measuring the sky with instruments he paid for himself, and the positional astronomy he made exact: meridian observing, the equation of time, parallax, the Moon’s inequalities, comet orbits, precession and the mapping of the celestial sphere.

Born 19 August 1646 (O.S.), Denby, Derbyshire Died 31 December 1719 (O.S.), Greenwich Catalogue 2,935 stars, published 1725 Level advanced undergraduate
Orion and its neighbours on the sinusoidal (Sanson–Flamsteed) projection, x = (α − 90°) cos δ, y = δ. Parallels stay straight and equally spaced, meridians bend, and areas are preserved. Right ascension increases to the left, as on the sky.
Lesson 00

How to use this course

John Flamsteed is remembered mostly as the man whose observations Isaac Newton and Edmond Halley published against his will. That story is true, and Part III tells it. But Flamsteed was also the first professional astronomer in England, the founder of the Royal Observatory at Greenwich, and the maker of the first great star catalogue made with telescopic sights. Newton’s lunar theory and his treatment of comets rested in part on Flamsteed’s numbers. This course follows the life and then the astronomy, which is less a single theory than a set of exact techniques.

Prerequisites for Part II
  • Trigonometry, including small-angle approximations and the addition formulas.
  • Single-variable calculus and first-order Taylor expansions.
  • Kepler’s laws and Newtonian gravitation, at the level of the courses on Kepler and Newton.
  • Basic spherical astronomy: right ascension, declination, the ecliptic. Lesson 04 reviews what is needed.

Each technical lesson opens with a paragraph for the general reader, then gives precise statements in modern notation, derivations and exercises with folded solutions. England used the Julian calendar throughout Flamsteed’s life, ten days behind the Gregorian calendar until 1700 and eleven days after. The course gives the English (Old Style) dates, as Flamsteed himself wrote them.

Reference

Timeline

1646Born 19 August at Denby, near Derby, only son of Stephen Flamsteed, a businessman, and his first wife Mary Spadman.
1662Leaves Derby School in May, his health broken by a rheumatic illness that began at about 14. Observes a partial solar eclipse on 12 September.
1662–69Teaches himself astronomy at home, against his father’s wishes.
1670Enters his name at Jesus College, Cambridge. In London meets Henry Oldenburg and Sir Jonas Moore, who becomes his patron.
1672Measures the diurnal parallax of Mars with a micrometer in October.
1673His tract on the equation of time and his lunar theory appear in print.
1674MA by royal warrant from Cambridge; hears Newton’s Lucasian lectures.
16754 March: appointed the King’s Astronomical Observator at £100 a year. Ordained. 10 August: lays the foundation stone of the Royal Observatory at Greenwich.
1676Moves into the Observatory in July. Elected to the Royal Society about this time (1676 or 1677; sources differ).
1680–81Argues that the comets of November and December 1680 were one body.
1684Rector of Burstow, Surrey.
1688–89Abraham Sharp becomes his assistant; a new mural arc is in use from 1689.
1690December: records Uranus as a star, later numbered 34 Tauri.
1692Marries Margaret Cooke.
1694–95Newton visits Greenwich; Flamsteed supplies over 150 lunar positions for Newton’s lunar theory.
1704Prince George of Denmark agrees to pay for printing his observations.
170912 December: Queen Anne appoints the Royal Society’s President and Council as Visitors of the Observatory.
1712An unauthorized Historia coelestis, edited by Halley, is printed in 400 copies.
1716Obtains about 300 unsold copies and burns most of them.
1719Dies 31 December at Greenwich, aged 73; buried at Burstow.
1725Historia coelestis Britannica, in three volumes, completed by Margaret Flamsteed, Joseph Crosthwait and Abraham Sharp.
1729Atlas coelestis.
Part I · Formation
Lesson 01

Denby and Derby, 1646–1670

John Flamsteed was born on 19 August 1646 at Denby, a village a few miles north of Derby. His father, Stephen Flamsteed, was a prosperous businessman in Derby. His mother, Mary Spadman, died when he was a child, and he was the only son of the marriage. He attended the free school in Derby, which prepared boys for the universities.

At about fourteen he fell ill with a chronic rheumatic complaint that stayed with him for the rest of his life. His father decided against university, and Flamsteed left school in May 1662. He later travelled to Ireland in search of a cure. Confined at home, he taught himself astronomy from whatever books he could borrow. On 12 September 1662 he observed his first partial eclipse of the Sun. His father disapproved of these studies, and Flamsteed came to suspect that health was not his father’s only reason for keeping him from Cambridge.

Horrocks and the Derby years

In 1665 he wrote a set of Mathematical Essays on the construction and use of an astronomer’s quadrant, with tables for the latitude of Derby. In a local gentleman’s library he found the work of Jeremiah Horrocks, the young astronomer who had died in 1641 after observing the transit of Venus of 1639. Horrocks’s tables and his model of the Moon impressed Flamsteed deeply, as they had Newton, and they shaped his own work for decades (Lesson 07). He computed the solar eclipses of 1666 and 1668 accurately in advance. His letters on astronomical calculations soon brought him to the notice of Henry Oldenburg, secretary of the Royal Society, and drew the Derbyshire invalid into the London scientific world.

Lesson 02

London, Moore and Cambridge, 1670–1675

In 1670 Flamsteed visited London. There he met Oldenburg and Sir Jonas Moore, Surveyor-General of the Ordnance, a mathematician and an energetic patron of science, who remained his protector until Moore’s death in 1679. The same year Flamsteed entered his name at Jesus College, Cambridge, though he never took up full residence.

The early 1670s were his most productive period as a theorist. In October 1672, when Mars came close to the Earth, he used the micrometer to measure its diurnal parallax and concluded that the Sun was far more distant than most astronomers believed (Lesson 06). In 1673 his tract on the inequality of solar days, the first correct account of the equation of time, was printed with the posthumous works of Horrocks, edited by John Wallis (Lesson 05). He also corrected and extended Horrocks’s lunar theory.

In 1674 he spent about two months at Cambridge and heard the Lucasian lectures of Isaac Newton. At Moore’s urging, King Charles II issued a warrant, and the university granted him the degree of MA without the usual residence. He intended to become a clergyman and was ordained deacon in 1675. He made a barometer and a thermometer for the King and for the Duke of York, an introduction that mattered a few months later.

Lesson 03

Founding Greenwich, 1675–1676

In February 1675 Flamsteed was in London when a French proposal reached the court: a method of finding longitude at sea from the position of the Moon among the stars. A royal commission that included Christopher Wren and Robert Hooke examined it, and Flamsteed advised. His verdict was that the method was sound in principle and useless in practice, because neither the star positions nor the tables of the Moon were good enough. Britain needed a national observatory to supply both.

Charles II agreed. By warrant of 4 March 1675 he appointed Flamsteed his astronomical observator, charged with rectifying the tables of the motions of the heavens and the places of the fixed stars, so as to find the longitude at sea. The salary was £100 a year, from which taxes were deducted, and it provided no instruments. A second warrant in June 1675 ordered an observatory to be built in Greenwich Park, on the foundations of an old castle on the hill. Wren designed the building, probably with Hooke’s help. Flamsteed laid the foundation stone on 10 August 1675. The work cost £520, partly from recycled materials, and Flamsteed moved in during July 1676.

Equipment

Moore paid for two clocks by Thomas Tompion, with 13-foot pendulums beating two seconds, installed in the Octagon Room; they kept time to a few seconds a day. His first major instrument was a seven-foot sextant for measuring angles between stars. Much of the equipment Flamsteed paid for himself, and he regarded those instruments as his own property. Halley, then a student at Oxford, helped with observations in the first years and became a friend. The friendship did not last.

Part II · The astronomy
Lesson 04

Meridian astronomy: clock and arc

For the general reader. The sky turns once a day. If you watch where and when each star crosses a fixed north–south line, a good clock gives you one coordinate of the star and a graduated arc gives you the other. Flamsteed turned Greenwich into a machine for doing exactly this, star after star, night after night, for forty years.

Goals
  • Define the equatorial coordinates and sidereal time.
  • Derive right ascension and declination from meridian transits.
  • Find the latitude of the observatory without knowing any star’s position in advance.

Coordinates and time

A star’s position on the celestial sphere is given by its right ascension \(\alpha\), measured eastward along the celestial equator from the vernal equinox, and its declination \(\delta\), measured north or south of the equator. The sphere turns westward once per sidereal day, about \(23^{\mathrm h}56^{\mathrm m}4^{\mathrm s}\) of mean solar time. The local sidereal time \(\theta\) is the hour angle of the equinox; the hour angle of a star is \(H=\theta-\alpha\), measured westward from the meridian.

Meridian relations

Let the observer have latitude \(\varphi\). A star crosses the meridian above the pole (upper culmination) when \(H=0\), that is, when \(\theta=\alpha\). If it then lies south of the zenith at zenith distance \(z\), its declination is \(\delta=\varphi-z\). A circumpolar star with upper and lower culmination altitudes \(h_1\) and \(h_2\) (corrected for refraction) satisfies

\[ \varphi=\tfrac12(h_1+h_2),\qquad 90^\circ-\delta=\tfrac12(h_1-h_2). \]

Derivation. The pole stands at altitude \(\varphi\). A star at polar distance \(p=90^\circ-\delta\) circles it, reaching altitudes \(\varphi+p\) above the pole and \(\varphi-p\) below. Adding and subtracting gives both formulas. Neither requires a previously known star, which is why the method is self-starting.

Flamsteed’s practice

With a clock that keeps sidereal time, the interval between the transits of two stars equals the difference of their right ascensions. Right ascension thus becomes a measurement of time, and declination a measurement of angle on the meridian. Flamsteed’s first great instrument, the seven-foot sextant, measured angular distances between pairs of stars, and positions had to be reduced by spherical trigonometry. From 1689 he used a mural arc fixed to a wall in the plane of the meridian, with which he and Abraham Sharp read zenith distances at the moment of transit. He also checked his clocks against the stars and found that the Earth turns at a constant rate, the foundation of every timekeeping method that followed.

Exercise 4.1At Greenwich (\(\varphi=51^\circ28.6'\)) a star crosses the meridian south of the zenith at zenith distance \(35^\circ10'\), and the sidereal clock reads \(5^{\mathrm h}14^{\mathrm m}32^{\mathrm s}\). Find \(\alpha\) and \(\delta\), ignoring refraction.

Show solution
\(\alpha=\theta=5^{\mathrm h}14^{\mathrm m}32^{\mathrm s}\). \(\delta=\varphi-z=51^\circ28.6'-35^\circ10.0'=+16^\circ18.6'\).

Exercise 4.2A circumpolar star culminates at altitudes \(53^\circ18'\) and \(49^\circ39'\) (after refraction). Find the latitude and the star’s declination.

Show solution
\(\varphi=\tfrac12(53^\circ18'+49^\circ39')=51^\circ28.5'\). The polar distance is \(\tfrac12(53^\circ18'-49^\circ39')=1^\circ49.5'\), so \(\delta=88^\circ10.5'\). This star circles very close to the pole, like Polaris in Flamsteed’s day.

Exercise 4.3A clock keeps mean solar time. By how much does a star’s transit time on this clock advance each day, and how much in a month of 30 days?

Show solution
The sidereal day is shorter than the mean solar day by about \(3^{\mathrm m}56^{\mathrm s}\), so each star transits that much earlier each day, about \(1^{\mathrm h}58^{\mathrm m}\) earlier after 30 days. In a year the gains add to a full day: there is one more sidereal day than solar days.
Lesson 05

The equation of time

For the general reader. A sundial and a good clock disagree, by up to about a quarter of an hour, in a pattern that repeats every year. Before pendulum clocks, few cared. Once the pendulum clock of Christiaan Huygens made timekeeping reliable, the disagreement mattered. Flamsteed gave the first correct explanation: two separate effects, the Earth’s elliptical orbit and the tilt of its axis, combine to produce it.

Goals
  • Define apparent and mean solar time.
  • Derive the equation of time to first order in the eccentricity and the obliquity.
  • Locate its zeros and extremes.

Two suns

Apparent solar time is the hour angle of the real Sun. Mean solar time is the hour angle of a fictitious mean sun that moves uniformly along the equator, with right ascension equal to the Sun’s mean longitude \(L\). The equation of time is \(E=\) apparent time \(-\) mean time. Since hour angle is \(\theta-\alpha\),

\[ E=(\theta-\alpha_\odot)-(\theta-L)=L-\alpha_\odot . \]

So \(E\) measures how far the true Sun’s right ascension lags or leads the uniform one. There are two reasons it does. The Sun moves along the ecliptic at a variable rate, because the Earth’s orbit is an ellipse (Kepler’s second law). And even uniform motion along the tilted ecliptic does not project to uniform motion in right ascension.

Theorem (first-order equation of time)

Let \(e\) be the eccentricity of the Earth’s orbit, \(\varepsilon\) the obliquity of the ecliptic, \(M\) the Sun’s mean anomaly and \(\lambda\) its true longitude, and put \(y=\tan^2(\varepsilon/2)\). To first order in \(e\) and \(y\),

\[ E \approx -2e\sin M + y\sin 2\lambda \quad\text{(radians)}. \]

With \(e=0.0167\), \(\varepsilon=23.44^\circ\) and \(1\ \text{rad}=229.2\) minutes of time, the two amplitudes are about \(7.7\) and \(9.9\) minutes.

Derivation. The equation of the centre gives \(\lambda=L+2e\sin M+O(e^2)\). For the reduction to the equator, \(\tan\alpha=\cos\varepsilon\tan\lambda\). Writing \(\cos\varepsilon=(1-y)/(1+y)\) one finds the exact series

\[ \alpha=\lambda-y\sin2\lambda+\tfrac12y^2\sin4\lambda-\cdots. \]

Then \(E=L-\alpha\approx L-\lambda+y\sin2\lambda=-2e\sin M+y\sin2\lambda\). The first term has a period of one year, the second of half a year, and their sum has four zeros: near mid-April, mid-June, the beginning of September and late December. The largest values are about \(+16\) minutes in early November and \(-14\) minutes in mid-February.

Earlier tables, such as those of Thomas Streete, had treated the problem incorrectly and were wrong by minutes. Flamsteed’s analysis, published in 1673, combined the two causes properly. At Greenwich he then used the Tompion clocks and observations of stars to show that the sidereal day is constant. This justified mean time as a uniform measure. It also mattered for the Moon: the Moon’s small annual inequality (Lesson 07) can only be separated from errors in the equation of time if the equation is right.

Exercise 5.1Show that \(\tan\alpha=\cos\varepsilon\tan\lambda\) implies \(\alpha=\lambda-y\sin2\lambda+O(y^2)\), with \(y=\tan^2(\varepsilon/2)\).

Show solution
Write \(\tan\alpha=\frac{1-y}{1+y}\tan\lambda\). Then \(\tan(\lambda-\alpha)=\frac{\tan\lambda-\tan\alpha}{1+\tan\lambda\tan\alpha}\). The numerator is \(\tan\lambda\cdot\frac{2y}{1+y}\); the denominator is \(1+\frac{1-y}{1+y}\tan^2\lambda\). Multiply through by \((1+y)\cos^2\lambda\): \(\tan(\lambda-\alpha)=\frac{2y\sin\lambda\cos\lambda}{1+y\cos2\lambda}=\frac{y\sin2\lambda}{1+y\cos2\lambda}\). To first order, \(\lambda-\alpha=y\sin2\lambda\).

Exercise 5.2Estimate \(E\) on 3 November, when \(\lambda\approx221^\circ\) and \(M\approx-60^\circ\). Is a sundial fast or slow?

Show solution
\(-2e\sin M=-0.0334\sin(-60^\circ)=+0.0289\) rad \(=6.6\) min. \(y\sin2\lambda=0.0430\sin442^\circ=0.0430\sin82^\circ=0.0426\) rad \(=9.8\) min. So \(E\approx+16.4\) min. The dial is about a quarter of an hour ahead of the clock.

Exercise 5.3What would the equation of time be for a planet with a circular orbit and zero obliquity? With a circular orbit but obliquity \(\varepsilon\)?

Show solution
In the first case the true and mean suns coincide: \(E\equiv0\). In the second, only the obliquity term remains, \(E\approx y\sin2\lambda\), with four equal zeros at the equinoxes and solstices and a half-year period.
Lesson 06

Diurnal parallax and the scale of the solar system

For the general reader. Kepler’s laws give the shape of the solar system but not its size: every distance is known in units of the Earth–Sun distance, whose value in miles was unknown. Measuring the distance to any one planet fixes them all. In 1672 Mars came unusually close, and astronomers in Paris and in England tried to measure its shift against the stars as seen from different points on the turning Earth.

Goals
  • Define horizontal parallax and relate it to distance.
  • Derive the diurnal shift in right ascension seen from one station.
  • Convert a planetary parallax into a solar parallax.

The horizontal parallax of a body at distance \(d\) is the angle \(\pi\) subtended by the Earth’s equatorial radius \(R\): \(\sin\pi=R/d\). Seen from the surface rather than the centre, the body is displaced away from the zenith by about \(\pi\sin z\).

Diurnal parallax in right ascension

For an observer at geocentric latitude \(\varphi'\) and distance \(\rho R\) from the Earth’s centre, a body at hour angle \(H\) and declination \(\delta\) with small horizontal parallax \(\pi\) has topocentric right ascension

\[ \alpha'\approx\alpha-\frac{\rho\,\pi\cos\varphi'\sin H}{\cos\delta}. \]

Derivation. In the equatorial frame, the observer sits at \(\rho R(\cos\varphi'\cos\theta,\cos\varphi'\sin\theta,\sin\varphi')\) and the body at \(d(\cos\delta\cos\alpha,\cos\delta\sin\alpha,\sin\delta)\). Subtract and take the component along the eastward unit vector \((-\sin\alpha,\cos\alpha,0)\). The observer’s contribution is \(-\rho R\cos\varphi'\sin(\theta-\alpha)=-\rho R\cos\varphi'\sin H\). Dividing by \(d\cos\delta\) converts this length into an angle in right ascension.

A single observer exploits the Earth’s rotation. Measure the planet’s offset from a nearby star when it rises in the east (\(H\approx-90^\circ\)) and again when it sets in the west (\(H\approx+90^\circ\)). The star has no measurable parallax, so after removing the planet’s own motion, the offset changes by \(2\rho\pi\cos\varphi'/\cos\delta\) in right ascension. This is the method Flamsteed used with his micrometer in 1672, while Giovanni Domenico Cassini in Paris compared observations made there with Jean Richer’s at Cayenne. Both concluded that the Sun’s parallax was no more than about ten seconds of arc, against the modern value of \(8.794''\). Flamsteed wrote that the Sun was more than ten times farther away than commonly supposed.

Scaling by Kepler’s third law

If a planet is at distance \(\Delta\) astronomical units, its parallax is \(\pi_{\text{planet}}=\pi_\odot/\Delta\). Hence \(\pi_\odot=\Delta\,\pi_{\text{planet}}\), and every distance in the solar system follows.

Exercise 6.1Take \(\pi_\odot=8.79''\) and let Mars be at \(0.38\) AU near declination \(0^\circ\). What total east–west swing would an observer at Greenwich (\(\varphi'\approx51.3^\circ\), \(\rho\approx1\)) see between rising and setting?

Show solution
\(\pi_{\text{Mars}}=8.79''/0.38=23.1''\). The swing is \(2\times23.1''\times\cos51.3^\circ\approx2\times23.1''\times0.625\approx28.9''\). Mars itself is about \(25''\) across at such an opposition, so the effect is comparable to the planet’s disc, and Mars meanwhile moves among the stars by many arcminutes a day. The measurement was at the limit of what a 1672 micrometer could do.

Exercise 6.2An astronomer finds that the solar parallax is at most \(10''\). What lower bound does this give for the Earth–Sun distance in Earth radii, and in kilometres (\(R=6378\) km)?

Show solution
\(d=R/\sin\pi\ge R/\sin10''\approx206\,265R/10\approx20\,600R\), about \(1.3\times10^8\) km. The true value is about \(23\,500R\), or \(1.496\times10^8\) km. Older estimates, going back to Ptolemy, had put the Sun near \(1200R\).
Lesson 07

The Moon’s inequalities

For the general reader. The Moon is the hardest body in the sky to predict. Its orbit is an ellipse, but the Sun constantly distorts that ellipse, stretching and turning it. A good lunar table was the key to finding longitude at sea, and the founding purpose of Greenwich. Flamsteed refined Jeremiah Horrocks’s model of the Moon and supplied the observations against which Newton tested his gravitational lunar theory.

Goals
  • Name the main periodic terms in the Moon’s longitude and their sizes.
  • Show that Horrocks’s variable ellipse reproduces the evection.
  • Understand why the annual equation depends on correct timekeeping.

Let \(L'\) be the Moon’s mean longitude, \(M'\) its mean anomaly, \(M\) the Sun’s mean anomaly, \(D\) the mean elongation of the Moon from the Sun, and \(F\) the Moon’s mean distance from its ascending node.

Main terms of the Moon’s longitude (modern values) \[ \lambda_{\text{Moon}}\approx L'+6.289^\circ\sin M'+1.274^\circ\sin(2D-M')+0.658^\circ\sin2D-0.186^\circ\sin M-0.114^\circ\sin2F+\cdots \]

The terms are, in order: the equation of the centre (Kepler ellipse), the evection (known to Ptolemy), the variation (found by Tycho Brahe), the annual equation, and the reduction to the ecliptic.

Horrocks’s variable ellipse

Horrocks represented the evection by letting the ellipse itself change. Its eccentricity oscillates and its line of apsides swings back and forth, both governed by the angle between the apse line and the Sun. Flamsteed published a corrected version of this model in 1673 and later tables based on it; Newton took it up in the Principia and tried to derive it from gravitation.

Proposition

Let the equation of the centre be \(2e\sin(L'-\varpi)\). Suppose \(e=e_0+\eta\cos\psi\) and \(\varpi=\varpi_0-(\eta/e_0)\sin\psi\), where \(\psi=2(\varpi_0-\lambda_\odot)\) and \(\eta\) is small. Then to first order in \(\eta\),

\[ 2e\sin(L'-\varpi)\approx2e_0\sin M'+2\eta\sin(2D-M'),\qquad M'=L'-\varpi_0,\ D=L'-\lambda_\odot. \]

Proof. Expand: \(2(e_0+\eta\cos\psi)\sin\big(M'+(\eta/e_0)\sin\psi\big)\approx2e_0\sin M'+2\eta\cos\psi\sin M'+2\eta\sin\psi\cos M'=2e_0\sin M'+2\eta\sin(M'+\psi)\). Finally \(M'+\psi=L'+\varpi_0-2\lambda_\odot=2D-M'\).

Matching \(2\eta=1.274^\circ=0.0222\) rad gives \(\eta\approx0.011\). With \(e_0=0.0549\), the eccentricity ranges from about \(0.044\) to \(0.066\), close to Horrocks’s values.

The annual equation and Newton

The annual equation, about \(11'\), arises because the Sun’s disturbing force is stronger when the Earth is nearer the Sun, in January. Tycho and Kepler had suspected it. Flamsteed measured it from Greenwich observations, which required an accurate equation of time: a ten-minute error in the clock correction moves the fast Moon by about \(5'\). In 1694 Newton visited Greenwich, and in the first half of 1695 Flamsteed sent him more than 150 lunar positions. Newton could not make gravitation reproduce every inequality, grew impatient, and blamed the observer. The relationship never recovered.

Exercise 7.1The Moon moves about \(13.18^\circ\) per day. How far does it move in 10 minutes of time? Why does this make the lunar method of longitude so demanding?

Show solution
\(13.18^\circ\times10/1440\approx0.0915^\circ\approx5.5'\). Conversely, an error of \(1'\) in the Moon’s predicted place corresponds to about 2 minutes of time, which is \(0.5^\circ\) of longitude, about 30 nautical miles at the equator.

Exercise 7.2Find the period of the evection term \(\sin(2D-M')\), given a synodic month of 29.531 days and an anomalistic month of 27.555 days.

Show solution
The frequency of \(2D-M'\) is \(2/29.531-1/27.555=0.067726-0.036291=0.031435\) per day, so the period is about \(31.81\) days.
Lesson 08

The comet of 1680 and parabolic orbits

For the general reader. In November 1680 a comet was seen in the morning sky, heading toward the Sun. In December a brilliant comet with an enormous tail appeared in the evening sky, moving away from it. Most astronomers saw two comets. Flamsteed argued they were one body that had turned near the Sun. He was right about that, though wrong about how it turned, and the episode helped lead Newton to treat comets as bodies obeying gravity.

Goals
  • Write the parabolic orbit in polar form.
  • Derive Barker’s equation for time of flight.
  • Estimate how fast and how close the 1680 comet passed the Sun.

The argument of 1681

Flamsteed proposed in 1681 that the comet had been attracted toward the Sun and then repelled, turning in front of it under a kind of magnetic action. Newton, in letters passed through an intermediary, objected that the Sun could not repel a body in that way and that the comet would have to go round behind the Sun; at first he also doubted that there was only one comet. By the time of the Principia Newton had accepted the single comet and fitted its path, using Flamsteed’s observations among others, with a parabola having the Sun at its focus (see the course on Newton).

Theorem (parabolic motion; Barker’s equation)

A body on a parabolic orbit about the Sun with perihelion distance \(q\) has \(r=q\sec^2(\nu/2)\), where \(\nu\) is the true anomaly. With \(\mu=GM_\odot\), the time since perihelion satisfies

\[ t-T=\sqrt{\frac{2q^3}{\mu}}\left(\tan\frac{\nu}{2}+\frac13\tan^3\frac{\nu}{2}\right). \]

In units of AU and days, \(\sqrt{\mu}=k=0.017\,202\,1\), and the speed is \(v=\sqrt{2\mu/r}\approx42.1/\sqrt{r_{\text{AU}}}\) km/s.

Derivation. A conic with focus at the Sun is \(r=p/(1+\cos\nu)\); for a parabola \(p=2q\), and \(1+\cos\nu=2\cos^2(\nu/2)\) gives \(r=q\sec^2(\nu/2)\). Kepler’s second law in Newtonian form reads \(r^2\dot\nu=h=\sqrt{\mu p}=\sqrt{2\mu q}\). So

\[ dt=\frac{r^2}{h}\,d\nu=\frac{q^2}{\sqrt{2\mu q}}\sec^4\frac{\nu}{2}\,d\nu . \]

Put \(s=\tan(\nu/2)\), so \(ds=\tfrac12\sec^2(\nu/2)\,d\nu\) and \(\sec^4(\nu/2)\,d\nu=2(1+s^2)\,ds\). Integrating from \(0\) gives \(t-T=\frac{2q^2}{\sqrt{2\mu q}}(s+s^3/3)=\sqrt{2q^3/\mu}\,(s+s^3/3)\).

Exercise 8.1The comet of 1680 had a perihelion distance of about \(0.0062\) AU. How long did it take to travel from perihelion out to 1 AU? What was its speed at perihelion?

Show solution
\(s=\tan(\nu/2)=\sqrt{r/q-1}=\sqrt{1/0.0062-1}\approx12.66\). Then \(s+s^3/3\approx12.66+676.4=689.1\), and \(\sqrt{2q^3}/k=\sqrt{2\times2.38\times10^{-7}}/0.0172=6.90\times10^{-4}/0.0172\approx0.0401\) days. So \(t-T\approx27.6\) days. Perihelion speed: \(42.1/\sqrt{0.0062}\approx535\) km/s. The comet passed only about a third of a solar radius above the Sun’s surface and swung from \(\nu=-90^\circ\) to \(+90^\circ\) in under three hours, so observers saw it only before and after.

Exercise 8.2Show that a body moving on a parabola has, at every point, exactly the escape speed from the Sun at that distance. Why could a long ellipse not be told from a parabola by observations near the Sun?

Show solution
The energy \(\tfrac12v^2-\mu/r\) of a conic is \(-\mu/(2a)\); a parabola is the limit \(a\to\infty\), so the energy is \(0\) and \(v=\sqrt{2\mu/r}\), the escape speed. For an ellipse of semi-major axis \(a\), \(v^2=\mu(2/r-1/a)\); near perihelion \(r\ll a\), so the speed and the path differ from the parabolic ones only by terms of order \(r/a\). This is why Halley needed a periodic return (see the course on Halley) to prove a comet was bound.
Lesson 09

The catalogue: precession, parallax and aberration

For the general reader. A star catalogue is only useful if every position refers to the same date, because the whole coordinate grid slowly slides. Flamsteed had to correct thousands of positions for this drift. He also searched for the tiny annual shift that would prove the Earth moves round the Sun. He found a shift in the Pole Star, but it was not the one he was looking for.

Goals
  • Apply the standard formulas for annual precession in right ascension and declination.
  • Distinguish the annual signatures of stellar parallax and of aberration.
  • Understand why Flamsteed’s Pole Star result could not be parallax.

Precession

The Earth’s axis precesses around the pole of the ecliptic, carrying the equinox westward along the ecliptic by about \(50.3''\) per year. Flamsteed adopted a rate of one degree in 72 years, which is \(50''\) per year, and rejected the old theory of trepidation, an oscillation of the equinoxes that had survived since the Middle Ages.

Annual precession in equatorial coordinates

To first order, over \(t\) years,

\[ \Delta\alpha=(m+n\sin\alpha\tan\delta)\,t,\qquad \Delta\delta=(n\cos\alpha)\,t, \]

with \(m\approx3.075^{\mathrm s}=46.1''\) and \(n\approx1.336^{\mathrm s}=20.04''\) per year. Here \(m=p\cos\varepsilon\) and \(n=p\sin\varepsilon\) for general precession \(p\) and obliquity \(\varepsilon\) (neglecting the small planetary term).

Parallax or aberration?

If the Earth orbits the Sun, a star should trace a small annual ellipse of semi-axis equal to its parallax, with its displacement always directed toward the Sun. Flamsteed reported in the late 1690s that the declination of Polaris varied over the year by an amount he attributed to parallax. The size was roughly right for what we now call aberration, but the timing was wrong for parallax.

Proposition

For a star at ecliptic longitude \(\lambda_*\) and latitude \(\beta\), with the Sun at longitude \(\lambda_\odot\), the displacements due to parallax \(\varpi\) and aberration \(\kappa\) are, in ecliptic coordinates,

\[ \begin{aligned} \text{parallax:}\quad &\Delta\beta=-\varpi\sin\beta\cos(\lambda_\odot-\lambda_*), \\ \text{aberration:}\quad &\Delta\beta=-\kappa\sin\beta\sin(\lambda_\odot-\lambda_*). \end{aligned} \]

The two have the same shape and are a quarter of a year out of phase. Here \(\kappa=v_\oplus/c\approx20.5''\).

Idea. Parallax displaces the star along the direction from Earth to Sun, which is \(-\hat r_\oplus\); aberration displaces it along the Earth’s velocity \(\hat v_\oplus\). On a circular orbit \(\hat v_\oplus\) is \(\hat r_\oplus\) turned through \(90^\circ\), so the second pattern lags the first by three months. James Bradley, Halley’s successor as Astronomer Royal, recognized aberration in 1728 and showed that the Polaris-type variations fitted it. The real stellar parallaxes, smaller than an arcsecond, were not measured until 1838.

Exercise 9.1Compute \(\kappa\) from \(v_\oplus=29.78\) km/s and \(c=299\,792\) km/s.

Show solution
\(\kappa=29.78/299\,792=9.934\times10^{-5}\) rad \(=9.934\times10^{-5}\times206\,265''\approx20.49''\).

Exercise 9.2Aldebaran has \(\alpha\approx69^\circ\), \(\delta\approx+16.5^\circ\). Estimate its change in right ascension and declination due to precession between 1690 and 1725.

Show solution
\(\sin\alpha\approx0.934\), \(\tan\delta\approx0.296\), so the annual change in \(\alpha\) is \(46.1''+20.04''\times0.934\times0.296\approx51.6''\), about \(3.44^{\mathrm s}\). Over 35 years: \(\approx30.1'\), or \(2.0^{\mathrm m}\) of time. In declination: \(20.04''\times\cos69^\circ\approx7.2''\) per year, about \(4.2'\) in 35 years. Both are far larger than Flamsteed’s observing errors, so every observation had to be reduced to the catalogue epoch.

Exercise 9.3A star at the north ecliptic pole is observed through a year. Describe its apparent path under parallax alone and under aberration alone. When is each displacement toward the vernal equinox direction?

Show solution
Both make circles (radii \(\varpi\) and \(\kappa\)). Parallax points toward the Sun, which lies in the direction of the vernal equinox (longitude \(0^\circ\)) around 20 March. Aberration points along the Earth’s velocity, toward longitude \(\lambda_\odot-90^\circ\); this equals \(0^\circ\) when \(\lambda_\odot=90^\circ\), around 21 June. The two circles are traced in the same sense but three months apart.
Lesson 10

Mapping the sky: the sinusoidal projection

For the general reader. A sphere cannot be flattened without distortion. A mapmaker must choose what to keep: angles, distances along some lines, or areas. The star charts of Flamsteed’s posthumous Atlas coelestis used a projection that keeps areas and keeps each parallel of declination straight and true to length. It is still called the Sanson–Flamsteed projection.

Goals
  • Define the sinusoidal projection and prove that it preserves area.
  • Compute its angular distortion.

The projection is older than Flamsteed. It appears on a world map by Jean Cossin of Dieppe dated 1570 and was used by Nicolas Sanson for maps of the continents. Flamsteed applied it to the sky. The MacTutor biography also credits him with a conical projection.

Definition and theorem

The sinusoidal projection with central meridian \(\lambda_0\) sends latitude \(\varphi\) and longitude \(\lambda\) on a unit sphere to

\[ x=(\lambda-\lambda_0)\cos\varphi,\qquad y=\varphi. \]

It is equal-area: every region of the sphere has the same area as its image. Scale is exact along the central meridian and along every parallel.

Proof. The area element on the unit sphere is \(\cos\varphi\,d\lambda\,d\varphi\). The Jacobian of the map is

\[ \det\begin{pmatrix}\partial x/\partial\lambda & \partial x/\partial\varphi\\ \partial y/\partial\lambda & \partial y/\partial\varphi\end{pmatrix}=\det\begin{pmatrix}\cos\varphi & -(\lambda-\lambda_0)\sin\varphi\\ 0 & 1\end{pmatrix}=\cos\varphi, \]

so \(dx\,dy=\cos\varphi\,d\lambda\,d\varphi\). A parallel at latitude \(\varphi\) has true length \(\cos\varphi\,\Delta\lambda\), and its image is a horizontal segment of the same length. Along \(\lambda=\lambda_0\), \(x=0\) and \(y=\varphi\), so distances are exact.

Exercise 10.1At a point with \(\lambda-\lambda_0=u\) and latitude \(\varphi\), find the angle between the images of the meridian and the parallel. Where is the projection conformal?

Show solution
The parallel’s image has tangent \((1,0)\). The meridian’s image has tangent \((\partial x/\partial\varphi,\partial y/\partial\varphi)=(-u\sin\varphi,1)\). The angle \(\theta\) between them satisfies \(\cot\theta=-u\sin\varphi\). It is \(90^\circ\) only when \(u\sin\varphi=0\): on the central meridian and on the equator. Away from them, shapes shear, badly near the corners of the map.

Exercise 10.2Show that the whole sphere maps to a region bounded by the curves \(x=\pm\pi\cos y\), and verify that its area is \(4\pi\).

Show solution
With \(|\lambda-\lambda_0|\le\pi\), the boundary is \(x=\pm\pi\cos\varphi=\pm\pi\cos y\) for \(|y|\le\pi/2\). The area is \(\int_{-\pi/2}^{\pi/2}2\pi\cos y\,dy=2\pi\cdot2=4\pi\), the area of the unit sphere.
Part III · Life and legacy
Lesson 11

Greenwich years and the great quarrel, 1676–1719

For forty-three years Flamsteed observed at Greenwich with little help and less money. He took private pupils to supplement his salary, and from 1684 held the living of Burstow in Surrey, where he became rector. In 1681 he contributed a Doctrine of the Sphere to Moore’s New Systeme of the Mathematicks, applying Kepler’s first two laws to the motions of the Sun and Moon. He served on the Council of the Royal Society. In 1688 Abraham Sharp, a gifted Yorkshire instrument-maker and calculator, became his assistant, and the mural arc they completed in 1689 transformed the accuracy of the work. In 1692 he married Margaret Cooke, then 22; he was 46. She copied, wrote his letters when his hand shook, and sometimes observed with him.

Halley and Newton

Flamsteed’s friendship with Halley turned to hostility in the 1680s. He thought Halley careless and irreligious, and in 1691 he opposed Halley’s candidacy for the Savilian chair of astronomy at Oxford. With Newton the break was over the Moon. After the exchanges of 1694–95 (Lesson 07), Newton wanted Flamsteed’s observations, and above all his star catalogue, for his lunar theory and his astronomy. Flamsteed refused to publish until the work was complete and checked. He also felt that Newton had not properly acknowledged his help, for example his report that Jupiter’s satellites obey Kepler’s third law, which Newton used in Book III of the Principia.

In 1704, with Newton now President of the Royal Society, Prince George of Denmark, the Queen’s husband, agreed to pay for printing Flamsteed’s observations, and a committee of referees took charge. Printing proceeded slowly and angrily, and stopped when the Prince died in 1708. On 12 December 1709 Queen Anne made the President and Council of the Royal Society Visitors of the Observatory, with power to demand its annual results and inspect its instruments. Flamsteed resisted. In 1712, using the incomplete catalogue Flamsteed had deposited under seal, Halley edited and the referees printed 400 copies of a Historia coelestis in a form Flamsteed denounced as corrupt (see also the courses on Newton and Halley).

After George I came to the throne, Flamsteed obtained the unsold copies in 1716, about 300 of them. He kept the sextant observations he had approved, set aside a few copies, and burned the rest of Halley’s edition as a sacrifice to heavenly truth, as he put it. He spent his last years, with Sharp and Joseph Crosthwait, preparing his own edition. He died at Greenwich on 31 December 1719, aged 73, and was buried in the chancel of St Bartholomew’s Church, Burstow.

DiscussionFlamsteed held that observations should not be released until complete and checked; Newton held that public money and the needs of theory required them sooner. Using Lessons 07 and 09, argue each side. How do modern observatories and data archives resolve the same tension?

Lesson 12

The Historia, the Atlas and Flamsteed’s legacy

Margaret Flamsteed saw his work through the press. The three-volume Historia coelestis Britannica appeared in 1725. Its third volume contained the British Catalogue of 2,935 stars, the first major catalogue made with telescopic sights and far more precise than Tycho Brahe’s naked-eye catalogue, whose best positions were good to about an arcminute. The Atlas coelestis followed in 1729, again with Crosthwait and Sharp doing the technical work. Margaret died in 1730.

She also removed from Greenwich the instruments her husband had bought, as was her right, and Halley, his successor as Astronomer Royal, found the Observatory nearly empty. Flamsteed’s papers were later returned; most of the instruments were lost.

Legacy

  • Flamsteed numbers. The 1712 edition numbered stars within each constellation in order of right ascension. Joseph Jérôme de Lalande adopted the scheme in a French edition of the catalogue in 1783, and designations such as 61 Cygni are still in use for about 2,554 stars.
  • Uranus and a supernova. In December 1690 he recorded Uranus as a star, later catalogued as 34 Tauri, ninety years before William Herschel recognized it as a planet. A star he recorded in Cassiopeia in 1680 has been proposed as an early sighting of the supernova whose remnant is Cassiopeia A, though the identification is disputed.
  • Greenwich. The Observatory he founded became the reference point for longitude and time. His house is still called Flamsteed House.
  • Method. His insistence on systematic, repeated, instrumentally controlled observation set the pattern for national observatories, and his lunar observations underpinned Newton’s lunar theory and later lunar tables.

Final projectChoose one star in Flamsteed’s catalogue with a Flamsteed number. Find its 1690 position as he would have measured it, apply precession (Lesson 09) to bring it to the year 2000, and compare with a modern catalogue. Explain the remaining difference in terms of proper motion, aberration, nutation and observing error.

Reference

Connections

People linked to John Flamsteed across this collection. A name without a link has no page here yet. See John Flamsteed in the network of all 337 lives →

Influenced by

Influenced

Colleagues and collaborators

Rivals

Family

  • Margaret Flamsteed

Also mentioned in this life

Fields

On CoursesHub

Reference

Sources and further reading

Biography

  1. J. J. O’Connor and E. F. Robertson, John Flamsteed, MacTutor History of Mathematics.
  2. Flamsteed, John, Biographical Encyclopedia of Astronomers (via MacTutor).
  3. John Flamsteed and Margaret Flamsteed, Wikipedia.
  4. Royal Observatory, Greenwich, Wikipedia.
  5. Flamsteed, John, 1646–1719, astronomer, AIM25 archive description.
  6. F. Willmoth (ed.), Flamsteed’s Stars: New Perspectives on the Life and Work of the First Astronomer Royal, Boydell, 1997.
  7. E. G. Forbes, L. Murdin, F. Willmoth (eds.), The Correspondence of John Flamsteed, 3 vols, Institute of Physics, 1995–2002.
  8. D. Howse, Greenwich Observatory, vol. 3: The Buildings and Instruments, Taylor & Francis, 1975.

Astronomy

  1. J. Flamsteed, Historia coelestis Britannica, 3 vols, London, 1725.
  2. J. Flamsteed, Atlas coelestis, London, 1729.
  3. J. Horrocks, Opera posthuma, ed. J. Wallis, London, 1673 (with Flamsteed’s tract on the equation of time).
  4. Flamsteed designation, Wikipedia.
  5. Sinusoidal projection, Wikipedia.
  6. Equation of time, Wikipedia.
  7. J. Meeus, Astronomical Algorithms, 2nd ed., Willmann-Bell, 1998 (precession, parallax, lunar series).
  8. A (not so) brief history of lunar distances, arXiv:2007.14504.