Lesson plan · Expert · Adults · 45 minutes
Mercury's perihelion advance
Mercury's perihelion turns about 575 arcseconds a century: the planets account for 532 of them and general relativity for 43, which the lesson derives from the Schwarzschild orbit equation and computes from JPL constants.
The lesson
The class sees Mercury's eccentric orbit at true scale, splits its perihelion advance into the planets' Newtonian share and the relativistic remainder, derives the 43 arcseconds a century from the Schwarzschild orbit equation and computes it from JPL constants. It ends with what the viewer's own ephemeris does with Mercury's perihelion, and what it cannot show.
What they take away
Mercury's perihelion advances 575.31 arcseconds per century in an inertial frame; the planets give 532.30, the Sun's oblateness 0.03 and general relativity 42.98, the value 6πGM/(c²a(1 − e²)) gives per orbit times 415.2 orbits. The viewer turns Mercury's ellipse by one rate fitted by JPL, close to the real total, and computes none of the parts.
Step by step
The caption is what the class reads on screen; the talking points are in your notes drawer (N) when you present. Steps marked Pupils only appear only in the pupil lesson.
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01
An orbit that turns
Presented and for pupilsCaption
Mercury's orbit is an ellipse whose perihelion slowly turns. Against the stars it advances about 575 arcseconds a century, and the Newtonian pull of the other planets accounts for all but about 43 of them.
Talking points
- Will (2014): in 1859 Le Verrier announced that once the planets' perturbations and the precession of the equinoxes had been removed, an unexplained advance of Mercury's perihelion remained. "The modern value for this discrepancy is 43 arcseconds per century." A planet Vulcan, a ring of planetoids, a solar quadrupole moment and a change to the inverse-square law were all proposed, and none succeeded.
- The classic observed figure, from observations of 1765 to 1937, is 574.10 ± 0.41 arcsec per century, against 531.513 from Newtonian theory, a difference of 42.587 ± 0.5 (the table Sigismondi 2011 reproduces from Sciama 1972). The modern total, from ranging to MESSENGER in orbit about Mercury, is 575.3100 ± 0.0015 arcsec per Julian century, measured along Mercury's mean orbit plane (Park et al. 2017).
- Seen from Earth the advance looks far larger, because the equinox the longitudes are counted from precesses too: JPL gives the general precession in longitude as 5028.83 ± 0.04 arcsec per century. Every rate in this lesson is in a fixed frame, with that removed.
- The scene opens at 07:00 UTC on 5 May 2027, at a perihelion passage. JPL Horizons (DE441) has it at 07:01:50 TDB, about 07:00:41 UTC, at 46.0017 million km from the Sun; the viewer's ephemeris has it at 07:03 UTC, at 46.0007 million km. The view is at true scale, from above the ecliptic, out to Earth's orbit.
- NASA's fact sheet: perihelion 46.000 and aphelion 69.818 million km, semi-major axis 57.909 million km, eccentricity 0.2056, sidereal period 87.969 days. The Sun sits (69.818 − 46.000)/2 = 11.909 million km from the centre of the ellipse.
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02
The Newtonian budget
Presented and for pupilsCaption
Venus gives 277 arcseconds a century, Jupiter 154, the Earth and Moon 91 and Saturn 7. In all the planets give 532.30, about 92 per cent of the 575.31 total.
Talking points
- Park et al. (2017), Table 3, in arcsec per Julian century along Mercury's mean orbit plane: Venus 277.4176, Earth and Moon 90.8881, Jupiter 153.9899, Saturn 7.3227, Mars 2.4814, Uranus 0.1425, Neptune 0.0424, asteroids 0.0012, Mercury itself 0.0050, pairwise interaction terms (the largest Jupiter with Saturn, 0.0411), solar oblateness 0.0286 ± 0.0011, gravitoelectric (Schwarzschild-like) 42.9799 ± 0.0009, Lense-Thirring −0.0020 ± 0.0002. Total 575.3100 ± 0.0015.
- Taking away the relativistic and solar terms: 575.3100 − 42.9799 − 0.0286 + 0.0020 = 532.3035 arcsec per century of Newtonian perturbation. Park et al.: "Most (∼92%) of this rate is due to perturbations on Mercury's orbit by the other planets, primarily Venus, Jupiter, and Earth", and the gravitoelectric term is "about 7.5% of the total" (42.9799 / 575.31 = 7.47 per cent).
- Park et al. found each planet's share by integrating the ephemeris with that body's mass set to zero and differencing the rates. The perihelion does not move linearly: they fitted 14 periodic terms of 0.5 arcsec or more (the largest 7.24 arcsec) and took the rate over a 2000-year integration.
- The older budget that Sigismondi (2011) reproduces from Sciama (1972) totals 531.513 arcsec per century. Its Venus row is printed as 227.856, which does not add up to that total (the rows then sum to 481.513); 277.856 does.
Ask the class
Put these contributions to Mercury's perihelion advance in order, largest first (Park et al. 2017).
They start in this order: Jupiter, General relativity (gravitoelectric term), Venus, Saturn, Earth and Moon.
Answer
- Venus
- Jupiter
- Earth and Moon
- General relativity (gravitoelectric term)
- Saturn
WhyVenus 277.42, Jupiter 153.99, Earth and Moon 90.89, relativity 42.98, Saturn 7.32 arcsec per Julian century (Park et al. 2017, Table 3).
Sources 1 3Present from this step
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03
The relativistic term
Presented and for pupilsCaption
In the Schwarzschild field the orbit equation gains a term in u². Treated as a small perturbation it turns the ellipse by Δφ = 6πGM/(c²a(1 − e²)) every orbit.
Talking points
- With u = 1/r and h the angular momentum per unit mass, a test body in the Schwarzschild field obeys d²u/dφ² + u = GM/h² + (3GM/c²)u². Without the last term the solution is the Kepler ellipse u₀ = (GM/h²)(1 + e cos φ).
- Put u₀ into the small term. (3GM/c²)u₀² contains (3G³M³/(c²h⁴))·2e cos φ, which drives the oscillator at its own frequency, so its particular solution grows: (3G³M³/(c²h⁴)) e φ sin φ. The constant and cos 2φ parts only shift and ripple the orbit.
- To first order the Kepler term and the secular term combine into u ≈ (GM/h²)[1 + e cos((1 − δ)φ)] with δ = 3G²M²/(c²h²). Perihelion comes back when (1 − δ)φ = 2π, at φ ≈ 2π(1 + δ), so the perihelion moves on by 2πδ = 6πG²M²/(c²h²) each orbit.
- With h² = GMa(1 − e²), Δφ = 6πGM/(c²a(1 − e²)). Will (2014, eq. 64) gives the parametrised post-Newtonian form, 6πm/p × (2 + 2γ − β)/3, plus preferred-frame terms that are zero in general relativity and a term for the Sun's quadrupole moment, with p = a(1 − e²) and G = c = 1; general relativity has γ = β = 1, so the factor is 1.
- Constants: GM☉ = 1.32712440041279419 × 10²⁰ m³ s⁻² and c = 299,792,458 m/s (JPL Astrodynamic Parameters); a = 57.909 × 10⁶ km and a sidereal period of 87.969 days (NASA Mercury Fact Sheet); e = 0.20563593 (JPL's elements, which the viewer uses; the fact sheet's 0.20563069 changes nothing at four figures).
Ask the class
Compute Δφ = 6πGM/(c²a(1 − e²)) for one orbit of Mercury, in arcseconds. Use GM☉ = 1.32712440041 × 10²⁰ m³ s⁻², c = 299,792,458 m/s, a = 57.909 × 10⁶ km and e = 0.20563593.
Answer0.10352 arcsec. Answers from 0.10302 to 0.10402 are marked right.
Whyc² = 8.98755 × 10¹⁶ m² s⁻²; a(1 − e²) = 5.7909 × 10¹⁰ m × 0.957714 = 5.54603 × 10¹⁰ m; 6πGM = 2.50157 × 10²¹ m³ s⁻². Δφ = 2.50157 × 10²¹ / (8.98755 × 10¹⁶ × 5.54603 × 10¹⁰) = 5.0187 × 10⁻⁷ rad, and × 206,264.8 arcsec per radian = 0.10352 arcsec. Leaving out (1 − e²) gives 0.0991, outside the tolerance.
Sources 2 4 5 6Present from this step
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04
Per century
Pupils onlyCaption
A Julian century is 36,525 days. Turn the advance per orbit into arcseconds per century and compare it with the 43 left over by the planets.
Talking points
- JPL Astrodynamic Parameters: Julian century = 36,525 d. Mercury's sidereal period is 87.969 days (NASA), so it makes 36,525 / 87.969 = 415.20 orbits a century.
- Park et al. (2017) get 42.9799 ± 0.0009 arcsec per Julian century for the gravitoelectric term by isolating it in the integrated ephemeris; Will (2014, eq. 65) writes the rate as 42.98 arcsec per century times (2 + 2γ − β)/3, plus a solar quadrupole term.
- In distance: the perihelion point moves q × Δφ = 46.000 × 10⁶ km × 5.0187 × 10⁻⁷ = 23.1 km along the orbit per orbit, about 9,600 km a century.
Pupil question
Using 0.10352 arcsec per orbit, a Julian century of 36,525 days and Mercury's sidereal period of 87.969 days, what is the relativistic advance in arcseconds per century?
Answer42.98 arcsec per century. Answers from 42.88 to 43.08 are marked right.
Why36,525 / 87.969 = 415.20 orbits per century; 0.103518 × 415.20 = 42.98 arcsec per century. Park et al. (2017) give 42.9799 ± 0.0009 and Will (2014) 42.98. Without the (1 − e²) factor you would get 41.16, which is wrong.
Sources 1 2 4 6Open this step as a pupil
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05
The Sun's shape and spin
Presented and for pupilsCaption
Two more terms are tiny. The Sun's oblateness adds about 0.03 arcseconds a century and its spin drags Mercury's perihelion back by 0.002; MESSENGER ranging gives the Sun's J2 as 2.25 × 10⁻⁷.
Talking points
- Will (2014, eq. 65): the rate is 42.98 arcsec per century × [(2 + 2γ − β)/3 + 3 × 10⁻⁴ (J2/10⁻⁷)]. Will quotes helioseismology inversions giving J2 = (2.2 ± 0.1) × 10⁻⁷.
- Park et al. (2017) separate β and J2 using the periodic as well as the secular motion of the perihelion, with γ from the Cassini Shapiro delay, and find J2 = (2.25 ± 0.09) × 10⁻⁷. Their Table 3 gives the oblateness term as 0.0286 ± 0.0011 and the Lense-Thirring (frame-dragging) term as −0.0020 ± 0.0002 arcsec per century.
- History, from Sigismondi (2011): in 1967 Dicke and Goldenberg measured a solar oblateness of (5.0 ± 0.7) × 10⁻⁵, which implied a discrepancy of 8 per cent of Einstein's value; their 1974 analysis gave J2 = (2.5 ± 0.2) × 10⁻⁵, a correction of 3.0 ± 0.3 arcsec per century. The modern J2 is about a hundred times smaller.
- The view is the Sun at true scale; its oblateness, about 2 × 10⁻⁷ in J2, is far below anything a screen can show.
Ask the class
Will's eq. 65 gives the rate as 42.98 × [(2 + 2γ − β)/3 + 3 × 10⁻⁴ (J2/10⁻⁷)] arcsec per century. With γ = β = 1, how much does a solar J2 of 2.25 × 10⁻⁷ add, in arcseconds per century?
Answer0.029 arcsec per century. Answers from 0.027 to 0.031 are marked right.
Why42.98 × 3 × 10⁻⁴ × (2.25 × 10⁻⁷ / 10⁻⁷) = 42.98 × 3 × 10⁻⁴ × 2.25 = 0.0290 arcsec per century. Park et al.'s value from the integrated ephemeris, 0.0286 ± 0.0011, agrees within the tolerance. It is under a thousandth of the relativistic term.
Sources 1 2 3Present from this step
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06
What the viewer computes
Presented and for pupilsCaption
The viewer moves Mercury on JPL's Keplerian elements with straight-line rates. Its perihelion longitude gains 0.16047689° a century, 577.7 arcseconds: one fitted number, with nothing relativistic computed in it.
Talking points
- src/orbits/ephemeris.js holds JPL's Table 1 elements for 1800 to 2050 unchanged. For Mercury, ϖ = 77.45779628° + 0.16047689° T and Ω = 48.33076593° − 0.12534081° T, with T in Julian centuries from J2000, and it solves Kepler's equation for each date. It contains no perturbation theory and no relativistic term.
- JPL on these elements: they "are not intended to represent any sort of mean; they are simply the result of being adjusted for a best fit", and they "are not valid outside the given time-interval". JPL's nominal error for Mercury over 1800 to 2050 is 15 arcsec in heliocentric longitude. The viewer's time machine runs to 2200, so after 2050 it extrapolates them.
- The fitted rate sits within about 1 per cent of the measured total (next step), so whatever really moves Mercury's perihelion, the 43 arcsec of relativity included, is folded into it. The viewer has no way to show the parts separately or to switch one off.
- Over the viewer's 400 years, 1800 to 2200, its perihelion longitude moves 4 × 0.16047689° = 0.64°; relativity's share of that is 4 × 42.98 = 172 arcsec, under 3 arcmin.
Ask the class
What does the turning of Mercury's perihelion in this viewer represent?
- General relativity alone, computed from the Schwarzschild metric
- Newtonian perturbations from the planets, integrated numerically
- One linear rate fitted by JPL, with every cause folded in and none computed separatelyRight answer
- Nothing: the viewer keeps Mercury's perihelion fixed
Whyephemeris.js advances JPL's fitted elements linearly, ϖ̇ = 0.16047689° per century. The viewer integrates neither the planets' pull nor relativity, so it cannot separate them.
Sources 5Present from this step
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07
A broken angle
Pupils onlyCaption
The longitude of perihelion ϖ = Ω + ω is counted partly along the ecliptic and partly along the orbit. While the node moves, ϖ̇ differs from the rate at which the perihelion turns within the orbit plane.
Talking points
- Park et al. (2017) write the rate along the orbit plane in terms of i, the node rate Ω̇ and the rate ω̇ of the argument of perihelion: Ω̇ cos i + ω̇. With ω̇ = ϖ̇ − Ω̇ this is ϖ̇ − Ω̇(1 − cos i).
- The viewer's rates (JPL Table 1): ϖ̇ = 0.16047689° = 577.717 arcsec per century, Ω̇ = −0.12534081° = −451.227 arcsec per century, i = 7.00497902°.
- Checked directly with the viewer's own code: the perihelion unit vector built from the elements at two dates a century apart, projected on the mean orbit plane, turns 581.09 arcsec per century, and the orbit's pole moves 59 arcsec per century (Park et al. note that the plane moves by under 60 arcsec per century).
- So the viewer's ellipse turns 581.1 arcsec per century against MESSENGER's 575.31, a 1 per cent difference. JPL fitted these elements to positions over 1800 to 2050; Park et al. find periodic terms of up to 7.24 arcsec in the real perihelion, so a 250-year linear fit need not reproduce a 2000-year mean rate.
Pupil question
The viewer's rates are ϖ̇ = 577.717 and Ω̇ = −451.227 arcsec per century, with i = 7.00498°. Compute ϖ̇ − Ω̇(1 − cos i), the rate at which its perihelion turns in the orbit plane, in arcseconds per century.
Answer581.08 arcsec per century. Answers from 580.98 to 581.18 are marked right.
Why1 − cos 7.00498° = 0.007464; −Ω̇(1 − cos i) = 451.227 × 0.007464 = 3.368; 577.717 + 3.368 = 581.085 arcsec per century. ϖ̇ on its own (577.7) or the wrong sign (574.3) falls outside the tolerance.
Sources 1 5Open this step as a pupil
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08
Your turn: the ellipse at true scale
Pupils onlyCaption
Switch the orbits on and look at Mercury's ellipse at true scale, the Sun well off its centre. Then work out how far relativity alone moves the perihelion point along the orbit each time round.
Talking points
- NASA: perihelion 46.000 and aphelion 69.818 million km. The centre of the ellipse is 11.909 million km from the Sun, a fifth of the semi-major axis (ae = 0.2056 × 57.909 = 11.906 million km), which the true-scale orbit shows plainly beside Earth's nearly circular one.
- The arc is q × Δφ with q the perihelion distance and Δφ in radians: 5.0187 × 10⁻⁷ from the derivation step.
Pupil question
Δφ is 5.0187 × 10⁻⁷ rad per orbit. How far along the orbit does that move the perihelion point, at q = 46.000 × 10⁶ km?
Answer23.1 km. Answers from 22.8 to 23.4 are marked right.
Why46.000 × 10⁶ km × 5.0187 × 10⁻⁷ = 23.09 km per orbit, and with 415.2 orbits about 9,600 km a century.
Pupil task
Switch the orbits on.
HintUse the Orbits button under this task.
The viewer checks the task as the pupil works and says when it is done.
Sources 2 4 6Open this step as a pupil
Sources
Every fact in this lesson comes from these sources, and each step lists the ones it uses.
- Park, R. S., Folkner, W. M., Konopliv, A. S. et al. (2017), Precession of Mercury's Perihelion from Ranging to the MESSENGER Spacecraft, The Astronomical Journal 153, 121, doi:10.3847/1538-3881/aa5be2
- Will, C. M. (2014), The Confrontation between General Relativity and Experiment, Living Reviews in Relativity 17, 4, doi:10.12942/lrr-2014-4
- Sigismondi, C. (2011), Relativistic implications of solar astrometry, International Journal of Modern Physics: Conference Series 3, 464, doi:10.1142/S2010194511000985 (arXiv:1106.2202)
- Astrodynamic Parameters, JPL Solar System Dynamics
- Approximate Positions of the Planets (Table 1, elements and rates for 1800 to 2050), JPL Solar System Dynamics
- Mercury Fact Sheet, NASA NSSDCA
- Horizons System, JPL Solar System Dynamics
