[{"data":1,"prerenderedAt":815},["ShallowReactive",2],{"layer:tides:extend":3},{"layer":4,"contentHash":790,"dependencyHashes":791,"approval":809,"releaseId":814},{"schemaVersion":5,"conceptId":6,"locale":7,"depth":8,"revision":5,"title":9,"subtitle":10,"summary":11,"objectives":12,"estimatedMinutes":17,"plate":18,"blocks":39,"sourceIds":785,"reviewStatus":786,"authoring":787},1,"tides","en","extend","Extend: deep time, deep space, and open questions","Tidal friction across hundreds of millions of years, tides on other worlds, and what is still unknown","Follow tidal friction from a subtle offset in Earth's bulge to a shorter Cretaceous day, a measurably receding Moon, tidal heating on Io, Europa and Enceladus, and a set of open questions and careers built on this one idea.",[13,14,15,16],"Explain, using angular momentum, why tidal friction slows Earth's spin and pushes the Moon outward at the same time.","Use fossil growth-ridge counts and today's laser-ranging rate to estimate ancient and future day lengths, while explaining the uncertainty in each method.","Describe tidal heating on Io, Europa and Enceladus as the same mechanism as Earth's ocean tides, acting at a different scale.","Identify at least two open questions in tidal science and two careers built on it.",45,{"title":19,"rows":20},"Lesson plate",[21,24,27,30,33,36],{"label":22,"value":23},"Depth","Extend",{"label":25,"value":26},"Reading time","≈ 45 minutes",{"label":28,"value":29},"Prior knowledge","Deepen: constituents, resonance; Gravity: orbits",{"label":31,"value":32},"Chapters","10",{"label":34,"value":35},"Labs","Tide lab, gravity drop, match, sort",{"label":37,"value":38},"Scope","Deep time, the Solar System, careers, open questions",[40,44,69,75,81,84,89,94,105,110,113,118,145,150,153,168,173,176,191,213,218,222,227,230,242,256,285,297,302,307,310,314,318,323,326,338,361,385,391,396,401,404,417,422,425,430,440,454,463,467,470,475,479,482,506,542,547,552,575,607,610,756,770],{"id":41,"type":42,"markdown":43},"e-intro","prose","Every earlier layer treated a day as 24 hours and the Moon's distance as fixed. Neither is quite true, and the reason is the tide itself: friction between the swirling ocean and the solid, spinning Earth is very slowly braking our planet's spin and pushing the Moon further away, a few centimetres at a time, every single year.\n\nThis layer follows that idea to the edges of what is known: fossil evidence of ancient, shorter days; laser beams bounced off mirrors astronauts left on the Moon; tidal heating that melts moons hundreds of millions of kilometres from here; and a few genuinely open questions nobody has finished answering.",{"id":45,"type":46,"tone":47,"items":48},"e-spec-glance","spec","amber",[49,53,57,61,65],{"label":50,"big":51,"value":52},"Moon's recession rate","≈ 3.8 cm\u002Fyr","Measured today by laser ranging.",{"label":54,"big":55,"value":56},"Cretaceous day (~70 Mya)","≈ 23.6 h","From 372 daily growth ridges per year in fossil shells.",{"label":58,"big":59,"value":60},"Ediacaran day (~620 Mya)","≈ 21.9 h","From tidal rhythmite layer counts.",{"label":62,"big":63,"value":64},"Io–Europa–Ganymede periods","1 : 2 : 4","The Laplace resonance, exact since at least 1743.",{"label":66,"big":67,"value":68},"Moon's own rotation","= 27.32 days","Locked to match its orbit long ago.",{"id":70,"type":71,"variant":72,"title":73,"markdown":74},"e-how-to","callout","observation","How to use this lesson","This is the least linear layer yet — pick a project, a puzzle, or an open question that grabs you and go deep on it. The two evidence-based chapters (fossils, laser ranging) are worth reading first, since the projects and puzzles lean on both.",{"id":76,"type":77,"title":78,"eyebrow":79,"navLabel":80},"e-ch1","chapter","The tide as a brake and a boost","Chapter 01","1 Brake and boost",{"id":82,"type":42,"markdown":83},"e-brake-boost","Understand's two-bulge picture was drawn as if the near bulge sat exactly under the Moon. In reality it does not, quite — and that small mismatch is the whole engine of this layer.\n\nEarth spins once every 24 hours; the Moon takes about a month to orbit. Because Earth spins so much faster than the Moon orbits, friction between the moving tidal bulge and the solid planet **drags the near bulge slightly ahead** of the Earth–Moon line, in the direction Earth is spinning. That small offset has two consequences at once, because gravity always pulls both ways:\n\n- The dragged-ahead bulge's extra mass pulls **back** on Earth's spin, very slightly slowing its rotation: **the brake**.\n- Earth's pull on that same offset bulge pulls the Moon **forward**, along its orbit, very slightly speeding it up: **the boost**.\n\nA satellite given more orbital speed does not spiral inward — it moves to a **higher** orbit instead (you will meet why in Gravity's orbit lab). So the 'boost' does not speed the Moon around faster in the everyday sense; it very gradually pushes the Moon into a wider, slower orbit. Earth loses spin; the Moon gains distance. Nothing is destroyed — it is all **angular momentum**, moving from one part of the system to the other.",{"id":85,"type":71,"variant":86,"title":87,"markdown":88},"e-def-momentum","definition","Angular momentum, in one line","**Angular momentum** is a measure of how much 'spinning or orbiting motion' a system has, and for an isolated system it cannot be created or destroyed — only moved from one part of the system to another. The Earth–Moon system's total angular momentum barely changes; tidal friction just slowly transfers it from Earth's spin to the Moon's orbit.",{"id":90,"type":71,"variant":91,"title":92,"markdown":93},"e-misc-friction","misconception","“Tidal friction is water rubbing on the sea floor”","That does happen and does dissipate energy as heat, but it is a side effect, not the main mechanism moving angular momentum around. The **gravitational pull between the offset bulge and the Moon** is what actually transfers angular momentum from Earth's spin to the Moon's orbit. Friction against the sea floor (especially in shallow seas, where currents are strongest) is what causes the bulge to lag behind, rather than sitting exactly under the Moon as the simplified picture in earlier layers assumed.",{"id":95,"type":96,"component":97,"componentVersion":5,"config":98,"objective":103,"textAlternative":104},"e-lab-bulges-drag","interactive","tide-lab",{"modes":99,"places":101,"challenges":102},[100],"bulges",[],2,"Revisit the two-bulge lab from Discover, and now imagine the near bulge dragged slightly ahead by friction, instead of sitting exactly under the Moon.","This is the same bulges lab from Discover: Earth, its ocean, and a Moon you can drag around. The lab itself still draws the simplified, perfectly-aligned bulges — that model is accurate enough for everything in Discover through Investigate.\n\nFor this layer, picture something the lab does not draw: because Earth spins faster than the Moon orbits, real friction drags the near bulge a little ahead of the Earth–Moon line, in the direction of Earth's spin. That tiny offset is the entire cause of the Moon's slow retreat and Earth's slowly lengthening day covered in this layer.",{"id":106,"type":77,"title":107,"eyebrow":108,"navLabel":109},"e-ch1b","The Moon already finished its half of this story","Chapter 01B","1b Moon's own lock",{"id":111,"type":42,"markdown":112},"e-moon-locked","Earth is only part-way through being tidally slowed by the Moon. The Moon itself finished the very same process long ago, on its own side of the relationship, which is why it always shows Earth the same face.\n\nEarth's pull raises tides on the Moon too — much bigger ones, since Earth is about 81 times more massive than the Moon. Long ago, when the Moon still spun faster than it circled Earth, friction from those Earth-raised tides braked the Moon's spin, exactly the way the Moon is now braking Earth's. Because the Moon is small, that braking finished completely: its spin slowed until its rotation period exactly matched its orbital period, at **27.32 days**, and it has stayed matched ever since. Astronomers call this **synchronous rotation**, and it is the one-way version of the mutual tidal locking Chapter 4 predicts, eventually, for Earth too.\n\nThis is also why there really is a 'far side of the Moon': not a permanently dark side (it gets just as much sunlight as the near side, over a full lunar month), but a side that human eyes never saw at all until a spacecraft photographed it in 1959.",{"id":114,"type":71,"variant":115,"title":116,"markdown":117},"e-aha-mass-matters","aha","Why the Moon finished first and Earth hasn't","Tidal locking happens faster for a **smaller** body being tugged by a **bigger** one, because the smaller body has less spin (less angular momentum) to remove in the first place, and the bigger body's pull on it is relatively stronger. The Moon is small and Earth is large, so the Moon locked first, long ago. Earth is large and the Moon is comparatively small, so braking Earth's spin to match the month would take vastly longer than the age of the Solar System — which is exactly why Chapter 4 says the Sun will change everything long before Earth could ever get there.",{"id":119,"type":120,"itemId":121,"prompt":122,"check":123,"hints":139,"feedback":142},"e-pr-locked","practice","tides.extend-locked","Why does the Moon always show Earth the same face?",{"kind":124,"options":125,"correct":138},"choice",[126,129,132,135],{"id":127,"label":128},"a","It does not spin at all",{"id":130,"label":131},"b","Its rotation period exactly matches its orbital period, because Earth's tides braked its spin long ago",{"id":133,"label":134},"c","It is held in place by Earth's magnetic field",{"id":136,"label":137},"d","Astronauts turned it to face us",[130],[140,141],"The Moon does spin — once every orbit.","Think about which body's tides act on which.",{"correct":143,"incorrect":144},"Right: the Moon spins exactly once per orbit (27.32 days), a state called synchronous rotation, reached because Earth's tides braked the Moon's spin long ago.","The Moon is not frozen; it turns once every orbit, which is precisely why the same face always points at Earth.",{"id":146,"type":77,"title":147,"eyebrow":148,"navLabel":149},"e-ch2","Evidence written in ancient shells and sand","Chapter 02","2 Fossil evidence",{"id":151,"type":42,"markdown":152},"e-fossil-evidence","If days used to be shorter, is there any way to check, hundreds of millions of years after the fact? Remarkably, yes — some living things keep a daily and a yearly calendar in their own growth.\n\nCertain corals and shellfish add a fine growth ridge to their shell or skeleton **every single day**, and a slightly thicker band at the same point **every year** (marking a season). Count the fine daily ridges between two yearly bands in a fossil and you have directly counted **how many days made up a year**, at the exact moment that fossil was alive.\n\nStudies of fossil rudist bivalves from the **Cretaceous period**, roughly 70 million years ago, have found around **372 daily growth ridges per year**. A year's actual length barely changes over that kind of timescale, so if there were 372 days packed into the same year that now holds 365.24, each of those days must have been **shorter**.",{"id":154,"type":155,"title":156,"problem":157,"steps":158,"help":163},"e-we-cretaceous","worked_example","How long was a Cretaceous day?","A year lasts 8765.8128 hours now and, to a very good approximation, always has (a year is set by Earth's orbit, which tidal friction barely touches). If the Cretaceous year contained 372 days instead of about 365.24, how many hours long was each Cretaceous day?",[159,160,161,162],"Hours per day = hours per year ÷ days per year = 8765.8 ÷ 372.","8765.8 ÷ 372 ≈ **23.56 hours**.","In hours and minutes, that is **23 h 34 min** — about 26 minutes shorter than today's 24-hour day.","Check the direction of the answer: **more** days had to fit into the same year, so each day must be **shorter** than 24 hours. 23.56 \u003C 24 ✓",{"simplerExplanation":164,"hints":165},"The year's length barely changes, so cramming more days into it means each day was shorter.",[166,167],"Divide the year's total hours by the number of days.","Sanity check: more days per year means each day is shorter, not longer.",{"id":169,"type":71,"variant":170,"title":171,"markdown":172},"e-example-wells","example","Where this whole method began: a coral called Wells's clock","The technique started with a real discovery. In **1963**, the geologist John Wells counted growth lines on well-preserved **Middle Devonian** rugose corals from central New York, about **380 million years old**. He found roughly **400 fine daily lines** inside each thicker annual band — corroborating, from nothing more than careful counting under a microscope, what physicists had already worked out from orbital mechanics: Earth's spin has slowed over deep time. His result, 400 days in a Devonian year, gives a day of 8766 ÷ 400 ≈ **21.91 hours** — remarkably close to the 21.9-hour figure this chapter already found for the much older Ediacaran rhythmites, a coincidence explained properly in the nuance box below.",{"id":174,"type":42,"markdown":175},"e-ediacaran","Push back further still, to the **Ediacaran period**, about 620 million years ago — before almost anything with a shell existed — and geologists instead read the story from **tidal rhythmites**: layers of sediment laid down grain by grain with every single tide, preserved in rock ever since. Counting layers between clear seasonal or monthly markers in rhythmite deposits has suggested a year of around **400 days** at that time.",{"id":177,"type":120,"itemId":178,"prompt":179,"check":180,"hints":185,"feedback":188},"e-pr-ediacaran","tides.extend-ediacaran","Using the same method as the worked example, and a year of 8766 hours, what was the length of an Ediacaran day if the year held 400 days? Give your answer in hours, to one decimal place.",{"kind":181,"answer":182,"tolerance":183,"unit":184},"number",21.9,0.05,"h",[186,187],"Hours per day = hours per year ÷ days per year.","8765.8 ÷ 400 = ?",{"correct":189,"incorrect":190},"Right: 8765.8 ÷ 400 ≈ **21.9 hours**, or about 21 h 55 min — nearly two hours shorter than today's day.","Divide the year's hours by the number of days: 8765.8 ÷ 400 ≈ 21.9 h.",{"id":192,"type":96,"component":193,"componentVersion":5,"config":194,"objective":211,"textAlternative":212},"e-lab-data-daylength","data-lab",{"datasets":195,"valueRange":202,"step":205,"challenges":206},[196],{"label":197,"unit":184,"values":198},"Day length through deep time",[199,200,201],21.91,23.56,24,{"min":203,"max":204},20,25,0.1,[207],{"measure":208,"target":209,"prompt":210},"range",2.09,"What is the range between the shortest and longest day length shown?","Plot three day lengths — Ediacaran, Cretaceous, today — as a tiny dataset and read off how much the day has grown.","This lab plots three points: about 21.9 hours (Ediacaran, ~620 million years ago), about 23.56 hours (Cretaceous, ~70 million years ago), and 24.0 hours (today). The one challenge asks for the range between the shortest and the longest — about 2.1 hours, nearly two whole hours of lengthening captured in just three data points spanning over half a billion years.",{"id":214,"type":71,"variant":215,"title":216,"markdown":217},"e-nuance-stall","nuance","Why Devonian and Ediacaran days come out so similar","The Ediacaran figure (21.9 h, 620 million years ago) and the Devonian figure (21.9 h, 380 million years ago) are suspiciously close, despite 240 million years between them — if the day had been lengthening at a steady rate the whole time, the Devonian day should be noticeably longer. Recent research suggests the reason is that the day's lengthening was **not steady**: for roughly a billion years in the middle of Earth's history, an **atmospheric thermal tide** — the same kind of Sun-driven air-pressure tide from Deepen — is thought to have pushed back against the Moon's slowing effect almost exactly hard enough to cancel it out, in a kind of accidental resonance. Only once the Moon had receded far enough for its effect to win outright did the day resume lengthening in earnest. This is active, ongoing research, not a settled textbook fact, and a good example of how a tidy story ('the day gets longer at a steady rate') can turn out to be more interesting than expected once enough evidence is gathered.",{"id":219,"type":71,"variant":215,"title":220,"markdown":221},"e-nuance-uncertainty","How sure are scientists of these numbers?","Less sure than a laboratory measurement, and honest sources say so plainly. Counting growth ridges in a 70-million- or 620-million-year-old fossil depends on the specimen being well-preserved, on correctly telling a daily ridge from noise, and on assuming growth was roughly steady through the year. Different fossils and different rhythmite beds give figures that agree in **rough shape** — shorter days, further back — but disagree by tens of days when you compare one study with another. Treat every number in this chapter as a **well-reasoned estimate**, not a direct measurement like a modern satellite laser ranging figure.",{"id":223,"type":77,"title":224,"eyebrow":225,"navLabel":226},"e-ch3","Measuring the Moon's retreat, today","Chapter 03","3 Measuring today",{"id":228,"type":42,"markdown":229},"e-laser-ranging","You do not have to dig up a fossil to measure this effect happening right now. Apollo astronauts (and robotic Soviet landers) left small mirror arrays called **retroreflectors** on the Moon's surface, designed to bounce a laser beam fired from Earth straight back the way it came.\n\nObservatories on Earth fire a laser pulse at one of these mirrors and time exactly how long the light takes to return. Multiply that time by the speed of light and you get the Earth–Moon distance at that instant, accurate to a few centimetres over a quarter of a million kilometres — one of the most precise distance measurements ever made by humans. Repeated over decades, these measurements show the Moon receding at about **3.8 cm per year**.",{"id":231,"type":155,"title":232,"problem":233,"steps":234,"help":238},"e-we-recession","How far has the Moon moved since your grandparents were born?","Using a recession rate of 3.8 cm per year, roughly how far has the Moon moved away from Earth over the last 70 years?",[235,236,237],"Distance = rate × time = 3.8 cm\u002Fyear × 70 years.","3.8 × 70 = **268 cm**, which is 2.7 m — about the length of two cars parked end to end.","Compared with the Earth–Moon distance of about 384,400 km, 2.7 m is utterly negligible — which is exactly why nobody notices the Moon 'shrinking' in the sky, even over a whole lifetime.",{"simplerExplanation":239,"hints":240},"Multiply the yearly rate by the number of years, the same as any steady-rate problem.",[241],"70 × 3.8 is a little over 260.",{"id":243,"type":120,"itemId":244,"prompt":245,"check":246,"hints":250,"feedback":253},"e-pr-recession-century","tides.extend-recession","At 3.8 cm per year, how many **kilometres** has the Moon receded over the last 100 million years? (Use 3.8 cm\u002Fyear × 100,000,000 years, then convert to km.)",{"kind":181,"answer":247,"tolerance":248,"unit":249},3830,50,"km",[251,252],"cm\u002Fyear × years = total cm.","Convert cm to km by dividing by 100,000.",{"correct":254,"incorrect":255},"Right: 3.8 × 100,000,000 = 383000000 cm = **3830 km** — this is only a rough extrapolation of today's rate; Chapter 4 explains why the real rate has not been constant.","3.8 cm\u002Fyear × 100,000,000 years = 383000000 cm, then ÷ 100,000 to get km ≈ 3830 km.",{"id":257,"type":258,"caption":259,"columns":260,"rows":265},"e-table-evidence","table","Four eras, four kinds of evidence for day length",[261,262,263,264],"Era","Age","Day length","Evidence",[266,271,275,280],[267,268,269,270],"Ediacaran","≈620 Mya","≈21.9 h","Tidal rhythmite layer counts, South Australia",[272,273,269,274],"Devonian","≈380 Mya","Wells's coral growth-line counts, New York",[276,277,278,279],"Cretaceous","≈70 Mya","≈23.56 h","Rudist bivalve growth-band counts",[281,282,283,284],"Today","now","24.00 h (by definition)","Atomic clocks and laser ranging",{"id":286,"type":155,"title":287,"problem":288,"steps":289,"help":293},"e-we-precision","How precise is a laser-ranging measurement, really?","A laser-ranging measurement is accurate to a few centimetres over a distance of about 384,400 km. Roughly what fraction is that, and how does it compare with the 3.8 cm a year the Moon actually recedes?",[290,291,292],"384,400 km = 384,400,000 m = 38,440,000,000 cm.","A precision of, say, 3 cm out of 38,440,000,000 cm is a fraction of about 3 ÷ 38,440,000,000 ≈ **8 parts in a hundred billion** — an almost unimaginably fine measurement.","Since the yearly recession of 3.8 cm is comfortably bigger than a few centimetres of measurement error, a handful of years of laser-ranging data is already enough to detect the trend clearly, long before decades have passed.",{"simplerExplanation":294,"hints":295},"The measurement error (a few cm) is smaller than one year's worth of actual recession (3.8 cm), so the trend shows up quickly.",[296],"Convert everything to the same unit (centimetres) before comparing.",{"id":298,"type":71,"variant":299,"title":300,"markdown":301},"e-model-limit-constant-rate","model_limit","Today's rate is not yesterday's rate","Multiplying today's 3.8 cm\u002Fyear by millions of years, as the practice question above does, is a useful **estimate**, not a fact about the past. Tidal friction depends heavily on the exact shapes and depths of the ocean basins doing the sloshing, and continents have drifted into completely different arrangements over hundreds of millions of years. The fossil evidence in Chapter 2 (shorter Cretaceous and Ediacaran days) broadly agrees with the direction of today's laser-ranging measurement, but the two methods measure different eras by different means, and a careful scientist keeps that distinction clear rather than pretending one constant rate explains everything.",{"id":303,"type":77,"title":304,"eyebrow":305,"navLabel":306},"e-ch4","Where is this heading?","Chapter 04","4 Where it's heading",{"id":308,"type":42,"markdown":309},"e-future","Run the process forward, and two things move towards each other: Earth's day lengthens, and the month (the Moon's orbital period) lengthens too, because a wider orbit takes longer to complete. In principle, they could eventually meet — Earth's day and the Moon's month becoming the same length. When that happens, Earth would always show the Moon the same face, exactly as the Moon already shows Earth today: this end state is called **mutual tidal locking**, and the Pluto–Charon system, where both bodies already keep the same faces to each other, is a real example of it having already happened.\n\nWill Earth and the Moon actually get there? Almost certainly **not**, for a much bigger reason than tides: the Sun is expected to swell into a red giant in a few billion years, engulfing or at least utterly transforming the Earth–Moon system long before the tidal-locking calculation could finish playing out. It is a good reminder that a real prediction has to be checked against every process at work, not just the one you happen to be studying.",{"id":311,"type":71,"variant":115,"title":312,"markdown":313},"e-aha-charon","You can already see the future, at Pluto","Pluto and its largest moon Charon are close enough in size, and have been tidally interacting for long enough, that they reached mutual tidal locking already: Pluto always shows Charon the same face, and Charon always shows Pluto the same face, both spinning in perfect step with their shared orbit. It is the Earth–Moon system's tidal story, played to its natural end, sitting a few billion kilometres away for anyone to check.",{"id":315,"type":316,"prompt":317},"e-reflect-future","reflection","Explain, in your own words, why 'the Moon will eventually stop moving away' is a more careful claim than 'the Moon will eventually stop', and why scientists usually prefer the more careful version even when it is less exciting to say.",{"id":319,"type":77,"title":320,"eyebrow":321,"navLabel":322},"e-ch5","Tides beyond Earth: moons melted by squeezing","Chapter 05","5 Tides elsewhere",{"id":324,"type":42,"markdown":325},"e-beyond-earth","Everything in this topic — differential gravity, flexing, friction, heat — works on any moon orbiting any planet, and a few places in the Solar System take it to spectacular extremes.\n\n**Io**, the innermost of Jupiter's four big moons, is the most volcanically active body ever found, anywhere. It is locked into an orbital rhythm with two neighbouring moons, Europa and Ganymede (each one takes exactly twice as long to orbit as the moon inside it — a pattern called a **Laplace resonance**, named after the same Laplace who worked out the dynamic theory of Earth's own tides). That resonance keeps forcing Io's orbit to stay slightly stretched instead of settling into a neat circle, so Jupiter's gravity never stops flexing it, generating enough heat to drive continuous volcanic eruptions.\n\n**Europa**, one step further out, feels a gentler version of the same squeezing — Jupiter's pull weakens roughly with the cube of distance, so Europa gets only a fraction of Io's flexing. That smaller but steady heat is thought to keep a deep ocean of liquid water flowing under Europa's icy shell, making it one of the most closely watched places in the search for life beyond Earth.\n\n**Enceladus**, a small moon of Saturn, shows the same idea from a different angle entirely: tidal flexing, driven by its own orbital resonance with the larger moon Dione, warms its south pole enough to blast geysers of water vapour and ice hundreds of kilometres into space — material that has since been found to feed one of Saturn's rings.",{"id":327,"type":155,"title":328,"problem":329,"steps":330,"help":334},"e-we-laplace-resonance","Checking the 1:2:4 resonance with real orbital periods","Io orbits Jupiter in about 1.77 days, Europa in about 3.55 days, and Ganymede in about 7.15 days. Show that these are close to a 1:2:4 ratio.",[331,332,333],"Europa ÷ Io = 3.55 ÷ 1.77 ≈ **2.01**, close to 2.","Ganymede ÷ Europa = 7.15 ÷ 3.55 ≈ **2.01**, close to 2 again.","So Europa takes about twice as long as Io, and Ganymede about twice as long as Europa (four times Io) — the **1:2:4 Laplace resonance**, first shown to be almost exactly exact from tables published by the astronomer Pehr Wilhelm Wargentin in **1743**.",{"simplerExplanation":335,"hints":336},"Divide each period by the one before it; both answers should come out close to 2.",[337],"3.55 ÷ 1.77 is a touch over 2.",{"id":339,"type":258,"caption":340,"columns":341,"rows":346},"e-table-beyond","Three moons, one mechanism",[342,343,344,345],"Moon","Planet","What tidal heating causes","Why it happens",[347,352,356],[348,349,350,351],"Io","Jupiter","Constant volcanic eruptions","Squeezed by an orbital resonance with Europa and Ganymede, so its orbit never settles into a circle",[353,349,354,355],"Europa","A liquid ocean under an icy shell","A gentler version of Io's squeezing, from further away",[357,358,359,360],"Enceladus","Saturn","Geysers of water vapour and ice at the south pole","Flexing driven by an orbital resonance with the moon Dione",{"id":362,"type":96,"component":363,"componentVersion":5,"config":364,"objective":383,"textAlternative":384},"e-lab-gravity-context","gravity-drop",{"worlds":365,"objects":370,"modes":380},[366,367,368,369],"earth","moon","jupiter","pluto",[371,375],{"id":372,"label":373,"massKg":5,"draggy":374},"ball","Ball",false,{"id":376,"label":377,"massKg":378,"draggy":379},"feather","Feather",0.01,true,[381,382],"drop","weigh","Compare plain surface gravity on Earth, the Moon, Jupiter and Pluto, then read the model-limit note on why that is a different question from tidal heating.","This lab drops or weighs objects on Earth, the Moon, Jupiter and Pluto, showing that Jupiter's surface gravity is enormous compared with tiny Pluto's.\n\nBut surface gravity alone is not what makes Io glow with volcanoes: Ganymede, further from Jupiter, feels weaker tidal flexing despite orbiting the very same giant planet, and Enceladus is tiny yet tidally heated because of Saturn's pull combined with its particular orbital resonance. This lab only shows plain surface gravity; the tidal heating in this chapter is about *how much that pull differs* across a small moon and *how the orbit keeps getting re-stretched*, exactly the 'difference, not strength' idea from Understand, now applied far from Earth.",{"id":386,"type":387,"conceptId":388,"relation":389,"explanation":390},"e-conn-phases","connection","phases-of-the-moon","related_to","The Moon's synchronous rotation (Chapter 1b) uses its 27.32-day sidereal orbit, while its phases follow the slightly longer 29.53-day synodic month — a good reminder that 'the Moon's period' is not one single number, but depends on what you are measuring it against.",{"id":392,"type":387,"conceptId":393,"relation":394,"explanation":395},"e-conn-gravity-2","gravity","helps_understand","Tidal heating on Io, Europa and Enceladus is the same 'difference in pull across a body' idea that raises Earth's ocean tides, just powerful enough at those distances and orbits to melt rock and ice instead of merely moving water.",{"id":397,"type":77,"title":398,"eyebrow":399,"navLabel":400},"e-ch6","Project: be a paleo-tide detective","Chapter 06","6 Project",{"id":402,"type":42,"markdown":403},"e-project-intro","Choose one of these projects and carry it through properly: a clear question, your method, your working (with real arithmetic, not guesses), and an honest statement of how confident you are in the answer.",{"id":405,"type":406,"title":407,"items":408},"e-steps-project","steps","Two project starting points",[409,413],{"title":410,"tag":411,"text":412},"Project A: count a real growth record","Shells or tree rings","Find a bivalve, coral, or tree-ring photo with clear fine and coarse growth bands. Count fine bands between two coarse ones and compare with this layer's method — a modern shell, with 365 daily bands a year, is a genuine test.",{"title":414,"tag":415,"text":416},"Project B: extrapolate carefully","Your own estimate","Using today's 3.8 cm\u002Fyear rate, estimate how long ago the Moon was twice as close as now (192,200 km). Then find out whether real geophysical models agree, and explain why they might not.",{"id":418,"type":71,"variant":419,"title":420,"markdown":421},"e-careful-project","careful","Keep your own uncertainty honest","Whichever project you choose, resist the temptation to report a single confident number. Real scientists working on this exact question report a range (such as \"400 ± 7 days a year\" for the Ediacaran figure), because their counting method has real limits. State your own range, however rough, rather than a single tidy figure — it is more honest, and it is also better science.",{"id":423,"type":316,"prompt":424},"e-reflect-project","Whichever project you choose, write two sentences on what would make your estimate more trustworthy: a second, independent method that should give a similar answer if your method is sound.",{"id":426,"type":77,"title":427,"eyebrow":428,"navLabel":429},"e-ch7","Harder problems: put the whole topic to work","Chapter 07","7 Harder problems",{"id":431,"type":155,"title":432,"problem":433,"steps":434,"help":438},"e-we-hard1","How many Cretaceous 'days' fit a modern week?","A modern week is 7 × 24 = 168 hours. Using the Cretaceous day length of 23.56 hours, how many Cretaceous days would fit into the same 168 hours?",[435,436,437],"Number of days = total hours ÷ hours per day = 168 ÷ 23.56.","168 ÷ 23.56 ≈ **7.1 days**.","So a Cretaceous 'week' of the same total length would have needed about **7.1** of its shorter days — noticeably more than 7, exactly as you'd expect since each of its days was shorter.",{"simplerExplanation":439},"Divide the same total number of hours by the shorter day length; you should get a number bigger than 7.",{"id":441,"type":120,"itemId":442,"prompt":443,"check":444,"hints":448,"feedback":451},"e-pr-hard-rate","tides.extend-rate","Using the Cretaceous figure (day length 23.56 h, 70 million years ago) and today's 24 h day, roughly how many **milliseconds** has the day lengthened per century, on average, since then? (Hint: convert the total lengthening to milliseconds, then divide by the number of centuries in 70 million years.)",{"kind":181,"answer":445,"tolerance":446,"unit":447},2.2,0.3,"ms\u002Fcentury",[449,450],"Total lengthening = (24 − 23.56) hours, converted to milliseconds.","70 million years = 700,000 centuries.",{"correct":452,"incorrect":453},"Right: (24 − 23.56) h = 0.44 h = 1569554 ms. Divide by 700,000 centuries: **≈ 2.2 ms per century**.","Convert the whole 24 − 23.56 hour lengthening to milliseconds, then divide by 700,000 (the number of centuries in 70 million years).",{"id":455,"type":120,"itemId":456,"prompt":457,"check":458,"hints":459,"feedback":460},"e-pr-devonian","tides.extend-devonian","Using Wells's figure of 400 days in a Devonian year and a year of 8765.8 hours, what was the length of a Devonian day, in hours, to two decimal places?",{"kind":181,"answer":199,"tolerance":183,"unit":184},[186,187],{"correct":461,"incorrect":462},"Right: 8765.8 ÷ 400 ≈ **21.91 hours** — matching the figure Wells's coral count implies.","Divide the year's total hours by the number of days: 8765.8 ÷ 400 ≈ 21.91 h.",{"id":464,"type":71,"variant":170,"title":465,"markdown":466},"e-example-leap-second","Where this shows up on your phone","The lengthening day is not just ancient history: it is the reason **leap seconds** have occasionally been added to official world time since the 1970s, to keep clocks (which tick at a fixed rate) in step with Earth's actual, very slowly lengthening rotation. The correction needed is tiny — a second every year or so at most, nowhere near the dramatic Cretaceous figure — because Earth's spin has already slowed for hundreds of millions of years since then.",{"id":468,"type":316,"prompt":469},"e-reflect-olympiad","An asteroid impact is thought to have ended the age of the dinosaurs about 66 million years ago, close to the age of the rudist-shell evidence in this layer. Explain why an asteroid impact would NOT be expected to change Earth's day length in the way tidal friction does, even though both are dramatic events in Earth's history.",{"id":471,"type":77,"title":472,"eyebrow":473,"navLabel":474},"e-ch8","Careers built on the tide","Chapter 08","8 Careers",{"id":476,"type":71,"variant":72,"title":477,"markdown":478},"e-observation-careers","One topic, five completely different jobs","Look back over this whole layer and notice how many different kinds of expert had to contribute: a geologist counting shell ridges, a physicist bouncing lasers off the Moon, a planetary scientist reading spacecraft images of Io, and an engineer weighing up a tidal barrage. Tides are a genuinely interdisciplinary subject — nobody masters the whole of it alone, and that is normal for real science, not a gap in this course.",{"id":480,"type":42,"markdown":481},"e-careers","Nearly everything in this topic is somebody's actual job.\n\n**Physical oceanographers** measure and model how oceans move, tides included, often working for INCOIS, the Survey of India, or a university. **Coastal and tidal-power engineers** design harbours, embankments and barrages that have to survive the biggest tide and the worst storm surge a coast can produce, not just an average day. **Geophysicists** use tools from GPS networks to laser ranging to study Earth tides, the planet's interior, and the slow evolution of the Earth–Moon system. **Planetary scientists** study tidal heating on Io, Europa and Enceladus, some of them specifically hunting for the conditions life might need. **Marine biologists** study how coastal life such as horseshoe crabs, mangroves and shorebirds time themselves to the tide, work that matters directly for conservation in places like the Sundarbans.",{"id":483,"type":96,"component":484,"componentVersion":5,"config":485,"objective":504,"textAlternative":505},"e-lab-match-careers","match-pairs",{"prompt":486,"mode":487,"pairs":488},"Match each career to the tide-related work it does.","connect",[489,492,495,498,501],{"a":490,"b":491},"Physical oceanographer","Measures and models how the ocean, including tides, actually moves",{"a":493,"b":494},"Coastal \u002F tidal-power engineer","Designs harbours, embankments and barrages to survive the worst tide and surge",{"a":496,"b":497},"Geophysicist","Studies Earth tides and the slow evolution of the Earth–Moon system",{"a":499,"b":500},"Planetary scientist","Studies tidal heating on moons such as Io, Europa and Enceladus",{"a":502,"b":503},"Marine biologist","Studies how coastal life times itself to the tide","Match five careers to the tide-related work each one does.","A matching game with five pairs. Physical oceanographer goes with measuring and modelling how the ocean, including tides, actually moves. Coastal\u002Ftidal-power engineer goes with designing harbours, embankments and barrages to survive the worst tide and surge. Geophysicist goes with studying Earth tides and the slow evolution of the Earth–Moon system. Planetary scientist goes with studying tidal heating on moons such as Io, Europa and Enceladus. Marine biologist goes with studying how coastal life times itself to the tide.",{"id":507,"type":96,"component":508,"componentVersion":5,"config":509,"objective":540,"textAlternative":541},"e-lab-sort-evidence","sort-game",{"prompt":510,"bins":511,"items":518,"seconds":539},"Direct measurement today, or an estimate from ancient evidence?",[512,515],{"id":513,"label":514},"direct","Direct measurement, today",{"id":516,"label":517},"estimate","Estimate from ancient evidence",[519,523,527,531,535],{"id":520,"label":521,"bin":513,"why":522},"ev1","Laser ranging off Apollo retroreflectors","A live measurement made with light and a clock, accurate to centimetres.",{"id":524,"label":525,"bin":516,"why":526},"ev2","Counting growth ridges on a fossil rudist shell","Depends on preservation and careful counting, so it carries real uncertainty.",{"id":528,"label":529,"bin":516,"why":530},"ev3","Counting layers in a 620-million-year-old tidal rhythmite","Ancient sediment evidence, interpreted rather than directly timed.",{"id":532,"label":533,"bin":513,"why":534},"ev4","Timing a laser pulse's round trip to the Moon and back","A precise, repeatable physical measurement made now.",{"id":536,"label":537,"bin":516,"why":538},"ev5","Extrapolating today's 3.8 cm\u002Fyear rate back 600 million years","An extrapolation, not a measurement — and Chapter 3 explained why it is not reliable that far back.",0,"Sort five pieces of evidence about the Moon's retreat into direct measurements or estimates from ancient evidence.","A sorting game with two bins. Direct measurements: laser ranging off Apollo retroreflectors; timing a laser pulse's round trip. Estimates from ancient evidence: counting growth ridges on a fossil shell; counting layers in an ancient tidal rhythmite; extrapolating today's rate back hundreds of millions of years.",{"id":543,"type":77,"title":544,"eyebrow":545,"navLabel":546},"e-ch9","What is still unknown","Chapter 09","9 Open questions",{"id":548,"type":71,"variant":549,"title":550,"markdown":551},"e-open-questions","question","Open questions for curious learners","Nobody has fully solved these yet. Pick one and dig in.\n\n- How precisely can fossil growth-ridge counting pin down day length hundreds of millions of years ago, and what would make the method more reliable?\n- If continents keep drifting, will Earth's tidal friction — and so the rate the day is lengthening — speed up, slow down, or stay roughly the same over the next 100 million years?\n- Could tidal heating like Europa's or Enceladus's support life, and how would we ever find out without physically drilling through kilometres of ice?\n- Should India build a tidal barrage in the Gulf of Kutch or Khambhat, given the trade-offs you weighed in Investigate? What extra evidence would change your answer?\n- Leap seconds are added irregularly, not on a fixed schedule. What would a fairer, more predictable system for keeping clocks matched to Earth's slowing spin look like?",{"id":553,"type":258,"caption":554,"columns":555,"rows":559},"e-table-careers","Careers built on tides, and a tool each one relies on",[556,557,558],"Career","Typical employer or setting","A tool of the trade",[560,563,566,569,572],[490,561,562],"INCOIS, universities, research ships","Tide gauges, harmonic analysis software",[493,564,565],"Port authorities, power companies","Tide tables, storm-surge models",[496,567,568],"Survey agencies, universities","GPS networks, gravimeters",[499,570,571],"Space agencies, universities","Spacecraft data (Galileo, Cassini)",[502,573,574],"Conservation bodies, universities","Field surveys timed to the tide",{"id":576,"type":577,"title":578,"terms":579},"e-glossary","glossary","Vocabulary for Extend",[580,583,586,589,592,595,598,601,604],{"term":581,"meaning":582},"Angular momentum","A measure of spinning or orbiting motion that cannot be created or destroyed, only transferred between parts of a system.",{"term":584,"meaning":585},"Tidal rhythmite","Layers of sediment laid down with the rhythm of the tide, preserved in rock and used to estimate ancient day length.",{"term":587,"meaning":588},"Retroreflector","A mirror array, including ones left on the Moon by Apollo astronauts, designed to bounce a laser beam straight back the way it came.",{"term":590,"meaning":591},"Mutual tidal locking","The end state where both bodies in a pair always show each other the same face, as Pluto and Charon already do.",{"term":593,"meaning":594},"Laplace resonance","A repeating pattern in which each of several orbiting moons takes exactly twice as long to orbit as the one just inside it, as with Io, Europa and Ganymede.",{"term":596,"meaning":597},"Tidal heating","Heat generated inside a moon or planet by repeated tidal flexing, powering Io's volcanoes and warming the oceans under Europa's and Enceladus's ice.",{"term":599,"meaning":600},"Synchronous rotation","A moon's spin period exactly matching its orbital period, so the same face always points at its planet — the Moon's own state today.",{"term":602,"meaning":603},"Orbital resonance","A simple whole-number ratio between two or more orbital periods, such as the Galilean moons' 1:2:4, which keeps orbits from settling into neat circles.",{"term":605,"meaning":606},"Atmospheric thermal tide","A daily air-pressure wobble driven mainly by the Sun heating the atmosphere, thought to have briefly stalled Earth's day lengthening by cancelling lunar tidal friction.",{"id":608,"type":316,"prompt":609},"e-reflect-stall","The 'stalled day length' idea in Chapter 2 was only proposed clearly in the 2020s, even though the Devonian and Ediacaran evidence behind it had been known for decades. Suggest one reason a puzzling pattern in old data might take a long time for anyone to properly explain.",{"id":611,"type":612,"title":613,"questions":614},"e-quiz","quiz","Check yourself: deep time, deep space, and open questions",[615,628,641,654,667,680,693,706,719,731,743],{"itemId":616,"prompt":617,"options":618,"correct":130,"why":627},"tides.extend-q-brake","What is 'the brake' in this layer's brake-and-boost picture?",[619,621,623,625],{"id":127,"label":620},"The Moon slowing down",{"id":130,"label":622},"Earth's spin very slowly slowing down",{"id":133,"label":624},"The Sun cooling",{"id":136,"label":626},"Ocean currents stopping","Friction drags the near tidal bulge ahead of the Moon, and Earth's pull on that offset bulge acts back on Earth as a brake on its own spin.",{"itemId":629,"prompt":630,"options":631,"correct":130,"why":640},"tides.extend-q-boost","What happens to the Moon because of 'the boost'?",[632,634,636,638],{"id":127,"label":633},"It falls towards Earth",{"id":130,"label":635},"It is pushed into a wider, slower orbit",{"id":133,"label":637},"It spins faster",{"id":136,"label":639},"Nothing; only Earth is affected","The offset bulge's gravity pulls the Moon slightly forward along its orbit, which raises it into a wider orbit rather than speeding it up in the everyday sense.",{"itemId":642,"prompt":643,"options":644,"correct":130,"why":653},"tides.extend-q-fossil","How do scientists estimate day length hundreds of millions of years ago?",[645,647,649,651],{"id":127,"label":646},"By measuring rocks with a stopwatch",{"id":130,"label":648},"By counting fine daily growth ridges between yearly bands in fossils or sediment layers",{"id":133,"label":650},"By asking how fast the Moon looks in old paintings",{"id":136,"label":652},"It cannot be estimated at all","Daily growth ridges, counted between clear yearly markers, directly count how many days made up a year at that time.",{"itemId":655,"prompt":656,"options":657,"correct":130,"why":666},"tides.extend-q-laser","How is the Moon's retreat measured today?",[658,660,662,664],{"id":127,"label":659},"By eye, comparing its size in the sky",{"id":130,"label":661},"By timing a laser pulse bounced off a mirror left on the Moon",{"id":133,"label":663},"By counting eclipses",{"id":136,"label":665},"It has never actually been measured","Laser ranging off Apollo-era retroreflectors gives the Earth–Moon distance to centimetres, repeated over decades to reveal the recession rate.",{"itemId":668,"prompt":669,"options":670,"correct":127,"why":679},"tides.extend-q-locking","What is 'mutual tidal locking', and where can you already see it?",[671,673,675,677],{"id":127,"label":672},"Two bodies always showing each other the same face, as Pluto and Charon do",{"id":130,"label":674},"The Moon crashing into Earth",{"id":133,"label":676},"A type of storm surge",{"id":136,"label":678},"A kind of tidal bore","Pluto and Charon are already mutually tidally locked; the Earth–Moon system is evolving the same way, extremely slowly, but the Sun will change dramatically long before it could finish.",{"itemId":681,"prompt":682,"options":683,"correct":130,"why":692},"tides.extend-q-io","Why does Io have constant volcanic eruptions?",[684,686,688,690],{"id":127,"label":685},"It is closest to the Sun",{"id":130,"label":687},"An orbital resonance with Europa and Ganymede keeps stretching its orbit, so Jupiter's gravity keeps flexing and heating it",{"id":133,"label":689},"It has a thick, insulating atmosphere",{"id":136,"label":691},"Its own spin is extremely fast","The Laplace resonance with Europa and Ganymede stops Io's orbit from settling into a neat circle, so Jupiter never stops flexing it.",{"itemId":694,"prompt":695,"options":696,"correct":130,"why":705},"tides.extend-q-europa","Why is Europa of such interest in the search for life?",[697,699,701,703],{"id":127,"label":698},"It has a thick breathable atmosphere",{"id":130,"label":700},"Tidal heating is thought to keep a liquid water ocean under its icy shell",{"id":133,"label":702},"It is the closest moon to Earth",{"id":136,"label":704},"It has active volcanoes like Io","A gentler version of Io's tidal flexing is thought to keep Europa's subsurface ocean liquid, making it a serious astrobiology target.",{"itemId":707,"prompt":708,"options":709,"correct":130,"why":718},"tides.extend-q-uncertainty","Why should you trust laser-ranging measurements of the Moon's retreat more than a 600-million-year extrapolation of today's rate?",[710,712,714,716],{"id":127,"label":711},"You shouldn't; both are equally certain",{"id":130,"label":713},"Laser ranging is a direct, repeatable measurement, while extrapolating ignores how much ocean basins and continents have changed",{"id":133,"label":715},"Ancient evidence is always more reliable than modern measurement",{"id":136,"label":717},"Laser ranging has never actually been done","Direct measurement today is solid; assuming today's exact rate held for hundreds of millions of years ignores real changes in ocean geography that affect tidal friction.",{"itemId":720,"prompt":122,"options":721,"correct":130,"why":730},"tides.extend-q-moonlocked",[722,724,726,728],{"id":127,"label":723},"It does not rotate",{"id":130,"label":725},"Its rotation period exactly matches its orbital period, a state reached because Earth's tides braked its spin long ago",{"id":133,"label":727},"Earth's magnetic field holds it in place",{"id":136,"label":729},"It is coincidence","This is synchronous rotation: the Moon spins exactly once per orbit, the end state of the very same tidal-braking process now slowly acting on Earth.",{"itemId":732,"prompt":733,"options":734,"correct":130,"why":742},"tides.extend-q-resonance-ratio","Io, Europa and Ganymede orbit Jupiter in a ratio close to…",[735,737,738,740],{"id":127,"label":736},"1 : 1 : 1",{"id":130,"label":63},{"id":133,"label":739},"1 : 3 : 9",{"id":136,"label":741},"2 : 3 : 5","The Laplace resonance: Europa takes about twice as long as Io, and Ganymede about twice as long as Europa — four times Io.",{"itemId":744,"prompt":745,"options":746,"correct":130,"why":755},"tides.extend-q-stall","Devonian corals (380 Mya) and Ediacaran rhythmites (620 Mya) imply almost the same day length. What does recent research suggest is the reason?",[747,749,751,753],{"id":127,"label":748},"Both measurements are simply wrong",{"id":130,"label":750},"An atmospheric thermal-tide resonance is thought to have cancelled the Moon's slowing effect for roughly a billion years in between",{"id":133,"label":752},"The Moon stopped moving during that time",{"id":136,"label":754},"Earth's orbit around the Sun changed instead","This is the 'stalled day length' hypothesis: a Sun-driven atmospheric tide is thought to have pushed back against lunar tidal friction almost exactly hard enough to cancel it for a long stretch of Earth's history.",{"id":757,"type":758,"title":759,"points":760},"e-cheat-sheet","summary","Cheat sheet: tides across deep time and deep space",[761,762,763,764,765,766,767,768,769],"Friction drags Earth's near tidal bulge slightly **ahead** of the Moon, which brakes Earth's spin and boosts the Moon into a wider, slower orbit — a transfer of angular momentum, not a loss of it.","Fossil evidence (rudist shells, 372 days\u002Fyear in the Cretaceous; tidal rhythmites, 400 days\u002Fyear in the Ediacaran) shows the day was shorter in the deep past — an **estimate**, not a direct measurement.","Laser ranging off Apollo-era mirrors directly measures the Moon receding at about **3.8 cm per year** today.","Extrapolating today's rate over hundreds of millions of years is a rough estimate at best, because ocean geography — and so tidal friction — has changed enormously over that time.","The **Moon** already finished its own tidal braking: its rotation exactly matches its 27.32-day orbit (synchronous rotation), which is why the same face always points at Earth.","Run far enough forward and Earth and the Moon would approach **mutual tidal locking**, as Pluto and Charon already show — but the Sun's own evolution will intervene first.","The same mechanism, applied to other moons, drives **Io's** volcanoes, keeps an ocean under **Europa's** ice, and powers **Enceladus's** geysers — tidal heating from orbital resonance, not surface gravity.","Careers built on this topic range from oceanography and coastal engineering to geophysics, planetary science and marine biology.","Leap seconds, added occasionally to world clocks, are a small, everyday trace of the same slow lengthening of the day documented across this whole layer.",{"id":771,"type":772,"sourceIds":773},"e-sources","sources",[774,775,776,777,778,779,780,781,782,783,784],"tides-nasa-moon-tides","tides-wikipedia-tidal-acceleration","tides-wikipedia-lunar-laser-ranging","tides-wikipedia-tidal-heating","tides-wikipedia-io","tides-wikipedia-europa","tides-wikipedia-enceladus","tides-wikipedia-charon","tides-wikipedia-leap-second","tides-wikipedia-john-wells","tides-nature-geoscience-stall",[774,775,776,777,778,779,780,781,782,783,784],"needs_review",{"generatedBy":788,"notes":789},"claude-code","Draft generated locally; pending owner review.","9b6a7822ae0fe743567b980cd1f9b9e5f06dc07cdbe3b87055bc0db70eb602cc",{"component:tide-lab@1":792,"logic:practice":793,"component:data-lab@1":794,"component:gravity-drop@1":795,"component:match-pairs@1":796,"component:sort-game@1":797,"source:tides-nasa-moon-tides":798,"source:tides-nature-geoscience-stall":799,"source:tides-wikipedia-charon":800,"source:tides-wikipedia-enceladus":801,"source:tides-wikipedia-europa":802,"source:tides-wikipedia-io":803,"source:tides-wikipedia-john-wells":804,"source:tides-wikipedia-leap-second":805,"source:tides-wikipedia-lunar-laser-ranging":806,"source:tides-wikipedia-tidal-acceleration":807,"source:tides-wikipedia-tidal-heating":808},"5fef8b331bba35d6df96a31b84dd1f98200bcd0242eeed843d22914391057b6c","3d6b0fe1b15255975a32b0fcd94e8019bc959ad45cbf12e136e86149549c6878","466896cc37735f48db03875fe9c9ce42fc8bcb7e5f937c9779d70513703b91bd","60b1b2628472895c3a36f45610b158c8d87f74e411a582d1c8293f4f80ecf53b","2a8ee4ac87460b4e1175a4bb13c96b03d577db06dde95670eb7fcfe4ad787899","b164f45a2c8ca08f26c450768ff0231e113e9fe45381eddb34dc6d0548596c38","d71a4932c5e1df6475bdbf45fedc9a0a348ce79ce9d887d3112d5b033fe39998","9e55ab69a13c4a04bd95ae70b6abcee67a3f18baf8fa4dbeaedfc206785b2654","23f1ac25ae02b26e396ed68e1e2e52545057a2726fd5ce3659b287f69179c07a","9cbf24efb1379ff7d3bf2343fc8e062526f6c51659d0a7438ff93921944578f3","2b2a434ca60a8592bea0ce7fdf2e3d6b7260820ddebf547661f9b69cc980d4d8","a6a3d0f584bd042019b95dc2b852d97a640f1cb62ffcaf2f95e0a343604fc5f3","4bf1d70ca7f75c603312224dccba323f994fd5a6ac92f3d42cbb01094241ead6","1158597dd43935c6ea4eee694f2e2852b8922d9305b61132d7819a9a96b037c0","fc70c0b27f890da598716fb9bb32885c3d06b6b79e4ad3e5b1271052983e1fb0","892b3b28f4a10515451eae991092c14ee1276c8074bf745241e15c5e6c826b8a","1422c3a2778bfe0991e623b2a8e779a4263b0c6134e6f01cd9bcbcc941280853",{"state":810,"reviewer":811,"selfReview":379,"reviewedAt":812,"method":813},"approved","The library owner","2026-09-20T10:18:37.581Z","owner_bulk","preview-7e1cbbcc4f",1789899597673]