[{"data":1,"prerenderedAt":895},["ShallowReactive",2],{"layer:gravity:investigate":3},{"layer":4,"contentHash":876,"dependencyHashes":877,"approval":889,"releaseId":894},{"schemaVersion":5,"conceptId":6,"locale":7,"depth":8,"revision":5,"title":9,"subtitle":10,"summary":11,"objectives":12,"estimatedMinutes":18,"plate":19,"blocks":43,"sourceIds":871,"reviewStatus":872,"authoring":873},1,"gravity","en","investigate","Test it: predictions, ramps, pendulums and Newton’s own proof","Predict, try, compare and ask \"is it always true?\" — with a ramp, a pendulum, a leaking cup and a spacecraft","Turn gravity into hands-on science: rebuild Galileo’s ramp, design fair tests for mass and shape, weigh the Earth with a pendulum, check whether Newton’s law survives the trip to the Moon, hunt for orbital speed by binary search, and see how ISRO climbs to the Moon and Mars one burn at a time.",[13,14,15,16,17],"Design a fair test that changes one variable at a time, and explain why the coin-vs-paper race is not one.","Explain why repeating an event (like 20 pendulum swings) shrinks a fixed timing error into a much smaller percentage.","Use T = 2π√(L\u002Fg) to predict a pendulum’s period and to calculate g from measured swings.","Describe Newton’s Moon test and explain why two independent methods agreeing is strong evidence.","Explain how repeated perigee burns raise a spacecraft’s orbit, using Mangalyaan and Chandrayaan-3 as real examples.",45,{"title":20,"rows":21},"Lesson plate",[22,25,28,31,34,37,40],{"label":23,"value":24},"Depth","Investigate",{"label":26,"value":27},"Reading time","≈ 45 minutes",{"label":29,"value":30},"Prior knowledge","Understand: weight, falling, orbits",{"label":32,"value":33},"Chapters","10",{"label":35,"value":36},"Labs","Ramp drop, drag comparison, orbit binary search, sort, match",{"label":38,"value":39},"You will need","String, a small weight, paper, coins, a phone stopwatch",{"label":41,"value":42},"Key method","Predict, then test, then compare",[44,48,54,60,63,76,94,115,139,144,149,154,179,184,187,200,205,209,214,217,240,244,257,262,265,278,298,341,355,371,375,380,383,396,424,429,434,437,457,461,474,480,485,488,517,539,544,547,560,581,586,589,602,606,611,614,627,631,674,679,712,737,844,858,861],{"id":45,"type":46,"markdown":47},"i-intro","prose","**Discover** told the story. **Understand** gave you the rules. This layer asks you to be the scientist: **predict** what should happen, **try** it (for real, or in a lab), **compare** your prediction with what actually happens, and ask **is it always true?**\n\nEvery experiment below is one that a real scientist could do — some are hundreds of years old, one used a spacecraft. You will time falls, swing a pendulum to weigh the Earth, test whether Newton's law survives a 384,400 km stretch to the Moon, and pull apart how ISRO actually gets to Mars.",{"id":49,"type":50,"variant":51,"title":52,"markdown":53},"i-how-to-read","callout","observation","How to use this lesson","Before every experiment, **write down your prediction first** — even just a sentence. The whole point of testing is that you might be wrong, and noticing that you were wrong is where the learning happens.\n\nNothing here needs anything you cannot find at home: paper, coins, string, a small weight and a phone stopwatch.",{"id":55,"type":56,"title":57,"eyebrow":58,"navLabel":59},"i-ch1","chapter","Rebuilding Galileo’s ramp","Chapter 01","1 Galileo’s ramp",{"id":61,"type":46,"markdown":62},"i-ramp-why","Galileo could not time a one-second fall accurately — nobody could, before accurate clocks existed. His trick was to **slow gravity down** by rolling a ball down a gentle slope instead of dropping it.\n\nA ball on a slope tilted at an angle still accelerates because of gravity, but only the part of gravity that points *along* the slope gets to act. Tilt the slope only 10° and the effective acceleration is far gentler than a straight drop: about **1.70 m\u002Fs²** instead of 9.8 m\u002Fs² — roughly a sixth as strong, easily slow enough to time with a heartbeat, or today, a phone.",{"id":64,"type":65,"items":66},"i-formulas-ramp","formulas",[67,70,73],{"expression":68,"caption":69},"a = g × sin(angle)","Effective acceleration down a frictionless slope.",{"expression":71,"caption":72},"10°: a = 9.8 × sin(10°) ≈ 1.70 m\u002Fs²","A gentle slope, easy to time by hand.",{"expression":74,"caption":75},"d = ½ a t²","The same square-law rule, just with a smaller a.",{"id":77,"type":78,"prompt":79,"options":80,"explanation":93},"i-predict-ramp","prediction","A ball is released from rest on a 10° ramp, with a = 1.70 m\u002Fs². It has rolled 0.213 m after 0.5 s. If the square law still holds, how far will it have rolled after **1.0 s** — twice the time?",[81,84,87,90],{"id":82,"label":83},"a","About 0.43 m — twice as far",{"id":85,"label":86},"b","About 0.85 m — four times as far",{"id":88,"label":89},"c","About 0.64 m — three times as far",{"id":91,"label":92},"d","About 0.21 m — the same, because the ramp is gentle","**About 0.85 m.** The ramp changes *how strong* the acceleration is, but not the *shape* of the rule. Distance still grows with time squared, so doubling the time still quadruples the distance: 0.213 × 4 = 0.852 m, which matches the computed value of 0.851 m almost exactly.",{"id":95,"type":96,"caption":97,"columns":98,"rows":102},"i-table-ramp","table","Testing the square law on a 10° ramp (a = 1.70 m\u002Fs², computed from a = g sin(10°))",[99,100,101],"Time rolling","Distance rolled","Ratio to the first value",[103,107,111],[104,105,106],"0.5 s","0.213 m","1×",[108,109,110],"1.0 s","0.851 m","4.0×",[112,113,114],"1.5 s","1.914 m","9.0×",{"id":116,"type":96,"caption":117,"columns":118,"rows":122},"i-table-ramp-angles","Choosing a ramp angle: steeper means less slowing-down, and less measuring time to play with",[119,120,121],"Angle","Effective a = g sin(angle)","Compared with a free fall (9.8)",[123,127,131,135],[124,125,126],"5°","0.854 m\u002Fs²","about 1\u002F11 — very gentle, easy to time, but a long ramp needed",[128,129,130],"10°","1.702 m\u002Fs²","about 1\u002F6 — Galileo’s rough range, a good balance",[132,133,134],"20°","3.352 m\u002Fs²","about 1\u002F3 — faster, needs a shorter reaction-time budget",[136,137,138],"30°","4.900 m\u002Fs²","exactly half of g — still much gentler than a vertical drop",{"id":140,"type":50,"variant":141,"title":142,"markdown":143},"i-nuance-ramp-choice","nuance","Why not just use the gentlest possible ramp?","A very gentle ramp (say 2°) gives you the most time to react and measure — but friction, which the simple formula ignores, becomes a bigger share of a very small push, and the ball may not even start rolling from rest at all if friction is stronger than gravity's small component along the slope. Galileo's own ramps were around 1–2° over several metres, chosen by trial and error to be the gentlest angle that reliably worked. There is a real, practical trade-off between \"slow enough to time\" and \"steep enough to move at all\".",{"id":145,"type":50,"variant":146,"title":147,"markdown":148},"i-aha-ramp-confirms","aha","Slower gravity, same law","0.213, 0.851, 1.914 — divide each by 0.213 and you get **1, 4.0, 9.0**, the same square-law pattern Galileo found on a straight drop. Slowing gravity down with a ramp changed the numbers but not the *rule*. That is exactly why the ramp is a fair test: if the square law is really a law, it must survive being slowed down.",{"id":150,"type":50,"variant":151,"title":152,"markdown":153},"i-model-limit-rolling","model_limit","A rolling ball is not quite a sliding block","The maths above treats the ball as if it slid frictionlessly. A real ball **rolls**, and some of its energy goes into spinning rather than moving forward. A solid ball rolling without slipping actually accelerates at only **5\u002F7 of g sin(angle)** — about **1.22 m\u002Fs²** on a 10° slope, not 1.70.\n\nThis does not break the experiment: the *square-law shape* (distance ∝ time²) still holds perfectly, because the rolling correction is just a constant multiplier. It is, however, why a careful lab report about a rolling-ball ramp should not be used to measure g directly without correcting for the spin — a mistake even careful students make.",{"id":155,"type":156,"component":157,"componentVersion":5,"config":158,"objective":175,"textAlternative":176,"help":177},"i-lab-ramp-drop","interactive","gravity-drop",{"worlds":159,"objects":162,"modes":172,"dropHeightM":174},[160,161],"earth","moon",[163,168],{"id":164,"label":165,"massKg":166,"draggy":167},"marble","Marble (rolled, not dropped)",0.02,false,{"id":169,"label":170,"massKg":171,"draggy":167},"steel-ball","Steel ball, dropped straight down",0.2,[173],"drop",2,"Compare a short, gentle fall with what the full drop from the same height would look like.","A short 2 m drop, timed on Earth and the Moon, to build intuition for how much gentler a ramp-style experiment feels compared with a full vertical fall — the timings are short enough to compare against your own reaction time, which sets up the next chapter.",{"simplerExplanation":178},"A short drop is over almost as fast as you can react — that is the whole problem the next chapter solves.",{"id":180,"type":56,"title":181,"eyebrow":182,"navLabel":183},"i-ch2","Testing \"heavy falls faster\" for yourself","Chapter 02","2 Test it yourself",{"id":185,"type":46,"markdown":186},"i-test-heavy","Chapter 4 of Discover told you the answer. Here is how to check it is not just something a book says.\n\n**Round 1.** Drop a coin and a flat sheet of paper together, from the same height, over a hard floor. Predict, then try it, five times.\n\n**Round 2.** Crumple the *same* sheet of paper into a ball and race it against the coin again, five times.\n\n**Round 3 (the real test of the idea).** Take two coins of different value — say a ₹1 coin and a ₹10 coin, which have different masses but similar shapes — and drop them together. Predict, then try it.",{"id":188,"type":78,"prompt":189,"options":190,"explanation":199},"i-predict-two-coins","A ₹1 coin (about 3.09 g) and a ₹10 coin (about 7.74 g) are dropped together from the same height, indoors, over a short distance. What will you most likely see?",[191,193,195,197],{"id":82,"label":192},"The ₹10 coin arrives clearly first",{"id":85,"label":194},"The ₹1 coin arrives clearly first",{"id":88,"label":196},"They land together, or so close it is hard to tell",{"id":91,"label":198},"The lighter coin floats","**They land together.** Both coins are dense and compact, so air resistance barely touches either of them over a metre or two — neither gets close to its terminal velocity. This is the fairest home test of \"does mass affect falling speed\", because unlike paper, both coins have almost the same shape.",{"id":201,"type":50,"variant":202,"title":203,"markdown":204},"i-careful-coins","careful","A fair test needs one difference at a time","Good experiments change **one thing** and hold everything else the same. The coin-versus-paper race actually changes two things at once — mass *and* shape — so it cannot tell you which one matters. The two-coins test changes mass while keeping shape almost the same. The flat-versus-crumpled test keeps mass the same and changes shape.\n\nOnly by running **both** tests can you work out that shape (through air resistance), not mass, is doing the work.",{"id":206,"type":207,"prompt":208},"i-reflect-variables","reflection","Design a third fair test that isolates **size** from **mass** — for example, two balls of the same material and shape but different diameters. Predict what you think will happen, and say what result would surprise you.",{"id":210,"type":56,"title":211,"eyebrow":212,"navLabel":213},"i-ch3","Why a stopwatch is not good enough","Chapter 03","3 The timing problem",{"id":215,"type":46,"markdown":216},"i-reaction-time","Try timing a coin's one-metre fall with a phone stopwatch, starting and stopping it by hand. The fall itself takes about **0.45 seconds**. Your own reaction time — the delay between deciding to tap and your finger actually moving — is typically about **0.2 seconds**, for *each* tap.\n\nThat error is not a rounding problem. It is roughly **44 % of the entire fall time**. You could easily \"measure\" a fall as anywhere from 0.25 s to 0.65 s and never notice anything was wrong, because there is nothing to compare it with.",{"id":218,"type":96,"caption":219,"columns":220,"rows":225},"i-table-timing-error","Why direct stopwatch timing struggles with short, single events",[221,222,223,224],"What you are timing","True duration","Typical reaction-time error","Error as a share of the reading",[226,231,235],[227,228,229,230],"A coin falling 1 m","≈ 0.45 s","≈ 0.2 s","≈ 44 %",[232,233,229,234],"A coin falling from a first-floor balcony (5 m)","≈ 1.01 s","≈ 20 %",[236,237,238,239],"20 swings of a 1 m pendulum","≈ 40.1 s","≈ 0.2 s (once, not per swing)","≈ 0.5 %",{"id":241,"type":50,"variant":146,"title":242,"markdown":243},"i-aha-repeat","The trick: time many repeats, not one instant","You cannot make your reaction time smaller. You *can* make it matter less, by timing something that takes much longer — and the neatest way to stretch a few seconds into a few dozen is to count **repeats** of the same motion rather than one single event.\n\nTwenty swings of a pendulum take about 40 seconds. Your 0.2 s reaction-time slip is still there, once, at the start and once at the end — but now it is being compared with 40 seconds, not less than one. The percentage error collapses from about 44 % to about **0.5 %**. Same stopwatch, same fingers, seventy times better.",{"id":245,"type":78,"prompt":246,"options":247,"explanation":256},"i-predict-which-timing","You want to measure how long it takes a small steel ball to fall exactly 20 cm, as precisely as you can with an ordinary phone. Which method will give the smallest percentage error?",[248,250,252,254],{"id":82,"label":249},"Time one single 20 cm drop by hand",{"id":85,"label":251},"Time ten separate 20 cm drops and average them",{"id":88,"label":253},"Stack ten identical 20 cm drops end to end into one continuous 2 m fall and time that once",{"id":91,"label":255},"All three give the same error","**(c) is best**, for the same reason a pendulum beats a single fall: one longer event has the same one-off reaction-time error, but a much bigger total to divide it into. (b) helps too, because averaging cancels out *random* timing mistakes — but it does nothing about the fixed one-off delay at the start and end of each individual short timing, which (c) avoids by only starting and stopping the watch once.",{"id":258,"type":56,"title":259,"eyebrow":260,"navLabel":261},"i-ch4","Weighing the Earth with a piece of string","Chapter 04","4 The pendulum",{"id":263,"type":46,"markdown":264},"i-pendulum-intro","A **pendulum** is the oldest precision instrument in this whole topic, and you can build one from a metre of string and a small weight (a washer, a key, a small stone) tied at the end.\n\nSwing it through a small angle — no more than about 15° from vertical — and something remarkable happens: the time for one full swing, called the **period**, does not depend on how heavy the bob is, and barely depends on how wide you swing it. It depends on only two things: the **length of the string** and **g**.\n\n**T = 2π × √(length ÷ g)**\n\nNotice mass is not in the formula at all. This is the same \"mass cancels\" idea from Understand, showing up again in a completely different experiment.",{"id":266,"type":78,"prompt":267,"options":268,"explanation":277},"i-predict-pendulum-length","A 0.25 m pendulum has a period of 1.00 s. If you make the string **four times longer** (1.00 m), what happens to the period?",[269,271,273,275],{"id":82,"label":270},"It becomes 4 times longer: 4.00 s",{"id":85,"label":272},"It becomes 2 times longer: 2.00 s",{"id":88,"label":274},"It stays the same: 1.00 s",{"id":91,"label":276},"It becomes 16 times longer: 16.00 s","**It doubles, to about 2.01 s.** The formula has a **square root** of length in it, not length itself. Multiply the length by 4 and the square root of 4 is 2, so the period only doubles. This is worth testing for real: it is a genuinely surprising result the first time you see it.",{"id":279,"type":96,"caption":280,"columns":281,"rows":285},"i-table-pendulum-predict","Predicted periods for four string lengths (T = 2π √(L ÷ 9.8)) — measure these yourself before checking",[282,283,284],"Length","Predicted period","Time for 20 swings",[286,290,294],[287,288,289],"0.25 m","1.00 s","20.1 s",[291,292,293],"0.50 m","1.42 s","28.4 s",[295,296,297],"1.00 m","2.01 s","40.1 s",{"id":299,"type":156,"component":300,"componentVersion":5,"config":301,"objective":339,"textAlternative":340},"i-lab-pendulum-sort","sort-game",{"prompt":302,"bins":303,"items":313,"seconds":338},"Does changing this make the pendulum swing faster, slower, or does it make no difference?",[304,307,310],{"id":305,"label":306},"faster","Swings faster (shorter period)",{"id":308,"label":309},"slower","Swings slower (longer period)",{"id":311,"label":312},"same","No difference",[314,318,322,326,330,334],{"id":315,"label":316,"bin":305,"why":317},"p1","Making the string shorter","T is proportional to the square root of length, so a shorter string gives a shorter period.",{"id":319,"label":320,"bin":311,"why":321},"p2","Using a heavier bob","Mass does not appear in T = 2π√(L\u002Fg) at all.",{"id":323,"label":324,"bin":311,"why":325},"p3","Swinging it through a wider angle (still under about 15°)","For small angles the period is almost independent of amplitude.",{"id":327,"label":328,"bin":308,"why":329},"p4","Taking the same pendulum to the Moon","Smaller g makes the square root bigger, so the period grows — by about 2.46 times on the Moon.",{"id":331,"label":332,"bin":308,"why":333},"p5","Making the string longer","A longer string gives a bigger square root and a longer period.",{"id":335,"label":336,"bin":305,"why":337},"p6","Doing the experiment in a stronger gravity, like Jupiter’s cloud tops","Bigger g makes the square root smaller, so the pendulum swings faster.",0,"Sort six changes to a pendulum experiment by what they do to the period.","Six changes to sort into faster, slower or no difference.\n\n**No difference:** a heavier bob; a wider swing (while still under about 15°) — both surprising the first time you meet them, because \"heavier things should swing differently\" feels intuitive and is wrong here.\n\n**Faster (shorter period):** a shorter string; stronger gravity (Jupiter's cloud tops).\n\n**Slower (longer period):** a longer string; weaker gravity (the Moon, where the same pendulum takes about 2.46 times as long per swing).",{"id":342,"type":343,"title":344,"problem":345,"steps":346,"help":352},"i-we-pendulum-g","worked_example","Measuring g with a pendulum: a real class result","A class ties a small weight to a **0.80 m** string and times **20 complete swings**, getting a total of **35.9 seconds**. Use this to calculate their measured value of g, and compare it with the accepted 9.8 N\u002Fkg.",[347,348,349,350,351],"First find the period of one swing: T = 35.9 ÷ 20 = **1.795 s**.","Rearrange T = 2π√(L\u002Fg) for g: square both sides, T² = 4π²L\u002Fg, so **g = 4π²L ÷ T²**.","Put in the numbers: g = 4 × π² × 0.80 ÷ 1.795² = 31.58 ÷ 3.222 = **9.80 N\u002Fkg**.","Compare with the accepted value: 9.80 against 9.8 — a match to three significant figures.","This is a genuinely good school result. A real class experiment, timing carefully with a stopwatch and averaging several trials of 20 swings, regularly lands within 1–2 % of the true value — proof that you do not need a laboratory full of equipment to weigh the pull of an entire planet.",{"simplerExplanation":353,"anotherExample":354},"Time 20 swings, divide by 20 for one period, then g = 4 × π² × length ÷ period².","The same pendulum on the Moon would take about 4.94 s per swing instead of 2.01 s — almost two and a half times longer, because g there is so much smaller.",{"id":356,"type":357,"itemId":358,"prompt":359,"check":360,"hints":365,"feedback":368},"i-practice-pendulum-g","practice","gravity.investigate-pendulum-g","A different class uses a **1.00 m** string and times 20 swings at **40.5 s**. What value of g, in N\u002Fkg, does their experiment give? Round to two decimal places.",{"kind":361,"answer":362,"tolerance":363,"unit":364},"number",9.71,0.05,"N\u002Fkg",[366,367],"Period = 40.5 ÷ 20.","g = 4 × π² × 1.00 ÷ period².",{"correct":369,"incorrect":370},"Correct (to a small rounding tolerance): T = 40.5 ÷ 20 = 2.025 s, g = 4π² × 1 ÷ 2.025² ≈ **9.71 N\u002Fkg** — close to 9.8, with the gap explained by ordinary experimental error such as air resistance and timing.","Find the period first: 40.5 ÷ 20 = 2.025 s. Then g = 4 × π² × 1.00 ÷ 2.025² ≈ 9.71 N\u002Fkg.",{"id":372,"type":50,"variant":141,"title":373,"markdown":374},"i-nuance-pendulum-limits","What could make a real result drift from 9.8","A measured g a little off from 9.8 does not necessarily mean a mistake. Real pendulums are affected by air resistance (a small drag on the bob and the string), the string stretching slightly, swinging through too wide an angle, and — for a very short or very long timing — the same reaction-time error from Chapter 3, just smaller than before.\n\nScientists handle this by repeating the measurement many times and looking at the spread of results, not by trusting a single run.",{"id":376,"type":56,"title":377,"eyebrow":378,"navLabel":379},"i-ch5","Testing terminal velocity","Chapter 05","5 Testing drag",{"id":381,"type":46,"markdown":382},"i-terminal-test","Understand claimed that **shape** sets terminal velocity, and that a heavier object of the *same shape* falls faster in air. Both claims are testable.\n\n**Test A — same mass, different shape.** Take two identical sheets of paper. Leave one flat; fold the other into a tight paper aeroplane or a ball. Drop from the same height. Predict, then try it.\n\n**Test B — same shape, different mass.** Nest two paper coffee filters or cupcake cases inside each other so they keep the same shape and area, doubling the mass. Drop one filter, then the doubled pair, from the same height. Predict, then try it.",{"id":384,"type":78,"prompt":385,"options":386,"explanation":395},"i-predict-nested-filters","Two identical paper cupcake cases are nested together (doubling the mass, keeping the shape and area almost the same) and dropped against a single case from the same height. What do you expect?",[387,389,391,393],{"id":82,"label":388},"They fall at the same rate; shape is all that matters",{"id":85,"label":390},"The nested (heavier) pair falls faster",{"id":88,"label":392},"The single case falls faster, being lighter",{"id":91,"label":394},"Both float, because paper never falls fast","**The nested pair falls faster.** Same shape means the same drag at any given speed — but the doubled case has twice the weight for that drag to balance, so its terminal velocity is higher. This is the cleanest home version of the \"heavier things fall faster in air\" test, because it isolates mass while holding shape fixed — exactly the opposite isolation to the flat-versus-crumpled paper test.",{"id":397,"type":156,"component":157,"componentVersion":5,"config":398,"objective":419,"textAlternative":420,"help":421},"i-lab-drag-compare",{"worlds":399,"objects":400,"modes":417,"dropHeightM":418},[160],[401,406,410,414],{"id":402,"label":403,"massKg":404,"draggy":405},"filter-1","One paper case",0.003,true,{"id":407,"label":408,"massKg":409,"draggy":405},"filter-2","Two nested cases (double mass)",0.006,{"id":411,"label":412,"massKg":413,"draggy":405},"paper-flat","Flat sheet of paper",0.005,{"id":415,"label":416,"massKg":413,"draggy":167},"paper-ball","Same sheet, crumpled",[173],3,"Compare two fair tests side by side: same shape with different mass, and same mass with different shape.","Four light objects dropped from 3 m on Earth.\n\nThe two nested paper cases have almost the same shape but the doubled case has twice the mass, and it noticeably wins the race — confirming that, with shape held fixed, more mass means a higher terminal velocity.\n\nThe flat sheet and the crumpled version of the *same* sheet have identical mass but very different shape, and the crumpled one wins by a wide margin — confirming that, with mass held fixed, a smaller, more compact shape means less drag and a higher terminal velocity.\n\nTogether the two results show that both mass and shape affect terminal velocity, through the single idea of a balance between weight and drag.",{"hints":422},[423],"Run each pair separately and only compare within a pair — comparing across pairs mixes the two variables again.",{"id":425,"type":50,"variant":426,"title":427,"markdown":428},"i-question-parachute-size","question","A question worth investigating further","A skydiver of mass 70 kg needs a canopy that brings her down at about 5.5 m\u002Fs. A heavier skydiver of mass 100 kg wants the *same* landing speed. Should her canopy be bigger, smaller, or the same size — and roughly how much bigger or smaller? (You do not need the drag formula to reason about the direction of the answer; you do need it to say how much.)",{"id":430,"type":56,"title":431,"eyebrow":432,"navLabel":433},"i-ch6","Does the law reach the Moon? Newton’s own test","Chapter 06","6 Testing the Moon",{"id":435,"type":46,"markdown":436},"i-moon-test-intro","Here is the test that convinced Newton his law was truly **universal** — that the same rule pulling an apple off a tree also steers the Moon. It can be run with only two facts you already have: Earth's surface gravity, and the Moon's distance.\n\n**The prediction.** If gravity really fades as one over the distance squared, then at the Moon's distance — about **60.3 Earth radii** from the centre — the pull should be 60.3² ≈ 3,640 times weaker than at the surface. Predicted acceleration: 9.8 ÷ 3,640 ≈ **0.00269 m\u002Fs²**.\n\n**The independent check.** The Moon's acceleration can also be worked out *without assuming anything about gravity at all* — purely from how fast it moves and how tightly it curves. A body moving at speed v on a circle of radius r has a centripetal acceleration of v² ÷ r. The Moon's observed orbital speed is about **1.02 km\u002Fs**, and its distance is **384,400 km**, which gives an acceleration of about **0.00272 m\u002Fs²**.\n\nTwo completely different routes — one from extrapolating a law measured on the ground, one from watching the sky — land within **about 1 %** of each other.",{"id":438,"type":96,"caption":439,"columns":440,"rows":444},"i-table-moon-test","Newton’s Moon test, redone with today’s numbers",[441,442,443],"Method","What it uses","Result",[445,449,453],[446,447,448],"Predicted (inverse-square law)","Surface g (9.8) and distance ratio (60.3 Earth radii)","≈ 0.00269 m\u002Fs²",[450,451,452],"Observed (pure orbital geometry)","Moon’s speed (1.02 km\u002Fs) and distance (384,400 km), via v² ÷ r","≈ 0.00272 m\u002Fs²",[454,455,456],"Agreement","—","within about 1.2 %",{"id":458,"type":50,"variant":146,"title":459,"markdown":460},"i-aha-moon-test","This is what \"the law is universal\" actually means","It would have been entirely reasonable, before Newton, to guess that gravity is a purely local, Earth-bound thing that has nothing to do with the heavens — after all, nobody had ever touched the Moon.\n\nThe test above is why that guess lost. A law built from watching things fall in an orchard, when stretched out 60 times further than anyone had tested it, correctly predicted how fast something 384,400 km away is curving. That kind of success, tested and not merely asserted, is what turns a good guess into a scientific law.",{"id":462,"type":78,"prompt":463,"options":464,"explanation":473},"i-predict-moon-test-fails","Suppose the 1 % gap between the predicted and observed Moon accelerations had instead come out as a factor of 2 (100 % off), no matter how carefully the calculation was checked. What would that mean?",[465,467,469,471],{"id":82,"label":466},"Nothing; a 100 % gap is still close enough",{"id":85,"label":468},"It would suggest the inverse-square law is wrong, or incomplete, at large distances",{"id":88,"label":470},"It would prove the Moon has no gravity",{"id":91,"label":472},"It would mean the Moon is not really orbiting the Earth","**(b).** Science takes a big, unexplained gap seriously. A 1 % agreement, well within the accuracy of 17th-century measurements of the Earth's size and the Moon's distance, is a strong pass. A 100 % gap would have forced Newton (or anyone since) to change the law itself — perhaps the power should not be exactly 2, or something else is acting at that distance. This is genuinely how the law has been tested and retested for centuries, including by spacecraft.",{"id":475,"type":476,"conceptId":477,"relation":478,"explanation":479},"i-connect-phases","connection","phases-of-the-moon","related_to","The same orbit whose acceleration is tested here is what carries the Moon through its monthly cycle of phases.",{"id":481,"type":56,"title":482,"eyebrow":483,"navLabel":484},"i-ch7","Hunting for the orbit speed","Chapter 07","7 Hunt the orbit speed",{"id":486,"type":46,"markdown":487},"i-orbit-hunt","Understand told you the orbital speed at the surface is about 7.91 km\u002Fs and escape velocity is about 11.19 km\u002Fs. Rather than take those numbers on trust, use the cannon lab as a proper investigation: a **binary search**.\n\nFire at a speed you are confident is too slow (it lands). Fire at a speed you are confident is too fast (it escapes or makes a huge ellipse). Then narrow the gap by trying the middle each time, exactly like guessing a number between 1 and 100.",{"id":489,"type":156,"component":490,"componentVersion":5,"config":491,"objective":511,"textAlternative":512,"help":513},"i-lab-cannon-hunt","orbit-lab",{"speedKmS":492,"presets":495,"showMoon":167},{"min":5,"max":493,"initial":494},13,5,[496,499,502,505,508],{"label":497,"speedKmS":498},"Definitely too slow",4,{"label":500,"speedKmS":501},"Halfway guess",6,{"label":503,"speedKmS":504},"Getting close",7.5,{"label":506,"speedKmS":507},"Circular orbit",7.9,{"label":509,"speedKmS":510},"Definitely escapes",12,"Use a binary search — too slow, too fast, then the middle — to home in on the exact orbital and escape speeds.","Start with 4 km\u002Fs (lands quickly) and 12 km\u002Fs (clearly escapes). Try the midpoint, 8 km\u002Fs: it makes a wide ellipse but still comes back, so it is *just* above orbital speed. Try 7 km\u002Fs: it lands, so it is *just* below. Narrow again between 7 and 8, and you converge on the true value of about **7.91 km\u002Fs** within a few tries.\n\nRepeat the same hunt between 10 km\u002Fs (a huge ellipse, still returns) and 12 km\u002Fs (escapes) to close in on escape velocity, about **11.19 km\u002Fs**.\n\nThis is exactly how you would search for any unknown threshold: bracket it, then halve the gap, again and again.",{"hints":514},[515,516],"Keep a small table of \"speed tried\" and \"landed \u002F orbited \u002F escaped\" as you go — it makes the pattern obvious.","Notice the ratio of your two thresholds: 11.19 ÷ 7.91 ≈ 1.414, which is √2, exactly as Understand predicted.",{"id":518,"type":357,"itemId":519,"prompt":520,"check":521,"hints":533,"feedback":536},"i-practice-binary-search","gravity.investigate-binary-search","You have tried 6 km\u002Fs (lands) and 9 km\u002Fs (escapes into a huge ellipse). What speed should you try next to narrow the gap fastest?",{"kind":522,"options":523,"correct":532},"choice",[524,526,528,530],{"id":82,"label":525},"6.5 km\u002Fs",{"id":85,"label":527},"7.5 km\u002Fs",{"id":88,"label":529},"8.9 km\u002Fs",{"id":91,"label":531},"12 km\u002Fs",[85],[534,535],"A binary search always tries the midpoint of the current bracket.","The midpoint of 6 and 9 is 7.5.",{"correct":537,"incorrect":538},"Right: **7.5 km\u002Fs**, the midpoint of the bracket [6, 9]. Whatever it does, you can then halve the bracket again.","A binary search tries the midpoint of the current range to halve it each time. Halfway between 6 and 9 is 7.5, not any of the other options.",{"id":540,"type":56,"title":541,"eyebrow":542,"navLabel":543},"i-ch8","How ISRO actually gets to the Moon and Mars","Chapter 08","8 ISRO’s strategy",{"id":545,"type":46,"markdown":546},"i-isro-strategy","India's Moon and Mars missions did not fire straight there in one giant push. Both used a strategy you can test the logic of yourself, in the orbit lab: **raise the orbit gradually, one careful burn at a time.**\n\nHere is the idea in miniature. Put a spacecraft in a stretched, oval (elliptical) orbit around the Earth. Every time it swings back to its closest point (**perigee**), fire the engine briefly, forward, in the direction it is already moving. Predict what happens to the *far* point of the orbit (the **apogee**).",{"id":548,"type":78,"prompt":549,"options":550,"explanation":559},"i-predict-perigee-burn","A spacecraft is in an elliptical orbit. Each time it passes its closest point to Earth, its engine fires briefly, speeding it up a little more in the direction it is already travelling. What happens to the orbit over several such burns?",[551,553,555,557],{"id":82,"label":552},"Nothing changes; a small burn cannot affect a whole orbit",{"id":85,"label":554},"The closest point stays the same but the farthest point (apogee) gets higher each time",{"id":88,"label":556},"The whole orbit shrinks",{"id":91,"label":558},"The spacecraft immediately escapes Earth after one burn","**(b).** A short burn at the closest point adds speed right there, which raises how far the spacecraft can coast out to on the opposite side of the orbit — while the closest point, where the next burn will happen, stays fixed by geometry. Repeat this at every perigee and the far point climbs, lap by lap, like pushing a swing a little higher on each pass. This is exactly ISRO's strategy for both Chandrayaan-3 and Mangalyaan (the Mars Orbiter Mission).",{"id":561,"type":562,"title":563,"items":564},"i-timeline-mom","timeline","Mangalyaan: raising an orbit until it reaches Mars",[565,569,573,577],{"time":566,"title":567,"text":568},"5 Nov 2013","Launch","A PSLV rocket — not powerful enough to send the spacecraft straight to Mars — puts it into a stretched Earth orbit instead.",{"time":570,"title":571,"text":572},"Nov 2013","Six orbit-raising burns","The spacecraft fires its own engine at perigee, six separate times, each one stretching the far point of its orbit further out.",{"time":574,"title":575,"text":576},"1 Dec 2013","Trans-Mars injection","A final burn, now that the orbit is stretched far enough, sends the spacecraft out of Earth orbit entirely, on a curving path towards Mars.",{"time":578,"title":579,"text":580},"24 Sep 2014","Mars orbit insertion","The engine fires again, this time to slow down and be captured by Mars’s gravity — 323 days after launch.",{"id":582,"type":50,"variant":583,"title":584,"markdown":585},"i-example-mom-cost","example","Why bother with six small burns instead of one big one?","A single giant burn straight out of Earth orbit needs a much bigger, heavier, more expensive rocket, because a rocket has to carry all its fuel with it from the start — including the fuel needed to lift the *rest* of the fuel. Small burns, spread across several orbits, let a smaller, cheaper rocket do the initial launch, with the spacecraft's own modest engine doing the rest of the work gradually, using Earth's own gravity to help build up speed efficiently at each pass. This patient approach is a large part of why Mangalyaan reached Mars for a famously low cost.",{"id":587,"type":46,"markdown":588},"i-ch3-landing","Landing is the same idea in reverse — and far more urgent, because there is no atmosphere on the Moon to help slow a spacecraft down the way Earth's atmosphere helps a returning capsule.\n\n**Chandrayaan-3's** lander, Vikram, arrived at the Moon travelling at roughly **1.68 km\u002Fs** sideways. To land safely it had to shed nearly all of that speed using only its own engines, arriving at the surface at under **2 metres per second** — its speed cut by roughly a factor of **840**, entirely under its own control, with no second attempt possible once the descent began. It launched on 14 July 2023 and landed on 23 August 2023, 40 days later.",{"id":590,"type":357,"itemId":591,"prompt":592,"check":593,"hints":596,"feedback":599},"i-practice-mom-days","gravity.investigate-mom-transit","Mangalyaan launched on 5 November 2013 and entered Mars orbit on 24 September 2014. How many days after launch did it arrive, including the time spent raising its Earth orbit?",{"kind":361,"answer":594,"tolerance":174,"unit":595},323,"days",[597,598],"Count the days from 5 November 2013 to 24 September 2014.","The answer is a little under a year.",{"correct":600,"incorrect":601},"Correct: **323 days** — the six orbit-raising burns, the trans-Mars cruise and the final capture burn all together.","From 5 Nov 2013 to 24 Sep 2014 is 323 days: about a month of orbit-raising around Earth followed by roughly ten months coasting to Mars.",{"id":603,"type":476,"conceptId":604,"relation":478,"explanation":605},"i-connect-body-systems-2","body-systems","The precise, no-second-attempt engine burns of a landing are a reminder of how unforgiving a world without air can be — the same reason astronauts must exercise hard to counter the effects of long, weightless journeys.",{"id":607,"type":56,"title":608,"eyebrow":609,"navLabel":610},"i-ch9","A kitchen-table test of weightlessness","Chapter 09","9 Falling together",{"id":612,"type":46,"markdown":613},"i-falling-cup","Here is a genuinely surprising experiment you can do with a paper or plastic cup, some water and a bit of care over a sink.\n\nPunch a small hole near the bottom of the cup. Fill it with water and hold it up: water squirts out of the hole, pulled by gravity and pushed by the weight of water above it.\n\nNow predict what happens to that squirting jet at the exact instant you **let the whole cup fall** — drop it a short, safe distance into a bucket or over a sink.",{"id":615,"type":78,"prompt":616,"options":617,"explanation":626},"i-predict-falling-cup","You let go of the leaking cup and it falls freely (with the hole open) for the half-second or so before it lands. What happens to the jet of water coming out of the hole during the fall?",[618,620,622,624],{"id":82,"label":619},"It squirts out faster, because falling adds extra push",{"id":85,"label":621},"It slows to a trickle but does not stop",{"id":88,"label":623},"It stops almost completely while the cup is falling",{"id":91,"label":625},"Nothing changes; the water keeps squirting exactly as before","**(c).** While the cup is falling freely, the water inside is falling too, at exactly the same rate as the cup. There is no longer any extra push from the weight of water above the hole relative to the falling cup — water and cup are weightless *relative to each other* — so the jet all but stops until the cup hits the bucket. This is a hands-on version of exactly what happens to a drifting pen on the ISS: nothing pushes on anything, because everything is falling together.",{"id":628,"type":50,"variant":202,"title":629,"markdown":630},"i-careful-cup","Do this over a sink or outside","Water and a falling cup make a mess. Do this test somewhere that is fine to get wet, and only drop the cup a short, safe distance.",{"id":632,"type":633,"title":634,"prompt":635,"options":636},"i-explorer-drop-tower","explorer","Real laboratories that manufacture free fall","Pick a method scientists actually use to get a few seconds of weightlessness for testing.",[637,650,662],{"id":638,"label":639,"chain":640,"badge":646,"note":649},"drop-tower","A drop tower",[641,642,643,644,645],"Air pumped out of a tall shaft","Sealed capsule released","Capsule falls freely","A few seconds of microgravity","Caught by a cushion below",{"text":647,"tone":648},"Free fall on the ground","yes","Some laboratories have built very tall, evacuated shafts so that a sealed capsule can fall for several seconds with almost no air resistance to interfere, giving researchers a short but genuine and repeatable window of microgravity without leaving the ground — far cheaper than a rocket, though far shorter than a space station.",{"id":651,"label":652,"chain":653,"badge":659,"note":661},"parabolic","A parabolic flight",[654,655,656,657,658],"Aircraft climbs steeply","Engines throttle back","Plane flies a falling arc","Everyone inside floats","Pilot pulls out of the dive",{"text":660,"tone":648},"Free fall in the sky","A large aircraft flies a carefully calculated arc in which the whole plane, briefly, is essentially in free fall, giving around 20 seconds of floating before the pilot must pull out. Many astronauts train this way before ever reaching orbit.",{"id":663,"label":664,"chain":665,"badge":671,"note":673},"iss","The space station",[666,667,668,669,670],"Orbits at 7.7 km\u002Fs","Falls continuously","Falls for months at a time","Crew and station fall together","Long-duration microgravity",{"text":672,"tone":648},"Free fall for months","The most extreme version of the same idea: instead of seconds, the ISS provides months of continuous free fall, which is exactly why it is so useful for studying the longer-term effects of weightlessness on the body — and exactly why those effects (bone and muscle loss) are so much bigger there than in a 20-second flight.",{"id":675,"type":56,"title":676,"eyebrow":677,"navLabel":678},"i-ch10","Bringing the investigations together","Chapter 10","10 Putting it together",{"id":680,"type":156,"component":681,"componentVersion":5,"config":682,"objective":710,"textAlternative":711},"i-lab-match-methods","match-pairs",{"prompt":683,"mode":684,"pairs":685},"Match each investigation to what it actually tested.","connect",[686,689,692,695,698,701,704,707],{"a":687,"b":688},"Ramp at 10°","Whether the square law survives slower gravity",{"a":690,"b":691},"Two coins, same shape","Whether mass alone affects falling speed",{"a":693,"b":694},"Nested paper cases","Whether mass affects terminal velocity when shape is fixed",{"a":696,"b":697},"20-swing pendulum","Turning a small timing error into a tiny percentage error",{"a":699,"b":700},"Newton’s Moon test","Whether the inverse-square law reaches 384,400 km",{"a":702,"b":703},"Binary search on the cannon","Narrowing in on the exact orbital and escape speeds",{"a":705,"b":706},"Six perigee burns","Raising an orbit’s far point gradually with a small engine",{"a":708,"b":709},"The leaking, falling cup","Showing weightlessness is about falling together","Match eight investigations in this lesson to the question each one actually answered.","Eight matches: the ramp tested whether the square law survives slower gravity; two coins of the same shape tested whether mass alone changes falling speed; nested paper cases tested whether mass affects terminal velocity when shape is fixed; the 20-swing pendulum tested how to shrink a timing error into a tiny percentage; Newton's Moon test checked whether the inverse-square law reaches the Moon; the binary search on the cannon narrowed in on the exact orbital and escape speeds; six perigee burns showed how a small engine gradually raises an orbit; and the leaking, falling cup showed weightlessness is about falling together, not an absence of gravity.",{"id":713,"type":714,"title":715,"terms":716},"i-glossary","glossary","New words from this investigation",[717,721,725,729,733],{"term":718,"meaning":719,"example":720},"Perigee","The point in an orbit around Earth that is closest to the planet.","Mangalyaan fired its engine at perigee, six times.",{"term":722,"meaning":723,"example":724},"Apogee","The point in an orbit around Earth that is farthest from the planet.","Each perigee burn raised Mangalyaan’s apogee.",{"term":726,"meaning":727,"example":728},"Binary search","Narrowing in on an unknown value by repeatedly testing the midpoint of a bracket known to contain it.","Used to home in on the exact orbital speed in the cannon lab.",{"term":730,"meaning":731,"example":732},"Fair test","An experiment that changes only one variable at a time, so the result can be trusted.","Nested paper cases changed mass while keeping shape fixed.",{"term":734,"meaning":735,"example":736},"Period (of a pendulum)","The time for one complete swing, out and back.","A 1 m pendulum on Earth: about 2.01 s.",{"id":738,"type":739,"title":740,"questions":741},"i-quiz","quiz","Investigate: check your reasoning",[742,755,768,781,793,806,818,831],{"itemId":743,"prompt":744,"options":745,"correct":85,"why":754},"gravity.investigate-q-ramp","Why did Galileo use a ramp instead of dropping objects straight down?",[746,748,750,752],{"id":82,"label":747},"Ramps remove air resistance completely",{"id":85,"label":749},"Ramps slow the motion down enough to time it accurately",{"id":88,"label":751},"Ramps make heavy objects fall faster",{"id":91,"label":753},"Ramps were the only surface available to him","Slowing gravity’s effect with a gentle slope turns a fall that lasts a fraction of a second into one that lasts several seconds — measurable with the clocks of Galileo’s time.",{"itemId":756,"prompt":757,"options":758,"correct":85,"why":767},"gravity.investigate-q-fair-test","Why is racing two coins of different value a fairer test of \"does mass affect falling speed\" than racing a coin against a flat sheet of paper?",[759,761,763,765],{"id":82,"label":760},"Coins are heavier, so they are easier to see falling",{"id":85,"label":762},"The coins keep the shape roughly constant while mass changes",{"id":88,"label":764},"Paper cannot be used in experiments",{"id":91,"label":766},"There is no difference; both tests are equally fair","A fair test changes one variable at a time. Two coins differ mainly in mass, with similar shape; a coin and a sheet of paper differ in both mass and shape at once.",{"itemId":769,"prompt":770,"options":771,"correct":88,"why":780},"gravity.investigate-q-pendulum-mass","A class doubles the mass of the bob on their pendulum, keeping the length the same. What happens to the period?",[772,774,776,778],{"id":82,"label":773},"It doubles",{"id":85,"label":775},"It halves",{"id":88,"label":777},"It stays the same",{"id":91,"label":779},"It becomes zero","Mass does not appear in T = 2π√(L\u002Fg). The period depends only on length and g.",{"itemId":782,"prompt":783,"options":784,"correct":85,"why":792},"gravity.investigate-q-pendulum-calc","A pendulum of length 0.50 m completes 20 swings in 28.4 s. What period does that give for one swing?",[785,787,788,790],{"id":82,"label":786},"0.71 s",{"id":85,"label":292},{"id":88,"label":789},"2.84 s",{"id":91,"label":791},"14.2 s","Period = total time ÷ number of swings = 28.4 ÷ 20 = 1.42 s, matching the predicted value for a 0.50 m pendulum.",{"itemId":794,"prompt":795,"options":796,"correct":85,"why":805},"gravity.investigate-q-moon-test","What makes Newton’s Moon test convincing?",[797,799,801,803],{"id":82,"label":798},"It uses a single measurement with no way to check it",{"id":85,"label":800},"Two independent methods — extrapolating surface gravity, and watching the Moon’s orbit — agree to about 1 %",{"id":88,"label":802},"It only works if you already believe the law is true",{"id":91,"label":804},"It was carried out using a modern space telescope","A predicted value (from the inverse-square law) and an observed value (from pure orbital geometry) come from completely different kinds of measurement, yet agree closely — strong evidence the law is genuinely universal.",{"itemId":807,"prompt":808,"options":809,"correct":85,"why":817},"gravity.investigate-q-binary","In a binary search for orbital speed between a \"lands\" result at 6 km\u002Fs and an \"escapes\" result at 10 km\u002Fs, what should you try next?",[810,811,813,815],{"id":82,"label":525},{"id":85,"label":812},"8 km\u002Fs",{"id":88,"label":814},"9.5 km\u002Fs",{"id":91,"label":816},"10.5 km\u002Fs","The midpoint of 6 and 10 is 8, which narrows the bracket as efficiently as possible.",{"itemId":819,"prompt":820,"options":821,"correct":85,"why":830},"gravity.investigate-q-perigee","Firing an engine briefly at the perigee (closest point) of an elliptical orbit, in the direction of travel, mainly changes...",[822,824,826,828],{"id":82,"label":823},"the perigee itself, making it lower",{"id":85,"label":825},"the apogee (farthest point), making it higher",{"id":88,"label":827},"nothing, unless the burn is very long",{"id":91,"label":829},"the direction of Earth’s spin","Extra speed at one point of an orbit raises how far out the spacecraft can coast on the opposite side, so the apogee climbs while the perigee — where the burn happened — stays fixed.",{"itemId":832,"prompt":833,"options":834,"correct":88,"why":843},"gravity.investigate-q-cup","Why does the jet from a leaking, falling cup almost stop during the fall?",[835,837,839,841],{"id":82,"label":836},"The water evaporates instantly",{"id":85,"label":838},"Gravity switches off while the cup falls",{"id":88,"label":840},"The water and the cup fall at the same rate, so nothing pushes the water out relative to the cup",{"id":91,"label":842},"Air pressure outside the cup increases","This is a tabletop version of astronaut weightlessness: everything involved is falling together, so there is no relative push left to squeeze the water out of the hole.",{"id":845,"type":846,"title":847,"points":848},"i-cheat-sheet","summary","Cheat sheet: how these investigations were run",[849,850,851,852,853,854,855,856,857],"**Ramps slow gravity down** without changing the rule: a = g sin(angle), and distance still grows with time squared.","**A fair test changes one variable at a time.** Two coins of different mass, same shape, isolate mass; flat versus crumpled paper (same mass) isolates shape.","**Repeating an event beats timing it once.** A 0.2 s reaction-time error is about 44 % of a single 1 s fall, but only about 0.5 % of a 20-swing pendulum timing.","**A pendulum measures g:** T = 2π√(L\u002Fg), so g = 4π²L ÷ T². Mass and (small) amplitude do not affect the period.","**Same shape, more mass → higher terminal velocity.** Same mass, smaller\u002Fdenser shape → also higher terminal velocity. Both were tested by isolating one variable at a time.","**Newton’s Moon test:** predicted acceleration from the inverse-square law (≈0.00269 m\u002Fs²) matches the acceleration implied by the Moon’s actual orbit (≈0.00272 m\u002Fs²) to about 1 %.","**A binary search** (try the midpoint, then halve the bracket) is an efficient way to hunt for an unknown threshold, such as orbital or escape speed.","**ISRO raises orbits gradually:** repeated perigee burns lift the apogee lap by lap, letting a smaller rocket do a bigger job — used for both Mangalyaan and Chandrayaan-3.","**Weightlessness is testable at home:** a leaking cup’s jet nearly stops in free fall, because the water and the cup fall together.",{"id":859,"type":207,"prompt":860},"i-reflect-close","Pick one experiment from this lesson that you have not actually tried. Write a short plan: what you will measure, what you predict, and what result would make you say \"the theory is wrong, or I made a mistake\" rather than \"close enough\".",{"id":862,"type":863,"sourceIds":864},"i-sources","sources",[865,866,867,868,869,870],"gravity-hyperphysics-gravity","gravity-physicsclassroom-free-fall","gravity-physicsclassroom-universal-gravitation","gravity-nasa-planetary-factsheet","gravity-isro-chandrayaan3","gravity-wiki-mars-orbiter-mission",[865,866,867,868,869,870],"needs_review",{"generatedBy":874,"notes":875},"claude-code","Draft. All numbers (ramp accelerations and distances, timing-error percentages, pendulum periods and measured g, Newton’s Moon test figures, mission day counts) computed and asserted in Python. Pending owner review.","74b4ae0b5ae8f04917605490e11830caaa7f2c0ee9d8bf89b4cd2beaada5c073",{"component:gravity-drop@1":878,"component:sort-game@1":879,"logic:practice":880,"component:orbit-lab@1":881,"component:match-pairs@1":882,"source:gravity-hyperphysics-gravity":883,"source:gravity-isro-chandrayaan3":884,"source:gravity-nasa-planetary-factsheet":885,"source:gravity-physicsclassroom-free-fall":886,"source:gravity-physicsclassroom-universal-gravitation":887,"source:gravity-wiki-mars-orbiter-mission":888},"60b1b2628472895c3a36f45610b158c8d87f74e411a582d1c8293f4f80ecf53b","b164f45a2c8ca08f26c450768ff0231e113e9fe45381eddb34dc6d0548596c38","3d6b0fe1b15255975a32b0fcd94e8019bc959ad45cbf12e136e86149549c6878","e59aa1a3427977ca02681e15776a720cd37d878bd774ea8c5531e2116616b667","2a8ee4ac87460b4e1175a4bb13c96b03d577db06dde95670eb7fcfe4ad787899","ad0b66634225092960d8f417463e2d74017822199cc2886992381487d6b8cbd9","4a3fd27e9999dffc0827fe18aa9f0049368cf20460f5ac72888566b7373d5a10","9c57a129761cacac5b2946c90eff33acba841871c7bbd0b69bdd1ec642680db9","8c311b8ddd919ef66d07a631e86b5823f4c525e1b8bf33ae94ffd5f066a7d201","a331b2d122b65b125c1004d73ef7cb3806c57d8dcb358b67a4f6cc16473bfe0f","a2bd9540c2f98b549ba5a3f8a01d306df9c3db545b3d8572d93f8f6bed5c6998",{"state":890,"reviewer":891,"selfReview":405,"reviewedAt":892,"method":893},"approved","The library owner","2026-09-20T10:18:37.581Z","owner_bulk","preview-7e1cbbcc4f",1789899597780]