[{"data":1,"prerenderedAt":811},["ShallowReactive",2],{"layer:gravity:deepen":3},{"layer":4,"contentHash":793,"dependencyHashes":794,"approval":805,"releaseId":810},{"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":788,"reviewStatus":789,"authoring":790},1,"gravity","en","deepen","The mathematics behind every number in this topic","G, orbits derived from first principles, Newton’s Moon test in full, and the coincidence Einstein could not ignore","Meet Newton’s law with its constant G, derive orbital and escape speed from scratch, redo Newton’s Moon test in full, explore why gravitational and inertial mass are equal, see why g is not uniform on Earth, and look at the mechanics behind ISRO’s orbit-raising missions.",[13,14,15,16,17],"State and use F = G m1 m2 ÷ r², and show it reduces to weight = mass × g at a world’s surface.","Derive orbital speed and escape velocity from force and energy arguments, and explain why their ratio is always √2.","Explain Kepler’s third law as a consequence of Newton’s law, and use it to predict an orbital period.","Describe the equivalence of gravitational and inertial mass and why it puzzled Newton but inspired Einstein.","Explain why g varies slightly across Earth, and how a Hohmann transfer minimises fuel between two orbits.",50,{"title":20,"rows":21},"Lesson plate",[22,25,28,31,34,37,40],{"label":23,"value":24},"Depth","Go deeper",{"label":26,"value":27},"Reading time","≈ 50 minutes",{"label":29,"value":30},"Prior knowledge","Understand and Investigate",{"label":32,"value":33},"Chapters","8",{"label":35,"value":36},"Labs","Orbit transfer, true\u002Ffalse sort, formula match",{"label":38,"value":39},"Key constant","G = 6.674 × 10⁻¹¹ N·m²\u002Fkg²",{"label":41,"value":42},"Big idea","g = G M ÷ r² is Newton’s law, not a separate rule",[44,48,54,60,63,68,80,93,106,131,146,151,154,165,178,183,195,206,231,236,249,254,257,269,273,291,296,299,304,317,322,325,330,355,359,400,405,408,438,443,455,468,473,476,480,502,508,513,516,528,533,559,564,591,635,757,774,778],{"id":45,"type":46,"markdown":47},"d-intro","prose","**Understand** gave you weight = mass × g and treated g as a fixed number for each world. This layer asks the harder question: **where does g itself come from?**\n\nThe answer is Newton's full law of universal gravitation, with an actual number, **G**, that lets you calculate the force between *any* two masses anywhere — a pencil and a planet, two people on a bus, or the Sun and the Earth. From that one equation you can derive g on any world, the exact speed of any orbit, escape velocity, Kepler's laws, and the reason nobody has ever fully explained why gravity and inertia use the same mass.",{"id":49,"type":50,"variant":51,"title":52,"markdown":53},"d-how-to-read","callout","observation","How to use this lesson","This is the most mathematical of the five layers. Work through the algebra with a pencil; every derivation here is short enough to redo from memory once you have done it twice. Nothing requires more than the arithmetic and square roots you already have.",{"id":55,"type":56,"title":57,"eyebrow":58,"navLabel":59},"d-ch1","chapter","Newton’s law, with an actual number","Chapter 01","1 Newton’s law",{"id":61,"type":46,"markdown":62},"d-newton-full","Newton's law of universal gravitation, in full:\n\n**F = G × m₁ × m₂ ÷ r²**\n\nF is the force in newtons, m₁ and m₂ are the two masses in kilograms, r is the distance between their **centres** in metres, and **G** is the **universal gravitational constant** — the same number everywhere in the universe, for every pair of masses that has ever existed.\n\n**G = 0.0000000000667, or 6.674 × 10⁻¹¹ N·m²\u002Fkg².**\n\nG is almost unimaginably small. That smallness is *why* gravity between everyday objects is too weak to notice, and why it takes a mass as big as a planet before the force becomes obvious.",{"id":64,"type":50,"variant":65,"title":66,"markdown":67},"d-def-g-constant","definition","The universal gravitational constant, G","**G** fixes the strength of gravity itself, the way the speed of light fixes how fast light travels. It was first measured in a laboratory by **Henry Cavendish** in 1798, using a delicate torsion balance — two small lead balls on a suspended beam, twisted ever so slightly by the pull of two much larger fixed lead spheres nearby. Cavendish did not set out to find G directly; he was measuring the density of the Earth, but his result is exactly equivalent to a value of G, and it was startlingly close to today's accepted figure.",{"id":69,"type":70,"title":71,"problem":72,"steps":73,"help":78},"d-we-two-friends","worked_example","The pull between two people, calculated exactly","Two friends, each of mass **50 kg**, stand **1 metre** apart. Use Newton's law to find the gravitational force between them.",[74,75,76,77],"F = G × m₁ × m₂ ÷ r².","F = 6.674 × 10⁻¹¹ × 50 × 50 ÷ 1² = 6.674 × 10⁻¹¹ × 2,500 ÷ 1.","F = 1.6685 × 10⁻⁷ N — about **0.00000017 newtons**.","For comparison, one newton is roughly the weight of a small apple. This force is about 6 million times smaller than that — utterly real, precisely predicted, and completely undetectable without laboratory equipment as sensitive as Cavendish’s.",{"simplerExplanation":79},"Multiply the two masses, multiply by G, then divide by the distance squared.",{"id":81,"type":70,"title":82,"problem":83,"steps":84,"help":91},"d-we-kid-earth","Checking the law against something familiar: your own weight","Use Newton’s law directly — not weight = mass × g — to calculate the force between the Earth (mass 5.9722 × 10²⁴ kg) and a 40 kg child standing on its surface (radius 6.371 × 10⁶ m). Compare with the Understand-layer answer of 392 N.",[85,86,87,88,89,90],"F = G × M_Earth × m ÷ R².","F = 6.674 × 10⁻¹¹ × 5.9722 × 10²⁴ × 40 ÷ (6.371 × 10⁶)².","Numerator: 6.674 × 10⁻¹¹ × 5.9722 × 10²⁴ × 40 ≈ 1.5945 × 10¹⁶.","Denominator: (6.371 × 10⁶)² ≈ 4.0590 × 10¹³.","F ≈ 1.5945 × 10¹⁶ ÷ 4.0590 × 10¹³ ≈ **392.8 N**.","This matches the familiar weight = mass × g answer (392 N) almost exactly — because weight = mass × g **is** Newton’s law, with g standing in for G × M ÷ r². They were never two different rules.",{"simplerExplanation":92},"This is the same calculation as weight = mass × g, just written out with G and the Earth’s actual mass and radius instead of the shortcut number 9.8.",{"id":94,"type":95,"items":96},"d-formulas-derive-g","formulas",[97,100,103],{"expression":98,"caption":99},"g = G × M ÷ r²","Surface gravity comes directly from Newton’s law with m₁ = M (the world) and r = its radius.",{"expression":101,"caption":102},"G × M_Earth ÷ R_Earth² ≈ 9.82 N\u002Fkg","Using Earth’s real mass and radius, matching the measured value to three figures.",{"expression":104,"caption":105},"G × M_Mars ÷ R_Mars² ≈ 3.73 N\u002Fkg","Every g value in this topic comes from exactly this calculation.",{"id":107,"type":108,"tone":109,"items":110},"d-spec-constants","spec","amber",[111,115,119,123,127],{"label":112,"big":113,"value":114},"G","6.674 × 10⁻¹¹","N·m²\u002Fkg². Measured by Cavendish, 1798, using a torsion balance.",{"label":116,"big":117,"value":118},"G × M_Earth","3.986 × 10¹⁴","This combination, \"standard gravitational parameter\", appears in every Earth-orbit formula.",{"label":120,"big":121,"value":122},"Two 50 kg friends, 1 m apart","≈ 1.67 × 10⁻⁷ N","Real, calculable, and about 6 million times smaller than the weight of an apple.",{"label":124,"big":125,"value":126},"Force, Sun on Earth","≈ 3.54 × 10²² N","Enough to bend a planet’s path into a year-long orbit.",{"label":128,"big":129,"value":130},"Equivalence precision, 2017","better than 1 in 10¹⁴","The MICROSCOPE satellite’s test of gravitational vs inertial mass.",{"id":132,"type":133,"itemId":134,"prompt":135,"check":136,"hints":140,"feedback":143},"d-practice-force-sun-earth","practice","gravity.deepen-force-sun-earth","The Sun’s mass is about 1.9885 × 10³⁰ kg, the Earth’s is about 5.9722 × 10²⁴ kg, and the distance between their centres is about 1.496 × 10¹¹ m. Using F = G M m ÷ r², the force between them works out to about 3.54 × 10²² N. Roughly how many **billion billion newtons** is that — enter the number in front of 10²¹ N, to 1 decimal place? (3.54 × 10²² = 35.4 × 10²¹)",{"kind":137,"answer":138,"tolerance":139},"number",35.4,0.5,[141,142],"Convert 3.54 × 10²² into ×10²¹ form: move the decimal point one place and reduce the power by one.","3.54 × 10 = 35.4.",{"correct":144,"incorrect":145},"Correct: 3.54 × 10²² N = **35.4 × 10²¹ N**, an almost incomprehensibly large force — and yet it only accelerates the enormous Earth into a gentle year-long orbit.","Move the decimal point: 3.54 × 10²² is the same as 35.4 × 10²¹. The digits in front of 10²¹ are 35.4.",{"id":147,"type":56,"title":148,"eyebrow":149,"navLabel":150},"d-ch2","Deriving orbital speed from first principles","Chapter 02","2 Deriving orbits",{"id":152,"type":46,"markdown":153},"d-derive-orbit","Understand gave you the orbital speed at Earth’s surface as a fact: about 7.91 km\u002Fs. Here is where that number actually comes from.\n\nFor an object moving in a circle, staying on the circle requires a **centripetal force** pointing towards the centre, equal to m × v² ÷ r. For an orbiting object, gravity **is** that force. Setting the two equal:\n\n**G × M × m ÷ r² = m × v² ÷ r**\n\nNotice m — the orbiting object's own mass — appears on both sides and cancels immediately. This is the same \"mass cancels\" idea from Understand, showing up a third time, in orbits. Rearranging what is left:\n\n**v = √(G × M ÷ r)**\n\nThis is the exact formula behind every orbital speed in this topic.",{"id":155,"type":70,"title":156,"problem":157,"steps":158,"help":163},"d-we-derive-v-surface","Using the orbit formula to get 7.91 km\u002Fs exactly","Use v = √(G M ÷ r) to find the circular orbital speed just above Earth’s surface (r ≈ 6.371 × 10⁶ m, using G M = 3.986 × 10¹⁴, the standard value for Earth).",[159,160,161,162],"v = √(G M ÷ r) = √(3.986 × 10¹⁴ ÷ 6.371 × 10⁶).","3.986 × 10¹⁴ ÷ 6.371 × 10⁶ ≈ 6.256 × 10⁷.","v = √(6.256 × 10⁷) ≈ 7,910 m\u002Fs = **7.91 km\u002Fs**.","This is exactly the number quoted in Understand, now derived rather than stated.",{"simplerExplanation":164},"Divide G×M by the radius, then take the square root.",{"id":166,"type":70,"title":167,"problem":168,"steps":169,"help":176},"d-we-derive-escape","Deriving escape velocity from energy, not force","Escape velocity comes from a different argument: **energy**, not force. An object escapes if its kinetic energy at launch is enough to climb out of the planet’s gravity well entirely, ending with (just barely) zero speed infinitely far away. Setting kinetic energy equal to the gravitational potential energy it must overcome: ½ m v² = G M m ÷ r. Solve for v.",[170,171,172,173,174,175],"Start from ½ m v² = G M m ÷ r.","The mass m cancels again, on both sides: ½ v² = G M ÷ r.","Multiply both sides by 2: v² = 2 G M ÷ r.","Take the square root: **v = √(2 G M ÷ r)**.","Compare with the orbital-speed formula v = √(G M ÷ r): escape velocity is exactly **√2 times** the orbital speed, for any world at all — not a coincidence, but a direct algebraic consequence.","Numerically for Earth: √2 × 7.91 = **11.19 km\u002Fs**, matching the value used throughout this topic.",{"simplerExplanation":177},"Escape velocity squared is exactly double orbital velocity squared, so escape velocity is √2 times bigger, always.",{"id":179,"type":50,"variant":180,"title":181,"markdown":182},"d-aha-root2","aha","The same √2, everywhere in the universe","This ratio — escape velocity = √2 × orbital velocity — holds for **every** planet, moon, star or black hole, because the derivation never used a specific value of G, M or r. It is a property of circles and energy, not of Earth in particular. Jupiter's escape velocity (about 59.5 km\u002Fs) is √2 times its low-orbit speed, and the same will be true of a planet in another solar system that nobody has ever visited.",{"id":184,"type":70,"title":185,"problem":186,"steps":187,"help":193},"d-we-jupiter-orbit","The same two formulas, on a much bigger world","Jupiter's cloud tops are 7.1492 × 10⁷ m from its centre, and Jupiter's mass is 1.8982 × 10²⁷ kg. Find the orbital and escape speeds just above the cloud tops.",[188,189,190,191,192],"G M for Jupiter = 6.674 × 10⁻¹¹ × 1.8982 × 10²⁷ ≈ 1.267 × 10¹⁷.","Orbital speed: v = √(G M ÷ r) = √(1.267 × 10¹⁷ ÷ 7.1492 × 10⁷) = √(1.772 × 10⁹) ≈ 42,100 m\u002Fs = **42.1 km\u002Fs**.","That alone is more than five times Earth’s 7.91 km\u002Fs — because Jupiter has about 318 times Earth’s mass, and even though it is also much bigger, the mass wins by far.","Escape speed: multiply by √2. 42.1 × 1.4142 ≈ **59.5 km\u002Fs**, over 190,000 km\u002Fh.","This is exactly why Jupiter is such a difficult planet to leave: any spacecraft that ever lands or flies close would need more than five times the speed budget of an Earth launch just to break free.",{"simplerExplanation":194},"Same two formulas as Earth, just with Jupiter’s much bigger mass and radius plugged in.",{"id":196,"type":70,"title":197,"problem":198,"steps":199,"help":204},"d-we-cannon-exact","Newton’s cannon, made exact","Understand argued informally that \"fall 5 m while going 8 km sideways\" gives an orbit. Make it exact: a circular orbit needs the sideways fall in one second, ½ g t², to equal the amount the round Earth’s surface drops away, v² t² ÷ (2R), in that same second. Show this leads to v = √(g R), and that it matches v = √(G M ÷ R).",[200,201,202,203],"Set the two \"falls\" equal for a one-second interval: ½ g (1)² = v² (1)² ÷ (2R).","Simplify: g ÷ 2 = v² ÷ (2R), so g = v² ÷ R, which rearranges to **v² = g R**, or v = √(g R).","Put in Earth’s numbers: v = √(9.82 × 6,371,000) = √(62,563,220) ≈ **7,910 m\u002Fs**, matching 7.91 km\u002Fs exactly.","Now check it is the same as v = √(G M ÷ R): since g = G M ÷ R², we have g R = G M ÷ R² × R = G M ÷ R, so √(g R) = √(G M ÷ R). Identical formula, arrived at two completely different ways — one purely geometric, one from force and circular motion.",{"simplerExplanation":205},"The \"falling matches the curve\" picture and the \"force equals centripetal force\" picture give exactly the same number, because g is secretly G M \u002F R² all along.",{"id":207,"type":208,"caption":209,"columns":210,"rows":215},"d-table-kepler","table","Kepler’s third law tested: period squared grows with distance cubed",[211,212,213,214],"Orbit","Radius (Earth radii or AU)","Predicted period (from r^1.5 scaling)","Actual period",[216,221,226],[217,218,219,220],"ISS","≈ 1.06 Earth radii","(reference orbit)","≈ 92.4 min",[222,223,224,225],"Geostationary","≈ 6.62 Earth radii","(1.06 ÷ 6.62)⁻¹·⁵ × 92.4 min ≈ 23.93 h","23 h 56 min, by design",[227,228,229,230],"Mars around the Sun","1.524 AU (Earth = 1 AU)","1.524^1.5 × 365.25 days ≈ 687.2 days","≈ 687 days",{"id":232,"type":50,"variant":233,"title":234,"markdown":235},"d-nuance-kepler-history","nuance","Kepler found the pattern; Newton found the reason","**Johannes Kepler** worked out, between 1609 and 1619, purely by patiently fitting curves to decades of naked-eye observations (mostly Tycho Brahe’s), that planets move on ellipses and that the square of a planet’s period is proportional to the cube of its distance from the Sun. He had no idea *why*.\n\nNewton’s law explains it completely: apply v = √(GM\u002Fr) and T = 2πr\u002Fv together, and T² ∝ r³ falls straight out of the algebra. A pattern found by decades of careful measurement, and a law derived from pure reasoning about falling apples, turned out to be the same fact seen from two directions.",{"id":237,"type":133,"itemId":238,"prompt":239,"check":240,"hints":243,"feedback":246},"d-practice-mars-period","gravity.deepen-mars-period","A newly discovered dwarf planet orbits the Sun at 4 AU (four times Earth’s distance). Using Kepler’s third law (period in years = distance in AU, to the power 1.5), roughly how many **Earth years** does one orbit take? Round to the nearest whole year.",{"kind":137,"answer":241,"tolerance":139,"unit":242},8,"years",[244,245],"Raise 4 to the power 1.5: that is 4 × √4.","√4 = 2, so 4 × 2 = 8.",{"correct":247,"incorrect":248},"Correct: 4^1.5 = 4 × √4 = 4 × 2 = **8 years**. Four times further out takes eight times as long to go round.","4 to the power 1.5 means 4 × √4 = 4 × 2 = 8. So the orbit takes about 8 Earth years.",{"id":250,"type":56,"title":251,"eyebrow":252,"navLabel":253},"d-ch3","Ellipses: why a spacecraft speeds up and slows down","Chapter 03","3 Speed on an ellipse",{"id":255,"type":46,"markdown":256},"d-ellipses-intro","Every orbit lab so far has shown circles alongside stretched ellipses without dwelling on one detail: on an ellipse, **speed is not constant**. A spacecraft on an elliptical path moves fastest at its closest point to Earth (**perigee**) and slowest at its farthest point (**apogee**).\n\nThis is **Kepler's second law**, found (like his third) purely from observation, decades before Newton: a line from the planet to the Sun sweeps out **equal areas in equal times**. Close to the Sun, that line is short, so the planet must move quickly to sweep the same area; far away, the line is long, so it can move slowly and still sweep the same area.",{"id":258,"type":70,"title":259,"problem":260,"steps":261,"help":267},"d-we-vis-viva","How much faster at perigee than at apogee?","A spacecraft is in an elliptical orbit with perigee 300 km above Earth’s surface and apogee 200,000 km above the surface. Using the vis-viva equation v = √(G M × (2 ÷ r − 1 ÷ a)), where a is the semi-major axis, find the speed at each point and their ratio.",[262,263,264,265,266],"Semi-major axis a = (r_perigee + r_apogee) ÷ 2 = (6,671 + 206,371) ÷ 2 ≈ **106,521 km**.","At perigee (r = 6,671 km): v = √(GM × (2 ÷ 6,671,000 − 1 ÷ 106,521,000)) ≈ **10.76 km\u002Fs**.","At apogee (r = 206,371 km): v = √(GM × (2 ÷ 206,371,000 − 1 ÷ 106,521,000)) ≈ **0.348 km\u002Fs**.","Ratio: 10.76 ÷ 0.348 ≈ **30.9** — the spacecraft is almost 31 times faster at perigee than at apogee.","Check against the distance ratio: r_apogee ÷ r_perigee = 206,371 ÷ 6,671 ≈ 30.9 as well. This is Kepler’s second law in numbers: speed and distance from the centre are in exactly inverse proportion at these two special points, so that the \"sweep rate\" stays the same throughout the orbit.",{"simplerExplanation":268},"Near the Earth, an elliptical orbit moves fast; far away, it moves slowly — by exactly the same ratio as the two distances, flipped upside down.",{"id":270,"type":50,"variant":180,"title":271,"markdown":272},"d-aha-mangalyaan-ellipse","Why this matters for a mission like Mangalyaan","Mangalyaan spent its early weeks in exactly this kind of stretched ellipse around Earth, swinging in close to perigee (where its engine fired, because a burn is most effective where you are moving fastest and closest) and coasting slowly out near apogee, far from Earth, before the trans-Mars injection burn. Every \"orbit-raising\" step in Investigate and Deepen relies on this same fact: a burn at the *fast*, *close* point of an orbit is the most efficient place to add energy.",{"id":274,"type":275,"prompt":276,"options":277,"explanation":290},"d-predict-ellipse-speed","prediction","A comet follows a very stretched ellipse around the Sun, spending most of its time far out in the outer Solar System and swinging in close to the Sun only briefly every few decades. Where is it moving fastest?",[278,281,284,287],{"id":279,"label":280},"a","Far from the Sun, because there is more room to accelerate",{"id":282,"label":283},"b","Close to the Sun, at its perihelion (closest point)",{"id":285,"label":286},"c","Its speed never changes",{"id":288,"label":289},"d","Exactly halfway between its closest and farthest points","**(b), at perihelion.** Exactly like a satellite at perigee, a comet moves fastest at the point of its orbit closest to the massive body it circles. This is why comet-hunters get only a short, dramatic window to observe a comet at its brightest and fastest, followed by decades of it crawling slowly through the outer darkness.",{"id":292,"type":56,"title":293,"eyebrow":294,"navLabel":295},"d-ch4","Newton’s Moon test, in full","Chapter 04","4 The Moon test, in full",{"id":297,"type":46,"markdown":298},"d-moon-test-full","Investigate showed you the two numbers side by side. Here is the reasoning behind each one, precisely.\n\n**The predicted acceleration** comes from taking the surface value g = 9.8 N\u002Fkg and scaling it down by the inverse square of how much further away the Moon is than the Earth’s surface: a distance ratio of 384,400 ÷ 6,371 ≈ **60.3**.\n\npredicted a = 9.8 ÷ 60.3² ≈ 9.8 ÷ 3,640 ≈ **0.00269 m\u002Fs²**\n\n**The observed acceleration** treats the Moon as a satellite on a circular path and uses pure kinematics: for circular motion, centripetal acceleration = v² ÷ r. The Moon’s orbital speed, from its known circumference and 27.32-day period, is about 1.02 km\u002Fs, and its distance is 384,400 km:\n\nobserved a = v² ÷ r = (1,023)² ÷ 384,400,000 ≈ **0.00272 m\u002Fs²**\n\nThe two agree to within about **1.2 %** — well inside the accuracy Newton could have expected from 17th-century measurements of the Earth’s size and the Moon’s distance.",{"id":300,"type":50,"variant":301,"title":302,"markdown":303},"d-model-limit-moon-test","model_limit","What this simplified test leaves out","The real historical calculation, and a fully rigorous modern one, correct for a few subtleties this version skips: the Moon and Earth actually orbit their common **centre of mass** (the barycentre), which sits inside the Earth but not at its centre; the Moon’s orbit is not perfectly circular; and precise modern distances come from bouncing laser light off mirrors astronauts left on the Moon, accurate to centimetres. None of these change the conclusion — they tighten the agreement even further, from about 1 % to a great many more decimal places.",{"id":305,"type":275,"prompt":306,"options":307,"explanation":316},"d-predict-mars-moon-test","Suppose you wanted to run the same style of test on Mars’s moon Phobos, which orbits much closer to Mars than our Moon orbits Earth. What would you need to know to predict Phobos’s acceleration using the inverse-square method?",[308,310,312,314],{"id":279,"label":309},"Only Phobos’s mass",{"id":282,"label":311},"Mars’s surface gravity and the ratio of Phobos’s orbital distance to Mars’s radius",{"id":285,"label":313},"Only the colour of Mars’s surface",{"id":288,"label":315},"Nothing; the test only works for Earth’s Moon","**(b).** The method is completely general: take the surface gravity of *whatever world you are orbiting*, and scale it by the inverse square of the distance ratio for *that* system. Phobos’s own mass does not enter into how strongly Mars pulls on it, for exactly the reason Understand explained: gravitational acceleration does not depend on the falling (or orbiting) object’s own mass.",{"id":318,"type":56,"title":319,"eyebrow":320,"navLabel":321},"d-ch5","The coincidence Newton could not explain","Chapter 05","5 Newton’s puzzle",{"id":323,"type":46,"markdown":324},"d-equivalence-deep","Go back to Understand’s Chapter 4: mass cancels in free fall because gravitational force and inertia both scale with the same mass. Stated as an equation:\n\n**gravitational mass (in F = G M m ÷ r²) = inertial mass (in F = m a)**\n\nThese come from two entirely different definitions. Gravitational mass measures how strongly a world pulls on an object. Inertial mass measures how stubbornly the object resists being pushed, by *anything* — a rocket engine, a cricket bat, a car crash — with no gravity involved at all. There is no obvious reason these should be the same number.\n\nEvery experiment ever performed says they are equal, to a precision better than **one part in 10¹⁴** in the most careful modern tests using satellites built specifically to check it. Newton knew about the equality and used it, but had no explanation for *why* it should be true. It sat, unexplained, for over two centuries.",{"id":326,"type":50,"variant":327,"title":328,"markdown":329},"d-example-galileo-proof","example","Galileo’s proof that needed no dropping at all","Before ever climbing a tower, Galileo is credited with a pure argument by **contradiction** that heavy and light objects must fall at the same rate — reasoning alone, no experiment required.\n\nSuppose, for the sake of argument, that a heavy cannonball genuinely falls faster than a light musket ball. Now tie the two together with a short cord and drop them as one object.\n\n**Argument A.** The light ball, falling slower on its own, should act as a brake on the heavy one, dragging on the cord and slowing the combination down. So the tied pair should fall **slower** than the cannonball alone.\n\n**Argument B.** Tied together, the two balls make one object **heavier** than the cannonball alone. If heavier objects fall faster, the tied pair should fall **faster** than the cannonball alone.\n\nBoth arguments follow from the same starting assumption, and they flatly contradict each other. The only way out of the contradiction is to reject the assumption: heavy and light objects **cannot** fall at different rates because of their weight. Nothing was dropped, and nothing was timed — the conclusion came from logic alone, decades before anyone in this topic mentions a stopwatch.",{"id":331,"type":332,"title":333,"items":334},"d-timeline-equivalence","timeline","From a puzzling coincidence to a foundation of physics",[335,339,343,347,351],{"time":336,"title":337,"text":338},"1687","Newton, Principia","Uses the equality of gravitational and inertial mass freely, and tests it with pendulums of different materials, but offers no explanation for why it holds.",{"time":340,"title":341,"text":342},"1889-1908","Eötvös","A Hungarian physicist tests the equality with a sensitive torsion balance, to about one part in a billion, finding no difference at all.",{"time":344,"title":345,"text":346},"1907","Einstein’s \"happiest thought\"","Einstein realises that someone falling freely feels no gravity at all — the seed of the equivalence principle.",{"time":348,"title":349,"text":350},"1915","General relativity","Einstein builds a full theory of gravity from the idea that gravitational and inertial mass are equal for a deep reason: gravity is not really a force, but the curving of spacetime itself.",{"time":352,"title":353,"text":354},"2017","MICROSCOPE satellite","A dedicated French space mission tests the equivalence to about one part in 10¹⁴ — the most precise test yet, and still no measurable difference.",{"id":356,"type":50,"variant":180,"title":357,"markdown":358},"d-aha-clue-not-coincidence","A clue hiding in plain sight","Newton treated the equality of the two masses as a useful fact to be tested and relied upon. Einstein treated the same fact as a **clue** demanding an explanation — and the explanation he found (spacetime itself is curved by mass, and everything, regardless of what it is made of, follows the same curved paths) is the seed of general relativity, which Extend introduces properly. The lesson: sometimes the most important step in science is refusing to shrug at a coincidence.",{"id":360,"type":361,"component":362,"componentVersion":5,"config":363,"objective":398,"textAlternative":399},"d-lab-sort-deepen","interactive","sort-game",{"prompt":364,"bins":365,"items":372,"seconds":397},"Is this statement about the deeper theory true or false?",[366,369],{"id":367,"label":368},"true","True",{"id":370,"label":371},"false","False",[373,377,381,385,389,393],{"id":374,"label":375,"bin":367,"why":376},"ds1","F = G m1 m2 \u002F r^2 applies to any two masses, not just planets","Newton’s law is universal: every pair of masses, everywhere.",{"id":378,"label":379,"bin":370,"why":380},"ds2","g = 9.8 is a separate rule from Newton’s law","g = G M \u002F r^2 is Newton’s law applied at a world’s surface, not a different rule.",{"id":382,"label":383,"bin":367,"why":384},"ds3","Escape velocity is always exactly the square root of 2 times orbital speed","It follows directly from comparing the energy and force derivations, for any world.",{"id":386,"label":387,"bin":370,"why":388},"ds4","Kepler discovered his three laws using Newton’s equations","Kepler worked from decades of observation, decades before Newton was born; Newton later explained why Kepler’s patterns hold.",{"id":390,"label":391,"bin":370,"why":392},"ds5","Gravitational mass and inertial mass have been shown to differ under extreme conditions","Every test so far, including the 2017 MICROSCOPE satellite, finds them equal to extraordinary precision.",{"id":394,"label":395,"bin":367,"why":396},"ds6","Newton’s Moon test used two independent methods that happened to agree","One from extrapolating surface gravity, one from the Moon’s observed orbit — agreeing to about 1%.",0,"Sort six statements about the mathematics of gravity into true and false.","**True:** Newton’s law applies to any two masses; escape velocity is always √2 times orbital speed; the Moon test combined two independent methods.\n\n**False:** g = 9.8 is a separate rule (it is Newton’s law applied at a surface); Kepler used Newton’s equations (the reverse is true — Newton later explained Kepler’s patterns); gravitational and inertial mass have ever been shown to differ.",{"id":401,"type":56,"title":402,"eyebrow":403,"navLabel":404},"d-ch6","Why g is not quite the same everywhere on Earth","Chapter 06","6 g is not uniform",{"id":406,"type":46,"markdown":407},"d-g-variation","Newton’s law says g = G M ÷ r², which depends only on distance from the centre — so in principle g should be identical at every point on a perfectly spherical, non-spinning Earth. Real measurements show small but very real differences, for two reasons.\n\n**The Earth is not a perfect sphere.** Spinning for billions of years has flattened it very slightly, so the equator bulges outward: about 21 km further from the centre than the poles. Since the equator is further from the centre, gravity there is very slightly weaker.\n\n**The Earth is spinning.** Standing at the equator, you are travelling in a circle nearly 1,670 km\u002Fh fast, which requires a small centripetal force pointing inward — supplied by gravity, leaving slightly less \"spare\" force to press you onto a scale. At the poles, you are on the spin axis and feel none of this effect.\n\nBoth effects push the same way: **weaker gravity at the equator, stronger at the poles** — about 9.78 N\u002Fkg against 9.83 N\u002Fkg, a difference of roughly half a percent.",{"id":409,"type":208,"caption":410,"columns":411,"rows":415},"d-table-g-variation","How g changes with height, ignoring air: all differences are tiny compared with the 9.8 baseline",[412,413,414],"Location","Approximate g","Compared with sea level",[416,420,424,428,432,434],[417,418,419],"Sea level (mid-latitude average)","9.80 m\u002Fs²","reference",[421,422,423],"Equator","9.78 m\u002Fs²","about 0.2 % weaker",[425,426,427],"Poles","9.83 m\u002Fs²","about 0.3 % stronger",[429,430,431],"Summit of Mount Everest (8,849 m)","9.79 m\u002Fs²","about 0.28 % weaker",[433,430,431],"Cruising airliner altitude (11,000 m)",[435,436,437],"International Space Station (400 km)","8.69 m\u002Fs²","about 11.3 % weaker",{"id":439,"type":50,"variant":440,"title":441,"markdown":442},"d-careful-precision","careful","Precision measurements of g are a real scientific tool","These tiny variations are not just trivia. Geologists use extremely sensitive instruments called **gravimeters** to detect local changes in g caused by what lies underground — dense rock or ore bodies pull very slightly harder than average, while an underground cavity, an oil reservoir or a salt dome pulls slightly less. Mapping these differences is a genuine way to find resources and study the Earth’s structure, all built on the same law that explains a dropped apple.",{"id":444,"type":70,"title":445,"problem":446,"steps":447,"help":453},"d-we-g-orbit-comparison","Comparing g at the ISS with g on the ground, precisely","Using g = G M ÷ r², compare gravity at the ISS’s orbital radius (6,371 + 400 = 6,771 km) with gravity at Earth’s surface (6,371 km), and express the ISS value as a percentage of the surface value.",[448,449,450,451,452],"Ratio of accelerations = (r_surface ÷ r_ISS)², because both use the same G M.","r_surface ÷ r_ISS = 6,371 ÷ 6,771 ≈ 0.9409.","Square it: 0.9409² ≈ 0.8853.","So g at the ISS is about **88.5 %** of the surface value — matching the 8.69 m\u002Fs² figure in the table above (9.82 × 0.8853 ≈ 8.69).","This is the precise version of Understand’s \"about 89 %\" claim, and it is why astronauts are still firmly inside Earth’s gravity, not beyond it.",{"simplerExplanation":454},"Square the ratio of the two distances from Earth’s centre; that ratio squared is how gravity compares.",{"id":456,"type":133,"itemId":457,"prompt":458,"check":459,"hints":462,"feedback":465},"d-practice-everest-weight","gravity.deepen-everest-weight","A mountaineer with equipment has a mass of **70 kg**. Using g = 9.82 N\u002Fkg at sea level and g = 9.79 N\u002Fkg at the summit of Mount Everest, how many **newtons lighter** (to the nearest whole newton) does the same mountaineer become, purely from gaining height, between base camp at sea level and the summit?",{"kind":137,"answer":460,"tolerance":5,"unit":461},2,"N",[463,464],"Find the weight at each g value: weight = mass × g.","Subtract the two weights.",{"correct":466,"incorrect":467},"Correct: 70 × 9.82 = 687.4 N at sea level, and 70 × 9.79 = 685.3 N at the summit — a difference of about **2 N**, roughly the weight of a small lemon, spread across an 8,849 m climb that itself takes weeks and burns tens of thousands of calories.","Weight at sea level: 70 × 9.82 ≈ 687.4 N. Weight at the summit: 70 × 9.79 ≈ 685.3 N. The difference is about 2 N — tiny compared with the effort of the climb itself.",{"id":469,"type":56,"title":470,"eyebrow":471,"navLabel":472},"d-ch7","ISRO’s manoeuvres, with the mechanics behind them","Chapter 07","7 Mission mechanics",{"id":474,"type":46,"markdown":475},"d-isro-mechanics","Investigate showed *what* ISRO’s missions did — raise an orbit gradually, then transfer, then insert. Here is a little more of *why* it works, using the tools from this layer.\n\nEach perigee burn adds speed at a single point of the orbit. Extra speed at any point of an orbit increases the **total energy** of that orbit, and more orbital energy means the object can coast further out before gravity pulls it back — raising the apogee, exactly as predicted in Investigate. Because the burn happens at perigee, and an orbit’s perigee is fixed by where you were and how fast you were going *at that exact point*, later burns keep returning to nearly the same perigee to push the apogee out again.",{"id":477,"type":50,"variant":327,"title":478,"markdown":479},"d-example-transfer-orbit","The Hohmann transfer: the fuel-cheapest route between two circular orbits","The general version of ISRO’s strategy has a name: a **Hohmann transfer**. To move between two circular orbits (say, low Earth orbit and a Mars-crossing path), you fire your engine once to stretch your circular orbit into an ellipse just touching the target orbit, coast most of the way around that ellipse using no fuel at all, then fire once more at the far end to circularise. It uses the least fuel of any two-burn transfer between two circular orbits, which is exactly why space agencies with limited budgets and modestly sized rockets, including ISRO for Mangalyaan, favour it over a faster, more expensive, more direct route.",{"id":481,"type":133,"itemId":482,"prompt":483,"check":484,"hints":496,"feedback":499},"d-practice-transfer","gravity.deepen-transfer-timing","A spacecraft is on a Hohmann transfer ellipse from Earth’s orbit toward Mars’s. At which point of that ellipse should it fire its engine to settle into a circular orbit around Mars?",{"kind":485,"options":486,"correct":495},"choice",[487,489,491,493],{"id":279,"label":488},"Wherever it happens to be when Mars is nearest",{"id":282,"label":490},"At the point of the ellipse farthest from the Sun, which meets Mars’s orbit",{"id":285,"label":492},"At the point closest to the Sun",{"id":288,"label":494},"It never needs a second burn",[282],[497,498],"A Hohmann transfer ellipse is designed so its farthest point from the Sun just touches the target orbit.","The second burn always happens where the transfer ellipse meets the destination orbit.",{"correct":500,"incorrect":501},"Right: the transfer ellipse is shaped so its far point exactly meets Mars’s orbit — that is the only point where a single, well-timed burn can circularise the path around Mars.","A Hohmann transfer’s outer point is specifically chosen to touch the target orbit. The second burn happens there, not at an arbitrary point.",{"id":503,"type":504,"conceptId":505,"relation":506,"explanation":507},"d-connect-four-ops","connection","four-operations","applied_in","Comparing orbital energies and transfer times uses the same arithmetic tools as any other physics calculation — multiplication, ratios and powers.",{"id":509,"type":56,"title":510,"eyebrow":511,"navLabel":512},"d-ch8","Tides: gravity that changes across an object","Chapter 08","8 Tides, briefly",{"id":514,"type":46,"markdown":515},"d-tides-derivation","Gravity pulling on a whole ocean, rather than a single point, produces an effect Understand did not need: a **differential** pull, stronger on the near side of the Earth than on the far side, because the near side is closer to the Moon.\n\nBecause gravity follows an inverse-square law, this difference can be estimated by comparing the pull at the near side (384,400 − 6,371 km from the Moon) with the pull at the far side (384,400 + 6,371 km away). The two are close enough that the difference is small — but not zero, and definitely not nothing.",{"id":517,"type":70,"title":518,"problem":519,"steps":520,"help":526},"d-we-tide-diff","How much stronger is the pull on the near side?","Using the Moon’s pull ∝ 1 ÷ distance², compare the pull at the Earth’s near side (384,400 − 6,371 = 378,029 km from the Moon) with the pull at the far side (384,400 + 6,371 = 390,771 km), as a percentage difference from the pull at the centre.",[521,522,523,524,525],"Pull near side ∝ 1 ÷ 378,029² ≈ 6.9976 × 10⁻¹² (in units of 1 ÷ km²).","Pull far side ∝ 1 ÷ 390,771² ≈ 6.5487 × 10⁻¹².","Pull at the centre ∝ 1 ÷ 384,400² ≈ 6.7676 × 10⁻¹².","Difference between near and far, as a share of the centre value: (6.9976 − 6.5487) ÷ 6.7676 × 100 ≈ **6.6 %**.","Small, but this 6.6 % difference, spread across an entire ocean, is exactly what stretches the water into a bulge on **both** the near and far sides of the Earth at once — which is why most coastlines see **two** high tides a day, not one. The complete mechanism is the subject of the tides topic.",{"simplerExplanation":527},"Compare 1\u002Fdistance² at the near side and the far side; the gap between them, divided by the pull at the centre, gives the tidal stretch.",{"id":529,"type":504,"conceptId":530,"relation":531,"explanation":532},"d-connect-tides-deep","tides","helps_understand","The differential pull across the Earth’s diameter, computed here from the inverse-square law, is the direct cause of the twice-daily tidal bulge explained fully in the tides topic.",{"id":534,"type":361,"component":535,"componentVersion":5,"config":536,"objective":554,"textAlternative":555,"help":556},"d-lab-orbit-transfer","orbit-lab",{"speedKmS":537,"presets":541,"showMoon":553},{"min":538,"max":539,"initial":540},6,12,7.9,[542,544,547,550],{"label":543,"speedKmS":540},"Circular low orbit",{"label":545,"speedKmS":546},"A stretched transfer ellipse",9.5,{"label":548,"speedKmS":549},"A wider transfer ellipse still",10.5,{"label":551,"speedKmS":552},"Escape entirely",11.2,true,"See how a single speed increase at one point of an orbit stretches the far point outward, the same idea behind a Hohmann transfer.","Starting from the 7.9 km\u002Fs circular orbit, increasing the speed at the same point stretches the orbit into a longer and longer ellipse — 9.5 km\u002Fs reaches roughly to the Moon’s distance, 10.5 km\u002Fs further still — while the point of the burn itself stays fixed as the orbit’s near point. This is the geometric heart of every orbit-raising manoeuvre in this lesson, from a single perigee burn to a full Hohmann transfer to Mars.",{"hints":557},[558],"Compare where the near point of the ellipse sits for every speed you try — it should barely move.",{"id":560,"type":56,"title":561,"eyebrow":562,"navLabel":563},"d-ch9","Putting the mathematics together","Chapter 09","9 Putting it together",{"id":565,"type":361,"component":566,"componentVersion":5,"config":567,"objective":589,"textAlternative":590},"d-lab-match-formulas","match-pairs",{"prompt":568,"mode":569,"pairs":570},"Match each formula to what it calculates.","connect",[571,574,577,580,583,586],{"a":572,"b":573},"F = G m1 m2 \u002F r²","The force between any two masses",{"a":575,"b":576},"g = G M \u002F r²","Surface gravity of a world",{"a":578,"b":579},"v = √(G M \u002F r)","Circular orbital speed at distance r",{"a":581,"b":582},"v = √(2 G M \u002F r)","Escape velocity at distance r",{"a":584,"b":585},"T² ∝ r³","Kepler’s third law",{"a":587,"b":588},"a = v² \u002F r","Centripetal acceleration on a circle","Match six formulas from this lesson to what each one calculates.","F = G m1 m2 ÷ r² gives the force between any two masses; g = G M ÷ r² gives a world’s surface gravity; v = √(G M ÷ r) gives circular orbital speed; v = √(2 G M ÷ r) gives escape velocity; T² ∝ r³ is Kepler’s third law; and a = v² ÷ r gives centripetal acceleration on any circular path.",{"id":592,"type":593,"title":594,"terms":595},"d-glossary","glossary","Deeper vocabulary",[596,599,603,607,611,615,619,623,627,631],{"term":112,"meaning":597,"example":598},"The universal gravitational constant, 6.674 × 10⁻¹¹ N·m²\u002Fkg², fixing the strength of gravity everywhere.","F = G m1 m2 \u002F r² uses it directly.",{"term":600,"meaning":601,"example":602},"Centripetal acceleration","The acceleration needed to keep an object moving on a circular path, always pointing towards the centre.","For an orbit, gravity supplies it.",{"term":604,"meaning":605,"example":606},"Barycentre","The common centre of mass that two orbiting bodies actually circle around.","The Earth–Moon barycentre sits inside the Earth, off-centre.",{"term":608,"meaning":609,"example":610},"Hohmann transfer","The lowest-fuel route between two circular orbits: one burn onto a connecting ellipse, one burn to circularise.","Used by Mangalyaan to reach Mars.",{"term":612,"meaning":613,"example":614},"Equivalence principle","The observation, later built into general relativity, that gravitational mass and inertial mass are always equal.","Tested to better than one part in 10¹⁴.",{"term":616,"meaning":617,"example":618},"Gravimeter","A precise instrument that detects tiny local changes in g, used to study what lies underground.","Can find dense ore or hidden cavities.",{"term":620,"meaning":621,"example":622},"Perigee \u002F apogee","The closest and farthest points of an orbit around the Earth.","Speed is highest at perigee, lowest at apogee.",{"term":624,"meaning":625,"example":626},"Perihelion \u002F aphelion","The closest and farthest points of an orbit around the Sun — the same idea as perigee\u002Fapogee, for a different centre.","A comet moves fastest at perihelion.",{"term":628,"meaning":629,"example":630},"Vis-viva equation","The formula v = √(G M (2\u002Fr − 1\u002Fa)) giving speed anywhere on an elliptical orbit, not just at perigee or apogee.","Used to compare speed at two points of the same ellipse.",{"term":632,"meaning":633,"example":634},"Torsion balance","A sensitive instrument that measures a tiny twisting force, used by Cavendish to weigh the Earth and by later scientists to test the equivalence principle.","A beam on a fine wire, twisted by a very small force.",{"id":636,"type":637,"title":638,"questions":639},"d-quiz","quiz","Go deeper: check your reasoning",[640,653,666,679,692,705,718,731,744],{"itemId":641,"prompt":642,"options":643,"correct":282,"why":652},"gravity.deepen-q-formula","Which formula is Newton’s law of universal gravitation?",[644,646,648,650],{"id":279,"label":645},"F = m × g",{"id":282,"label":647},"F = G × m1 × m2 ÷ r²",{"id":285,"label":649},"F = m × a",{"id":288,"label":651},"F = ½ × m × v²","F = G m1 m2 \u002F r² is the full law; F = m g and g = G M \u002F r² are what it becomes at a single world’s surface.",{"itemId":654,"prompt":655,"options":656,"correct":282,"why":665},"gravity.deepen-q-cancel-orbit","In deriving orbital speed from G M m \u002F r² = m v² \u002F r, why does the orbiting object’s own mass not appear in the final formula?",[657,659,661,663],{"id":279,"label":658},"It was never included in the equation",{"id":282,"label":660},"It cancels from both sides, just as it does in free fall",{"id":285,"label":662},"Only very light objects can orbit",{"id":288,"label":664},"The formula only works for objects with zero mass","m appears on both sides of the equation and cancels exactly, which is why a pebble and a space station follow the same orbit at the same speed at the same altitude.",{"itemId":667,"prompt":668,"options":669,"correct":282,"why":678},"gravity.deepen-q-escape-ratio","Why is escape velocity always exactly √2 times the circular orbital speed, for any planet?",[670,672,674,676],{"id":279,"label":671},"It is a coincidence that only happens to be true for Earth",{"id":282,"label":673},"Comparing the energy-based and force-based derivations shows the ratio is always √2, regardless of G, M or r",{"id":285,"label":675},"Escape velocity does not depend on distance, unlike orbital speed",{"id":288,"label":677},"It is only true for perfectly spherical planets","v_escape² = 2 G M \u002F r and v_orbit² = G M \u002F r, so their ratio squared is always 2, whatever the numbers, for every world in the universe.",{"itemId":680,"prompt":681,"options":682,"correct":279,"why":691},"gravity.deepen-q-kepler","What did Kepler contribute, and what did Newton add?",[683,685,687,689],{"id":279,"label":684},"Kepler discovered the laws from observation; Newton explained why they must be true",{"id":282,"label":686},"Newton discovered the laws first; Kepler only confirmed them",{"id":285,"label":688},"They worked together at the same time",{"id":288,"label":690},"Kepler’s laws have since been shown to be wrong","Kepler fitted patterns to decades of observations, decades before Newton was born. Newton later showed those patterns follow directly from F = G M m \u002F r².",{"itemId":693,"prompt":694,"options":695,"correct":282,"why":704},"gravity.deepen-q-equivalence","What is remarkable about gravitational mass and inertial mass?",[696,698,700,702],{"id":279,"label":697},"They have been shown to differ in extreme conditions",{"id":282,"label":699},"They come from unrelated definitions, yet every experiment finds them exactly equal",{"id":285,"label":701},"They are always different by a factor of two",{"id":288,"label":703},"Only Earth-based objects have both kinds of mass","One measures how hard gravity pulls, the other how hard an object is to move by any force at all. Tests find them equal to better than one part in 10¹⁴.",{"itemId":706,"prompt":707,"options":708,"correct":282,"why":717},"gravity.deepen-q-g-equator","Why is g slightly weaker at the equator than at the poles?",[709,711,713,715],{"id":279,"label":710},"The equator has less mass beneath it",{"id":282,"label":712},"The equator bulges further from Earth’s centre, and the spin needs some of gravity’s pull for centripetal force",{"id":285,"label":714},"The Sun is closer to the equator",{"id":288,"label":716},"g is not actually different anywhere on Earth","Both effects point the same way: greater distance from the centre at the bulging equator, plus some of the local gravity being used to keep you moving in Earth’s daily circle.",{"itemId":719,"prompt":720,"options":721,"correct":282,"why":730},"gravity.deepen-q-hohmann","What makes a Hohmann transfer fuel-efficient?",[722,724,726,728],{"id":279,"label":723},"It uses continuous thrust the whole way",{"id":282,"label":725},"It uses only two brief burns, coasting freely on an ellipse in between",{"id":285,"label":727},"It avoids gravity entirely",{"id":288,"label":729},"It is only possible for very small spacecraft","One burn shapes the transfer ellipse, gravity does all the work during the long coast, and a second burn circularises at the destination — the minimum-fuel route between two circular orbits.",{"itemId":732,"prompt":733,"options":734,"correct":282,"why":743},"gravity.deepen-q-ellipse-speed","A spacecraft on an elliptical orbit is 30 times closer to Earth at perigee than at apogee. Roughly how many times faster is it moving at perigee?",[735,737,739,741],{"id":279,"label":736},"About the same speed",{"id":282,"label":738},"About 30 times faster",{"id":285,"label":740},"About 30 times slower",{"id":288,"label":742},"About 5.5 times faster (the square root of 30)","At perigee and apogee, speed and distance from the centre are in exact inverse proportion, so 30 times closer means about 30 times faster — this is Kepler’s second law in numbers.",{"itemId":745,"prompt":746,"options":747,"correct":282,"why":756},"gravity.deepen-q-galileo-logic","What was special about Galileo’s \"tied cannonball\" argument?",[748,750,752,754],{"id":279,"label":749},"It required dropping objects from a great height",{"id":282,"label":751},"It used pure logic — a contradiction — with no experiment at all",{"id":285,"label":753},"It only worked for objects of exactly equal mass",{"id":288,"label":755},"It proved heavy objects fall faster","Assuming heavy objects fall faster leads to two contradictory predictions for a tied-together heavy-and-light pair, so the assumption itself must be false — reasoning alone, no stopwatch needed.",{"id":758,"type":759,"title":760,"points":761},"d-cheat-sheet","summary","Cheat sheet: the equations behind the numbers",[762,763,764,765,766,767,768,769,770,771,772,773],"**F = G m1 m2 ÷ r².** The full law. G = 6.674 × 10⁻¹¹ N·m²\u002Fkg², measured by Cavendish in 1798.","**g = G M ÷ r²** is this same law at a world’s surface. It is not a separate rule from F = G m1 m2 ÷ r².","**Orbital speed: v = √(G M ÷ r).** Derived by setting gravity equal to the centripetal force needed for a circle; the orbiting mass cancels.","**Escape velocity: v = √(2 G M ÷ r) = √2 × orbital speed**, always, for any world — derived from energy, not force.","**Kepler’s third law, T² ∝ r³**, was found from observation decades before Newton, and falls straight out of v = √(GM\u002Fr) combined with T = 2πr\u002Fv.","**Newton’s Moon test:** predicted 0.00269 m\u002Fs² (inverse square from surface g) against observed 0.00272 m\u002Fs² (pure orbital kinematics) — about 1.2 % apart.","**Gravitational mass = inertial mass**, to better than one part in 10¹⁴ in the best modern tests — an unexplained fact until Einstein used it as the seed of general relativity.","**g is not perfectly uniform on Earth:** about 9.78 N\u002Fkg at the equator, 9.83 N\u002Fkg at the poles, because of the equatorial bulge and the Earth’s spin.","**A Hohmann transfer** — one burn onto a connecting ellipse, one burn to circularise — is the minimum-fuel route between two orbits, and the strategy behind ISRO’s Mars and Moon missions.","**Kepler’s second law:** on an ellipse, speed and distance from the centre are in inverse proportion — about 31 times faster at Mangalyaan-style perigee than at its apogee.","**Galileo’s tied-cannonball argument** shows, by pure logic and no experiment, that weight cannot make one object fall faster than another.","**Tides come from a difference, not a total pull:** the Moon’s pull on Earth’s near side is only about 6.6 % stronger than on the far side — small, but enough to raise two ocean bulges at once.",{"id":775,"type":776,"prompt":777},"d-reflect","reflection","Newton used the equality of gravitational and inertial mass without explaining it; Einstein used the same fact as the foundation of a whole new theory. Describe, in your own words, one other \"it just works, nobody knows why\" fact you have met in maths or science (in this topic or elsewhere), and say what it might mean if someone eventually explained it.",{"id":779,"type":780,"sourceIds":781},"d-sources","sources",[782,783,784,785,786,787],"gravity-hyperphysics-gravity","gravity-physicsclassroom-universal-gravitation","gravity-nasa-planetary-factsheet","gravity-britannica-gravity","gravity-wiki-mars-orbiter-mission","gravity-isro-chandrayaan3",[782,783,784,785,786,787],"needs_review",{"generatedBy":791,"notes":792},"claude-code","Draft. Every derivation (orbital speed, escape velocity, Kepler check for Mars, Newton’s Moon test, g at ISS altitude, equatorial vs polar g) computed and asserted in Python. Pending owner review.","b09db45f8e00980c5c1296917119b0444cbb9c5eda398ac402da1cb53cd03255",{"logic:practice":795,"component:sort-game@1":796,"component:orbit-lab@1":797,"component:match-pairs@1":798,"source:gravity-britannica-gravity":799,"source:gravity-hyperphysics-gravity":800,"source:gravity-isro-chandrayaan3":801,"source:gravity-nasa-planetary-factsheet":802,"source:gravity-physicsclassroom-universal-gravitation":803,"source:gravity-wiki-mars-orbiter-mission":804},"3d6b0fe1b15255975a32b0fcd94e8019bc959ad45cbf12e136e86149549c6878","b164f45a2c8ca08f26c450768ff0231e113e9fe45381eddb34dc6d0548596c38","e59aa1a3427977ca02681e15776a720cd37d878bd774ea8c5531e2116616b667","2a8ee4ac87460b4e1175a4bb13c96b03d577db06dde95670eb7fcfe4ad787899","8a68b69eee24a181ce32e96e29a3f9d05221ee983d659a93152038805c82ae06","ad0b66634225092960d8f417463e2d74017822199cc2886992381487d6b8cbd9","4a3fd27e9999dffc0827fe18aa9f0049368cf20460f5ac72888566b7373d5a10","9c57a129761cacac5b2946c90eff33acba841871c7bbd0b69bdd1ec642680db9","a331b2d122b65b125c1004d73ef7cb3806c57d8dcb358b67a4f6cc16473bfe0f","a2bd9540c2f98b549ba5a3f8a01d306df9c3db545b3d8572d93f8f6bed5c6998",{"state":806,"reviewer":807,"selfReview":553,"reviewedAt":808,"method":809},"approved","The library owner","2026-09-20T10:18:37.581Z","owner_bulk","preview-7e1cbbcc4f",1789899598136]