{"version":1,"lectureId":"01M4NWBG348QD47R7KWRD7CQCP","attempt":0,"publication":{"slug":"gravitazione-universale-e-meccanica-celeste-teoria-derivazioni-e-applicazioni","title":"Gravitazione Universale e Meccanica Celeste: Teoria, Derivazioni e Applicazioni","subject":"physics","summary":"A rigorous university lecture on universal gravitation and celestial mechanics. Starting from Kepler's three empirical laws, it deduces the inverse-square law, separates inertial from gravitational mass through the equivalence principle, and shows how Cavendish measured G. It proves the shell theorem by ring integration, works through a layered planet, the gravity tunnel, escape speed and a binary star system, then solves the two-body problem with polar kinematics, reduced mass and Binet's equation, arriving at the conic orbit and its classification by energy, the exact third law and the limits of the three-body problem. It closes with surface gravity, tidal forces and the solar to lunar ratio, the Lagrange point L1 and an elliptical transfer solved by conservation laws.","metaDescription":"Newton's gravitation from Kepler's laws: shell theorem, two-body conic orbits, tides, Lagrange point L1 and an orbital transfer, derived step by step.","transcript":"Two lines of reasoning meet in this lecture. Kepler found, from years of careful observation, three rules that the planets obey. Newton showed that one force, falling as the inverse square of distance, explains all three. We will build that force, test it, and then derive every orbit it allows. Start with the observations. Here is one planetary orbit, and here is the Sun. The first law says the orbit is an ellipse, and the Sun sits not at its centre but at one focus. The second law concerns speed. Join the Sun to the planet. In one interval of time near the Sun, the line sweeps this green sector. Now let the planet travel round to the far side. There, in the same interval, it sweeps this thin sector, and the two areas are exactly equal. Far from the Sun the planet moves slowly; near it, quickly. The third law compares different planets. The square of the period, divided by the cube of the semi-major axis, is the same number for all of them. These three laws describe the motion; they do not explain it. Newton's argument is cleanest for a nearly circular orbit of radius r. The planet moves uniformly, so the only acceleration is centripetal, pointing at the Sun. So the force is m omega squared r. Writing omega as two pi over the period, the force becomes m times four pi squared over T squared, times r. Now use the third law: T squared equals k r cubed. Substitute, and the period disappears in a telling way. The force on the planet is proportional to its mass and falls as one over r squared. The third law of dynamics finishes the job. The planet pulls the Sun with an equal and opposite force, so the same expression must also be proportional to the mass M of the Sun. The constant four pi squared over k must therefore be G times M. This is the law of universal gravitation. Every pair of masses attracts along the line joining them, with a force G m M over r squared, and G is one universal constant. Look closely at one step we took for granted. On the left of the second law of dynamics sits the inertial mass, the resistance to changes of velocity. In the law of gravitation sits the gravitational mass, the charge that sources and feels the attraction. Dividing, the acceleration carries the ratio m g over m i. Only if these two masses are equal, for every material, does every body fall with the same acceleration, G M over r squared. That equality is the principle of equivalence. It can be tested with a plumb line. On the rotating Earth the bob feels gravity, proportional to m g, and a centrifugal pull, proportional to m i. If the ratio differed between materials, plumb lines of lead and of wood would hang at different angles. They do not, to better than one part in ten trillion in modern torsion experiments. With the two masses equal, a circular orbit follows at once. Gravity supplies the centripetal acceleration, v squared over r, so the orbital speed is the square root of G M over r. Closer orbits are faster. Measuring G is hard, because gravity between laboratory masses is tiny. Cavendish used a torsion balance. A light rod carrying two small masses m hangs from a thin fibre, and a mirror on the fibre reflects a light beam onto a distant scale, magnifying the smallest twist. Two large masses M are brought close, one on each side. Each attracts its neighbour with a force G m M over r squared, and the rod twists until the restoring torque of the fibre balances the torque of the two forces. The torque of the pair is two b times F, with b the lever arm. The fibre answers with kappa theta, where kappa is its calibrated torsion constant. Equating the two and solving gives G, about six point six seven times ten to the minus eleven, in newton metre squared per kilogram squared. Newton's law is written for point masses, yet planets are large. The shell theorem tells us how an extended sphere attracts. Take a thin uniform shell of mass M and radius R, and a point P at distance r from its centre. Slice the shell into rings around the line to P. The ring at angle theta has radius R sine theta, and all of it lies at the same distance s from P. Its share of the mass is M over two, times sine theta d theta. The ring contributes a potential minus G d M over s. The law of cosines links s to theta, and differentiating it gives s d s equal to R r sine theta d theta, which turns the integral over angles into an integral over s. Everything cancels except a plain integral of d s, running from the nearest point of the shell to the farthest, so the potential is minus G M over two R r, times the length of that range. Outside, the range runs from r minus R to r plus R, a length two R. The potential is minus G M over r: exactly that of a point mass at the centre. Now move P inside. The range runs from R minus r to R plus r, a length two r, and the potential becomes minus G M over R, the same everywhere. A constant potential means zero field. Inside a uniform shell there is no gravity at all. Any spherically symmetric body is a nest of such shells. At radius r, the shells outside pull nothing, and the shells inside act as a point mass. So g of r is G times the mass enclosed, over r squared. Consider a body with a hollow core of radius R one, a shell of density rho two out to R two, and an outer layer of density rho three out to R three. The enclosed mass is the integral of four pi r squared rho over the radius. Follow the field outward. In the hollow core it is exactly zero. Through the first shell it grows as r cubed minus R one cubed, over r squared. Through the second it keeps rising, but more gently, because rho three is smaller. Beyond the surface the whole mass acts from the centre, and the field falls as one over r squared. The potential energy is minus m times the integral of g from r out to infinity: continuous everywhere, and flat inside the cavity, where the field vanishes. Here is a striking consequence. Treat the Earth as a uniform sphere and drill a straight frictionless tunnel along a diameter. A body at distance x from the centre feels only the mass inside radius x, which grows as x cubed. So the force grows linearly, minus G M m over R cubed, times x. That is a spring law. The body oscillates harmonically about the centre, with angular frequency the square root of g over R. Putting in numbers, the period is about eighty-four minutes. Drop a stone in at one end and it reaches the other end some forty-two minutes later, at rest, then falls back. Curiously, this is also the period of a satellite skimming the surface. Energy answers the next question: how fast must a body leave the surface never to return? Take zero potential energy at infinity. The minimum launch arrives there with nothing to spare, so its total energy is zero. Solving, the escape speed is the square root of two G M over R, or the square root of two g R: about eleven point two kilometres per second for the Earth. It is exactly root two times the speed of a circular orbit grazing the surface. The same circular orbit reasoning weighs stars. Two stars of equal mass circle their common centre of mass, each at two hundred and twenty kilometres per second, with a period of fourteen point four days. Each star sits at radius r, but the other is two r away. So gravity, G M squared over two r squared, supplies the centripetal force M v squared over r. The radius comes from the motion itself: r is v T over two pi. Solving, M equals four r v squared over G, about one point three times ten to the thirty-two kilograms, some sixty-four solar masses each. This is how astronomers weigh stars they will never visit. To find every possible orbit we need a better description of motion under a central force. Use polar coordinates centred on the source: a distance r and an angle theta, with unit vectors u r pointing outward and u theta pointing along increasing angle. The unit vectors turn as the point moves. The derivative of u r is theta dot u theta, and the derivative of u theta is minus theta dot u r. With these, the velocity has a radial part r dot, and a transverse part r theta dot. Differentiate once more and collect terms. The radial acceleration is r double dot minus r theta dot squared, the second piece being the familiar centripetal term. The transverse acceleration is one over r times the time derivative of r squared theta dot. Now two bodies, m one and m two, attract each other. Their centre of mass moves uniformly, so we watch only the separation vector r, from one body to the other. Subtracting the two equations of motion gives a single one, for a fictitious particle of reduced mass mu, attracted to a fixed centre by the force minus k over r squared, with k equal to G m one m two. The force is central, so it exerts no torque, and the angular momentum L equals mu r squared theta dot, constant. The transverse acceleration vanishes, and the area swept per unit time, L over two mu, is constant. That is the second law of Kepler, now derived. The energy is conserved too. Kinetic energy splits into radial motion and a centrifugal term, L squared over two mu r squared, which acts like a repulsive barrier added to the gravitational potential minus k over r. The idea of Binet is to describe the orbit by r as a function of theta, and to use u equals one over r. The chain rule and constant L turn time derivatives into angle derivatives: r dot becomes minus L over mu, times d u by d theta. Differentiating again, r double dot is minus L squared over mu squared, u squared, times the second derivative of u. Insert this into the radial equation, and every power of u lines up. The result is astonishingly simple: the second derivative of u, plus u, equals the constant mu k over L squared. It is the equation of a harmonic oscillator in the angle, shifted by a constant. Its general solution is the constant plus a cosine. Choosing the orientation, u equals mu k over L squared, times one minus epsilon cosine theta. Invert it, and r of theta is D over one minus epsilon cosine theta, with D equal to L squared over mu k. This is the polar equation of a conic with its focus at the centre of force. The first law of Kepler appears as a theorem, and epsilon, the eccentricity, decides the shape. Evaluate the energy at closest approach, and the eccentricity turns out to be the square root of one plus two E L squared over mu k squared. The sign of the energy now classifies every orbit. At the minimum allowed energy, epsilon is zero: a circle. For negative energy, epsilon lies below one: a closed ellipse, a bound orbit like those of the planets. At exactly zero energy, epsilon equals one: a parabola, the borderline case of a body that just escapes. For positive energy, epsilon exceeds one: a hyperbola, an open path that swings once past the centre and leaves forever, like a comet or a probe on a flyby. The ellipse also yields the third law, now exactly. Its area is pi a b, swept at the constant rate L over two mu, so the period is pi a b divided by that rate. Square it, and use the geometric fact that b squared equals a times D. The angular momentum cancels, and T squared equals four pi squared a cubed over G times the total mass. The constant of the third law depends, slightly, on the mass of the planet itself. Add a third body and this beautiful solvability is lost. Poincaré showed that the three-body problem has no general solution in closed form, so the Sun, Earth and Moon are followed numerically, advancing positions and velocities in small time steps. Back on Earth, the measured g is not one fixed number. With height h, the distance to the centre grows, and expanding to first order, g falls by the fraction two h over R. Climb one kilometre and g drops by about three parts in ten thousand. Gravimeters detect far smaller changes, enough to map dense rock underground. Latitude matters too, because the Earth rotates. A point at latitude lambda circles the axis at distance R cosine lambda, and in the rotating frame it feels a centrifugal acceleration directed away from the axis. Its component along the vertical reduces the effective g by omega squared R cosine squared lambda. At the equator that is about three hundredths of a metre per second squared, a third of a percent; the flattening of the Earth adds a comparable effect. Tides arise not from the pull of the Moon itself, but from its variation across the Earth. The whole Earth falls toward the Moon with the acceleration at its centre. The near side is pulled a little more, the far side a little less. Subtract the central acceleration and expand for r much smaller than d. The difference is two G M r over d cubed: tidal effects fall as the inverse cube of distance, like the field of a dipole, because they are a difference of two nearly equal pulls. On the sides, the pull converges slightly inward. The result is two bulges, one facing the Moon and one opposite, and as the Earth turns beneath them each coast sees two high tides a day, about twelve hours apart. Now compare the Sun. Its mass is twenty-seven million times that of the Moon, but it is about three hundred and ninety times farther away, and the cube of that ratio wins. The solar tide is about forty-six percent of the lunar one: roughly half. At new and full moon, the syzygies, Sun, Earth and Moon line up, the two tides add, and we get spring tides. At the quarter moons they partly cancel. What we observe on a coast depends on geography. Each basin has its own natural period, so resonance amplifies or suppresses the forcing: barely half a metre at the Po delta, fifteen metres at Mont Saint-Michel. The rock itself is not rigid. The solid Earth rises and falls twice a day, and at CERN the twenty-seven kilometre ring of the collider stretches by about a millimetre, enough to shift the beam energy, so physicists correct for the Moon. One last equilibrium. Between the Sun and the Earth lies a point where a small body can orbit the Sun in step with the Earth. The pull of the Earth partly cancels that of the Sun, so the body needs less centripetal force than a free orbit at that radius would. Require the net pull to equal omega squared times the distance from the Sun, with the angular velocity of the Earth itself. Expanding to first order in x over d, the distance from the Earth is d times the cube root of M E over three M S. That is about one and a half million kilometres, one percent of the way to the Sun. Solar observatories are parked there, with an uninterrupted view of the Sun. Finally, a manoeuvre that uses everything at once. A ship circles a planet of radius R at radius two R. It releases a capsule at point A, which must land at B, diametrically opposite, on the surface. The free path of the capsule is half an ellipse, with A as apogee and B as perigee. Angular momentum fixes the speeds: v A times two R equals v B times R, so the capsule is twice as fast at B. Energy conservation between A and B adds a second equation. Substituting v B equals two v A, the capsule must leave A with speed the square root of G M over three R. But the ship moves at the square root of G M over two R, which is faster. So the capsule must be thrown backward, against the direction of the orbit, with a relative speed of about point one three root G M over R. Braking, not pushing, brings it down. Released this way, it falls along the ellipse and meets the surface at B with speed twice the square root of G M over three R. Every step used only the two conservation laws that the orbits of Kepler taught us to trust.","watch":{"version":1,"scenes":[{"title":"From Kepler to Newton","start":0,"end":261.8206458333333,"objects":{"big_l":"a Circle [blue] drawn in lab (center=(-1.2, -0.85), radius=0.28, filled=True)","big_r":"a Circle [blue] drawn in lab (center=(1.2, 0.85), radius=0.28, filled=True)","bob":"a Point [yellow] drawn in plumb (location=(0.6, -1.0))","card":"a Title that says \"Physics I: Classical Mechanics — Gravitazione Universale e Meccanica Celeste: Teoria, Derivazioni e Applicazioni\"","derivation":"a Derivation [text] that says \"$F &= m omega^2 r \\ &= m frac(4 pi^2, T^2) r \\ &= frac(4 pi^2, k) thin frac(m, r^2) \\ frac(4 pi^2, k) &= G M$\"","derivation_2":"a Derivation [text] that says \"$m_i thin a &= G frac(m_g M, r^2) \\ a &= frac(m_g, m_i) thin frac(G M, r^2) \\ a &= frac(G M, r^2) \\ frac(v^2, r) &= frac(G M, r^2) \\ v &= sqrt(frac(G M, r))$\"","derivation_3":"a Derivation [text] that says \"$tau &= 2 b F = 2 b thin G frac(m M, r^2) \\ tau &= kappa theta$\"","far":"a Polygon [green] drawn in sky (vertices=((1.3228756555322954, 0.0), (-1.9878682448351077, 0.16496563478…, fill_opacity=0.4)","fibre":"a Point [gray] labelled \"upright(\"fibre\")\" drawn in lab","force_l":"a Vector [red] labelled \"F\" drawn in lab (start=(-1.2, -0.25), end=(-1.2, -0.55))","force_r":"a Vector [red] labelled \"F\" drawn in lab (start=(1.2, 0.25), end=(1.2, 0.55))","g_form":"a Math [text] that says \"$G = frac(kappa theta r^2, 2 b m M)$\"","g_num":"a Math [text] that says \"$G approx 6.67 times 10^(-11) thin upright(\"N\") thin upright(\"m\")^2 thin upright(\"kg\")^(-2)$\"","head_cav":"a Heading that says \"Measuring G: the Cavendish Balance\"","head_eq":"a Heading that says \"Inertial and Gravitational Mass\"","head_kepler":"a Heading that says \"Kepler's Three Empirical Laws\"","head_law":"a Heading that says \"Why the Force Falls as $1/r^2$\"","kepler3":"a Math [text] that says \"$T^2 = k thin r^3$\"","lab":"a Figure (x_range=(-2.0, 2.0), y_range=(-1.4, 1.4), aspect=(4.0, 2.8))","law1":"a Text [text] that says \"First law: each planet moves on an ellipse with the Sun at one focus.\"","law2":"a Text [text] that says \"Second law: the line from Sun to planet sweeps equal areas in equal times.\"","law3":"a Text [text] that says \"Third law: $T^2 prop a^3$, with one constant for every planet.\"","mover":"a Point [red] labelled \"m\" drawn in ring (location=((0.0 + (1.4 * cos((phi * 0.017453292519943295)))), (0.0 + (1.4…)","near":"a Polygon [green] drawn in sky (vertices=((1.3228756555322954, 0.0), (1.7551651237807455, -0.71913830790…, fill_opacity=0.4)","newton":"a Math [text] that says \"$F = G frac(m M, r^2)$\"","orbit":"a ParametricCurve [blue] drawn in sky (function=<function>, t_range=(-1.5707963267948966, 4.71238898038469))","outward":"a Vector [green] labelled \"m_i omega^2 rho\" drawn in plumb (start=(0.6, -1.0), end=(0.8, -1.0))","path":"a Circle [blue] drawn in ring (radius=1.4)","phi":"a VariableNumber (initial_value=40.0)","planet":"a Point [red] labelled \"P\" drawn in sky (location=((2.0 * cos(t_planet)), (1.5 * sin(t_planet))))","plumb":"a Figure (x_range=(-1.4, 1.8), y_range=(-1.8, 1.0), aspect=(3.2, 2.8))","pull":"a Vector [red] labelled \"arrow(F)\" drawn in ring (start=((1.4 * cos(((phi * 3.141592653589793) / 180.0))), (1.4 * sin((…, end=((0.45 * (1.4 * cos(((phi * 3.141592653589793) / 180.0)))), (0.…)","ray":"a Line [red] drawn in sky (start=(1.3228756555322954, 0.0), end=((2.0 * cos(t_planet)), (1.5 * sin(t_planet))))","rest":"a Line [gray] drawn in lab (start=(-1.2, 0.0), end=(1.2, 0.0), dashed=True)","ring":"a Figure (x_range=(-2.0, 2.0), y_range=(-2.0, 2.0), aspect=(1.0, 1.0))","rod":"a Line [text] drawn in lab (start=(-1.2, -0.25), end=(1.2, 0.25))","sky":"a Figure (x_range=(-2.4, 2.4), y_range=(-1.8, 1.8), aspect=(4.8, 3.6))","small_l":"a Point [yellow] labelled \"m\" drawn in lab (location=(-1.2, -0.25))","small_r":"a Point [yellow] labelled \"m\" drawn in lab (location=(1.2, 0.25))","speed":"a Vector [green] labelled \"arrow(v)\" drawn in ring (start=((1.4 * cos(((phi * 3.141592653589793) / 180.0))), (1.4 * sin((…, end=(((1.4 * cos(((phi * 3.141592653589793) / 180.0))) - (0.43 * (1…)","star":"a Point [yellow] labelled \"M\" drawn in ring","sun":"a Point [yellow] labelled \"S\" drawn in sky (location=(1.3228756555322954, 0.0))","support":"a Line [gray] drawn in plumb (start=(-0.8, 0.8), end=(0.8, 0.8))","t_planet":"a VariableNumber (initial_value=-0.5)","thread":"a Line [text] drawn in plumb (start=(0.0, 0.8), end=(0.6, -1.0))","twist":"an Angle [cyan] labelled \"theta\" drawn in lab (sides=((1.2, 0.0), (1.2, 0.25)), radius=0.5)","vertical":"a Line [gray] drawn in plumb (start=(0.0, 0.8), end=(0.0, -1.3), dashed=True)","weight":"a Vector [red] labelled \"m_g g\" drawn in plumb (start=(0.6, -1.0), end=(0.6, -1.6))"},"beats":[{"start":0,"say":"Two lines of reasoning meet in this lecture. Kepler found, from years of careful observation, three rules that the planets obey. Newton showed that one force, falling as the inverse square of distance, explains all three. We will build that force, test it, and then derive every orbit it allows.","live":[],"does":[[0,"card is shown on the screen, written out."],[1.3126526806526808,"card: enter:write-left-to-right."],[19.3975,"card is hidden from the screen — left the board."]]},{"start":20.5975,"say":"Start with the observations. Here is one planetary orbit, and here is the Sun. The first law says the orbit is an ellipse, and the Sun sits not at its centre but at one focus.","live":null,"does":[[20.5975,"head_kepler is shown on the screen, written out."],[21.6,"sky is shown on the screen, written out."],[24.16,"orbit is shown on the screen, drawn."],[25.6,"sun is shown on the screen, written out."],[26.76,"law1 is shown on the screen, written out."]]},{"start":33.004999999999995,"say":"The second law concerns speed. Join the Sun to the planet. In one interval of time near the Sun, the line sweeps this green sector.","live":["law1","sky","head_kepler","orbit","sun"],"does":[[33.48,"law2 is shown on the screen, written out."],[34.16,"planet is shown on the screen, written out."],[35.64,"ray is shown on the screen, written out."],[40.199999999999996,"planet is redrawn as the numbers it depends on change."],[40.199999999999996,"ray is redrawn as the numbers it depends on change."],[40.199999999999996,"near is shown on the screen, written out."],[40.199999999999996,"t_planet ticks to 0.5."]]},{"start":42.584999999999994,"say":"Now let the planet travel round to the far side. There, in the same interval, it sweeps this thin sector, and the two areas are exactly equal. Far from the Sun the planet moves slowly; near it, quickly.","live":["law1","law2","sky","head_kepler","orbit","sun","planet","ray","near"],"does":[[43.839999999999996,"planet is redrawn as the numbers it depends on change."],[43.839999999999996,"ray is redrawn as the numbers it depends on change."],[43.839999999999996,"t_planet ticks to 3.0313926535897933."],[46.64,"planet is redrawn as the numbers it depends on change."],[46.64,"ray is redrawn as the numbers it depends on change."],[46.64,"far is shown on the screen, written out."],[46.64,"t_planet ticks to 3.251792653589793."],[51.12,"near is indicated — a transient flash."],[51.63266360700324,"far is indicated — a transient flash."]]},{"start":56.79749999999999,"say":"The third law compares different planets. The square of the period, divided by the cube of the semi-major axis, is the same number for all of them. These three laws describe the motion; they do not explain it.","live":["law1","law2","sky","head_kepler","orbit","sun","planet","ray","near","far"],"does":[[57.4,"law3 is shown on the screen, written out."],[67.24000000000001,"planet is redrawn as the numbers it depends on change."],[67.24000000000001,"ray is redrawn as the numbers it depends on change."],[67.24000000000001,"t_planet ticks to 4.4."],[70.10749999999999,"head_kepler is hidden from the screen — left the board."],[70.10749999999999,"law1 is hidden from the screen — left the board."],[70.10749999999999,"law2 is hidden from the screen — left the board."],[70.10749999999999,"law3 is hidden from the screen — left the board."],[70.10749999999999,"sky is hidden from the screen — left the board."],[70.10749999999999,"orbit is hidden from the screen — sky left the board."],[70.10749999999999,"sun is hidden from the screen — sky left the board."],[70.10749999999999,"planet is hidden from the screen — sky left the board."],[70.10749999999999,"ray is hidden from the screen — sky left the board."],[70.10749999999999,"near is hidden from the screen — sky left the board."],[70.10749999999999,"far is hidden from the screen — sky left the board."]]},{"start":71.30749999999999,"say":"Newton's argument is cleanest for a nearly circular orbit of radius r. The planet moves uniformly, so the only acceleration is centripetal, pointing at the Sun.","live":[],"does":[[71.30749999999999,"head_law is shown on the screen, written out."],[73.96000000000001,"ring is shown on the screen, written out."],[73.96000000000001,"path is shown on the screen, written out."],[73.96000000000001,"star is shown on the screen, written out."],[76.6,"mover is shown on the screen, written out."],[77.32,"mover is redrawn as the numbers it depends on change."],[77.32,"speed is redrawn as the numbers it depends on change."],[77.32,"pull is redrawn as the numbers it depends on change."],[77.32,"speed is shown on the screen, written out."],[77.32,"phi ticks to 100.0."],[79.52,"pull is shown on the screen, written out."]]},{"start":82.09249999999999,"say":"So the force is m omega squared r. Writing omega as two pi over the period, the force becomes m times four pi squared over T squared, times r.","live":["ring","head_law","path","star","mover","speed","pull"],"does":[[83.11999999999999,"ring moves to a new place on the board."],[83.11999999999999,"derivation is shown on the screen, written out."],[85.75999999999999,"derivation is shown on the screen, written out."]]},{"start":94.07249999999999,"say":"Now use the third law: T squared equals k r cubed. Substitute, and the period disappears in a telling way. The force on the planet is proportional to its mass and falls as one over r squared.","live":null,"does":[[95.24000000000001,"kepler3 is shown on the screen, written out."],[99.00000000000001,"derivation is shown on the screen, written out."],[105.56000000000002,"derivation (the \"frac(m, r^2)\" part) is emphasized."],[107.9425,"derivation (the \"frac(m, r^2)\" part) is no longer emphasized."]]},{"start":108.54249999999999,"say":"The third law of dynamics finishes the job. The planet pulls the Sun with an equal and opposite force, so the same expression must also be proportional to the mass M of the Sun. The constant four pi squared over k must therefore be G times M.","live":["kepler3","ring","head_law","path","star","mover","speed","pull"],"does":[[122.67999999999999,"derivation is shown on the screen, written out."]]},{"start":125.32749999999999,"say":"This is the law of universal gravitation. Every pair of masses attracts along the line joining them, with a force G m M over r squared, and G is one universal constant.","live":null,"does":[[127.39999999999999,"newton is shown on the screen, written out."],[131.39999999999998,"A box is drawn around newton."],[137.70999999999998,"derivation is hidden from the screen — left the board."],[137.70999999999998,"head_law is hidden from the screen — left the board."],[137.70999999999998,"kepler3 is hidden from the screen — left the board."],[137.70999999999998,"newton is hidden from the screen — left the board."],[137.70999999999998,"ring is hidden from the screen — left the board."],[137.70999999999998,"path is hidden from the screen — ring left the board."],[137.70999999999998,"star is hidden from the screen — ring left the board."],[137.70999999999998,"mover is hidden from the screen — ring left the board."],[137.70999999999998,"speed is hidden from the screen — ring left the board."],[137.70999999999998,"pull is hidden from the screen — ring left the board."]]},{"start":138.91,"say":"Look closely at one step we took for granted. On the left of the second law of dynamics sits the inertial mass, the resistance to changes of velocity. In the law of gravitation sits the gravitational mass, the charge that sources and feels the attraction.","live":[],"does":[[138.91,"head_eq is shown on the screen, written out."],[138.91,"plumb is shown on the screen, written out."],[138.91,"support is shown on the screen, written out."],[138.91,"vertical is shown on the screen, written out."],[138.91,"thread is shown on the screen, written out."],[138.91,"bob is shown on the screen, written out."],[139.56,"derivation_2 is shown on the screen, written out."],[144.92,"derivation_2 (the \"m_i\" part) is emphasized."],[150.28,"derivation_2 (the \"m_g\" part) is emphasized."],[150.28,"derivation_2 (the \"m_i\" part) is no longer emphasized."],[154.5075,"derivation_2 (the \"m_g\" part) is no longer emphasized."]]},{"start":155.1075,"say":"Dividing, the acceleration carries the ratio m g over m i. Only if these two masses are equal, for every material, does every body fall with the same acceleration, G M over r squared. That equality is the principle of equivalence.","live":["plumb","head_eq","support","vertical","thread","bob"],"does":[[155.59999999999997,"derivation_2 is shown on the screen, written out."],[165.51999999999998,"derivation_2 is shown on the screen, written out."]]},{"start":172.26,"say":"It can be tested with a plumb line. On the rotating Earth the bob feels gravity, proportional to m g, and a centrifugal pull, proportional to m i. If the ratio differed between materials, plumb lines of lead and of wood would hang at different angles. They do not, to better than one part in ten trillion in modern torsion experiments.","live":null,"does":[[176.44,"weight is shown on the screen, written out."],[179.31999999999996,"outward is shown on the screen, written out."],[185.56,"bob is indicated — a transient flash."]]},{"start":194,"say":"With the two masses equal, a circular orbit follows at once. Gravity supplies the centripetal acceleration, v squared over r, so the orbital speed is the square root of G M over r. Closer orbits are faster.","live":["plumb","head_eq","support","vertical","thread","bob","weight","outward"],"does":[[198.63999999999996,"derivation_2 is shown on the screen, written out."],[202.71999999999997,"derivation_2 is shown on the screen, written out."],[208.0475,"derivation_2 is hidden from the screen — left the board."],[208.0475,"head_eq is hidden from the screen — left the board."],[208.0475,"plumb is hidden from the screen — left the board."],[208.0475,"support is hidden from the screen — plumb left the board."],[208.0475,"vertical is hidden from the screen — plumb left the board."],[208.0475,"thread is hidden from the screen — plumb left the board."],[208.0475,"bob is hidden from the screen — plumb left the board."],[208.0475,"weight is hidden from the screen — plumb left the board."],[208.0475,"outward is hidden from the screen — plumb left the board."]]},{"start":209.2475,"say":"Measuring G is hard, because gravity between laboratory masses is tiny. Cavendish used a torsion balance. A light rod carrying two small masses m hangs from a thin fibre, and a mirror on the fibre reflects a light beam onto a distant scale, magnifying the smallest twist.","live":[],"does":[[209.2475,"head_cav is shown on the screen, written out."],[215.51999999999998,"lab is shown on the screen, written out."],[215.51999999999998,"fibre is shown on the screen, written out."],[217.27999999999997,"rod is shown on the screen, written out."],[218.39999999999998,"small_r is shown on the screen, written out."],[218.39999999999998,"small_l is shown on the screen, written out."]]},{"start":227.885,"say":"Two large masses M are brought close, one on each side. Each attracts its neighbour with a force G m M over r squared, and the rod twists until the restoring torque of the fibre balances the torque of the two forces.","live":["lab","head_cav","fibre","rod","small_r","small_l"],"does":[[228.5199999999999,"big_r is shown on the screen, written out."],[228.5199999999999,"big_l is shown on the screen, written out."],[232.43999999999988,"force_r is shown on the screen, written out."],[232.43999999999988,"force_l is shown on the screen, written out."],[236.99999999999991,"rest is shown on the screen, written out."],[238.43999999999988,"lab moves to a new place on the board."],[238.43999999999988,"derivation_3 is shown on the screen, written out."],[239.4799999999999,"twist is shown on the screen, written out."]]},{"start":242.20999999999998,"say":"The torque of the pair is two b times F, with b the lever arm. The fibre answers with kappa theta, where kappa is its calibrated torsion constant. Equating the two and solving gives G, about six point six seven times ten to the minus eleven, in newton metre squared per kilogram squared.","live":["lab","head_cav","fibre","rod","small_r","small_l","big_r","big_l","force_r","force_l","rest","twist"],"does":[[247.6799999999999,"derivation_3 is shown on the screen, written out."],[252.15999999999985,"g_form is shown on the screen, written out."],[254.95999999999984,"g_num is shown on the screen, written out."],[260.77897916666666,"derivation_3 is hidden from the screen — left the board."],[260.77897916666666,"g_form is hidden from the screen — left the board."],[260.77897916666666,"g_num is hidden from the screen — left the board."],[260.77897916666666,"head_cav is hidden from the screen — left the board."],[260.77897916666666,"lab is hidden from the screen — left the board."],[260.77897916666666,"fibre is hidden from the screen — lab left the board."],[260.77897916666666,"rod is hidden from the screen — lab left the board."],[260.77897916666666,"small_r is hidden from the screen — lab left the board."],[260.77897916666666,"small_l is hidden from the screen — lab left the board."],[260.77897916666666,"big_r is hidden from the screen — lab left the board."],[260.77897916666666,"big_l is hidden from the screen — lab left the board."],[260.77897916666666,"force_r is hidden from the screen — lab left the board."],[260.77897916666666,"force_l is hidden from the screen — lab left the board."],[260.77897916666666,"rest is hidden from the screen — lab left the board."],[260.77897916666666,"twist is hidden from the screen — lab left the board."]]}]},{"title":"Shells, Potentials and Classic Examples","start":261.8206458333333,"end":543.5196666666667,"objects":{"angle_b":"a VariableNumber (initial_value=20.0)","axis_r":"a Line [gray] labelled \"r\" drawn in space (end=(<VariableNumber p_x = 0.4>, 0.0))","b_x":"a VariableNumber (initial_value=0.9)","ball":"a Point [red] labelled \"m\" drawn in globe (location=(<VariableNumber b_x = 0.9>, 0.0))","band":"a Line [green] drawn in space (start=(0.5, -0.866), end=(0.5, 0.866), dashed=True)","centre":"a Point [gray] labelled \"C\" drawn in pair","chord":"a Line [red] labelled \"s\" drawn in space (start=(0.5, 0.866), end=(<VariableNumber p_x = 0.4>, 0.0))","derivation":"a Derivation [text] that says \"$dif M &= frac(M, 2) sin theta thin dif theta \\ dif V &= -frac(G thin dif M, s) \\ s thin dif s &= R r sin theta thin dif theta \\ V(r) &= -frac(G M, 2 R r) integral_(abs(r - R))^(r + R) dif s$\"","derivation_2":"a Derivation [text] that says \"$g(r) &= frac(G M(r), r^2) \\ M(r) &= integral_0^r 4 pi r'^2 rho(r') dif r' \\ R_1 < r < R_2: quad g &= frac(4 pi G rho_2, 3) thin frac(r^3 - R_1^3, r^2) \\ U(r) &= -m integral_r^infinity g(r') dif r'$\"","derivation_3":"a Derivation [text] that says \"$M(x) &= M frac(x^3, R^3) => g(x) = frac(G M, R^3) x \\ m frac(dif^2 x, dif t^2) &= -frac(G M m, R^3) x \\ omega &= sqrt(frac(G M, R^3)) = sqrt(frac(g, R)) \\ T &= 2 pi sqrt(frac(R, g)) approx 84 thin upright(\"min\")$\"","derivation_4":"a Derivation [text] that says \"$frac(1, 2) m v_f^2 - frac(G M m, R) &= 0 \\ v_f &= sqrt(frac(2 G M, R)) = sqrt(2 g R) approx 11.2 thin upright(\"km/s\")$\"","derivation_5":"a Derivation [text] that says \"$frac(G M^2, (2 r)^2) &= M frac(v^2, r) \\ r &= frac(v T, 2 pi) approx 4.36 times 10^10 thin upright(\"m\") \\ M &= frac(4 r v^2, G) approx 1.3 times 10^32 thin upright(\"kg\")$\"","earth":"a Circle [blue] drawn in globe","edge1":"a Line [gray] labelled \"R_1\" drawn in field (start=(1.0, 0.0), end=(1.0, 4.0), dashed=True)","edge2":"a Line [gray] labelled \"R_2\" drawn in field (start=(2.0, 0.0), end=(2.0, 4.0), dashed=True)","edge3":"a Line [gray] labelled \"R_3\" drawn in field (start=(3.0, 0.0), end=(3.0, 4.0), dashed=True)","escape_note":"a Text [text] that says \"Circular speed at the surface: $v_c = sqrt(g R) approx 7.9$ km/s, so $v_f = sqrt(2) thin v_c$.\"","far_res":"a Math [text] that says \"$r > R: quad V = -frac(G M, r)$\"","field":"an Axes (x_range=(0.0, 5.0), y_range=(0.0, 4.2), x_ticks_every=1.0)","g_curve":"a FunctionPlot [blue] drawn in field (function=<function>)","globe":"a Figure (x_range=(-1.4, 1.4), y_range=(-1.2, 1.2), aspect=(2.8, 2.4))","head_escape":"a Heading that says \"Escape Speed\"","head_layer":"a Heading that says \"A Layered Body\"","head_shell":"a Heading that says \"The Shell Theorem\"","head_tunnel":"a Heading that says \"A Tunnel Through the Earth\"","in_res":"a Math [text] that says \"$r < R: quad V = -frac(G M, R), quad g = 0$\"","mass_answer":"a Math [text] that says \"$M approx 64 thin M_(upright(\"sun\"))$\"","p_x":"a VariableNumber (initial_value=2.2)","pair":"a Figure (x_range=(-1.5, 1.5), y_range=(-1.3, 1.3), aspect=(3.0, 2.6))","probe":"a Point [red] labelled \"P\" drawn in space (location=(<VariableNumber p_x = 0.4>, 0.0))","problem":"a Text [text] that says \"Two equal stars orbit their common centre of mass with $v = 220$ km/s and $T = 14.4$ days. Find their mass.\"","r_probe":"a VariableNumber (initial_value=0.5)","radius_r":"a Line [gray] labelled \"R\" drawn in space (end=(0.5, 0.866))","reader":"a PlotPoint [yellow] drawn in field (target='g_curve', x=<VariableNumber r_probe = 4.5>)","shell":"a Circle [blue] drawn in space","space":"a Figure (x_range=(-1.4, 2.6), y_range=(-1.4, 1.4), aspect=(4.0, 2.8))","spring":"a Vector [red] drawn in globe (start=(<VariableNumber b_x = 0.9>, 0.0), end=((0.5 * b_x), 0.0))","star_a":"a Point [yellow] labelled \"M\" drawn in pair (location=((0.0 + (1.0 * cos((angle_b * 0.017453292519943295)))), (0.0 + …)","star_b":"a Point [yellow] labelled \"M\" drawn in pair (location=((0.0 + (1.0 * cos(((angle_b + 180.0) * 0.017453292519943295)))…)","theta_mark":"an Angle [yellow] labelled \"theta\" drawn in space (sides=((0.5, 0.866), (<VariableNumber p_x = 0.4>, 0.0)), radius=0.3)","track":"a Circle [gray] drawn in pair","tunnel":"a Line [gray] drawn in globe (start=(-1.0, 0.0), dashed=True)"},"beats":[{"start":261.8206458333333,"say":"Newton's law is written for point masses, yet planets are large. The shell theorem tells us how an extended sphere attracts. Take a thin uniform shell of mass M and radius R, and a point P at distance r from its centre.","live":[],"does":[[261.8206458333333,"head_shell is shown on the screen, written out."],[266.7006458333333,"space is shown on the screen, written out."],[270.8606458333333,"shell is shown on the screen, written out."],[274.7006458333333,"probe is shown on the screen, written out."],[275.7406458333333,"axis_r is shown on the screen, written out."]]},{"start":277.7956458333333,"say":"Slice the shell into rings around the line to P. The ring at angle theta has radius R sine theta, and all of it lies at the same distance s from P. Its share of the mass is M over two, times sine theta d theta.","live":["space","head_shell","shell","probe","axis_r"],"does":[[278.4206458333333,"band is shown on the screen, written out."],[282.1806458333333,"radius_r is shown on the screen, written out."],[283.4606458333333,"theta_mark is shown on the screen, written out."],[286.4206458333333,"chord is shown on the screen, written out."],[288.9006458333333,"space moves to a new place on the board."],[288.9006458333333,"derivation is shown on the screen, written out."]]},{"start":293.59564583333326,"say":"The ring contributes a potential minus G d M over s. The law of cosines links s to theta, and differentiating it gives s d s equal to R r sine theta d theta, which turns the integral over angles into an integral over s.","live":["space","head_shell","shell","probe","axis_r","band","radius_r","theta_mark","chord"],"does":[[294.4606458333333,"derivation is shown on the screen, written out."],[301.8206458333333,"derivation is shown on the screen, written out."]]},{"start":310.97564583333326,"say":"Everything cancels except a plain integral of d s, running from the nearest point of the shell to the farthest, so the potential is minus G M over two R r, times the length of that range.","live":null,"does":[[311.7006458333333,"derivation is shown on the screen, written out."]]},{"start":323.4356458333333,"say":"Outside, the range runs from r minus R to r plus R, a length two R. The potential is minus G M over r: exactly that of a point mass at the centre.","live":null,"does":[[324.3806458333333,"far_res is shown on the screen, written out."]]},{"start":335.9856458333333,"say":"Now move P inside. The range runs from R minus r to R plus r, a length two r, and the potential becomes minus G M over R, the same everywhere. A constant potential means zero field. Inside a uniform shell there is no gravity at all.","live":["far_res","space","head_shell","shell","probe","axis_r","band","radius_r","theta_mark","chord"],"does":[[336.9006458333333,"probe is redrawn as the numbers it depends on change."],[336.9006458333333,"axis_r is redrawn as the numbers it depends on change."],[336.9006458333333,"theta_mark is redrawn as the numbers it depends on change."],[336.9006458333333,"chord is redrawn as the numbers it depends on change."],[336.9006458333333,"p_x ticks to 0.4."],[344.0206458333333,"in_res is shown on the screen, written out."],[348.5806458333333,"A box is drawn around in_res."],[354.3431458333333,"derivation is hidden from the screen — left the board."],[354.3431458333333,"far_res is hidden from the screen — left the board."],[354.3431458333333,"head_shell is hidden from the screen — left the board."],[354.3431458333333,"in_res is hidden from the screen — left the board."],[354.3431458333333,"space is hidden from the screen — left the board."],[354.3431458333333,"shell is hidden from the screen — space left the board."],[354.3431458333333,"probe is hidden from the screen — space left the board."],[354.3431458333333,"axis_r is hidden from the screen — space left the board."],[354.3431458333333,"band is hidden from the screen — space left the board."],[354.3431458333333,"radius_r is hidden from the screen — space left the board."],[354.3431458333333,"theta_mark is hidden from the screen — space left the board."],[354.3431458333333,"chord is hidden from the screen — space left the board."]]},{"start":355.5431458333333,"say":"Any spherically symmetric body is a nest of such shells. At radius r, the shells outside pull nothing, and the shells inside act as a point mass. So g of r is G times the mass enclosed, over r squared.","live":[],"does":[[355.5431458333333,"head_layer is shown on the screen, written out."],[358.26064583333334,"field is shown on the screen, written out."],[365.7806458333333,"field moves to a new place on the board."],[365.7806458333333,"derivation_2 is shown on the screen, written out."]]},{"start":370.8256458333333,"say":"Consider a body with a hollow core of radius R one, a shell of density rho two out to R two, and an outer layer of density rho three out to R three. The enclosed mass is the integral of four pi r squared rho over the radius.","live":["field","head_layer"],"does":[[372.46064583333333,"edge1 is shown on the screen, written out."],[375.90064583333333,"edge2 is shown on the screen, written out."],[379.4206458333333,"edge3 is shown on the screen, written out."],[381.82064583333334,"derivation_2 is shown on the screen, written out."]]},{"start":387.0931458333333,"say":"Follow the field outward. In the hollow core it is exactly zero. Through the first shell it grows as r cubed minus R one cubed, over r squared. Through the second it keeps rising, but more gently, because rho three is smaller.","live":["field","head_layer","edge1","edge2","edge3"],"does":[[387.7806458333333,"g_curve is shown on the screen, drawn."],[388.50064583333335,"reader is shown on the screen, written out."],[389.70064583333334,"The segment (0.0, 0.0) to (1.0, 0.0) in field is lit up."],[392.6606458333333,"reader is redrawn as the numbers it depends on change."],[392.6606458333333,"field: retire a lit segment (unemphasize_line)."],[392.6606458333333,"r_probe ticks to 1.6."],[393.43364583333334,"derivation_2 is shown on the screen, written out."],[397.62064583333336,"reader is redrawn as the numbers it depends on change."],[397.62064583333336,"r_probe ticks to 2.6."]]},{"start":402.7231458333333,"say":"Beyond the surface the whole mass acts from the centre, and the field falls as one over r squared. The potential energy is minus m times the integral of g from r out to infinity: continuous everywhere, and flat inside the cavity, where the field vanishes.","live":["field","head_layer","edge1","edge2","edge3","g_curve","reader"],"does":[[403.50064583333335,"reader is redrawn as the numbers it depends on change."],[403.50064583333335,"r_probe ticks to 4.5."],[409.90064583333333,"derivation_2 is shown on the screen, written out."],[419.3806458333333,"derivation_2 is hidden from the screen — left the board."],[419.3806458333333,"field is hidden from the screen — left the board."],[419.3806458333333,"edge1 is hidden from the screen — field left the board."],[419.3806458333333,"edge2 is hidden from the screen — field left the board."],[419.3806458333333,"edge3 is hidden from the screen — field left the board."],[419.3806458333333,"g_curve is hidden from the screen — field left the board."],[419.3806458333333,"reader is hidden from the screen — field left the board."],[419.3806458333333,"head_layer is hidden from the screen — left the board."]]},{"start":420.5806458333333,"say":"Here is a striking consequence. Treat the Earth as a uniform sphere and drill a straight frictionless tunnel along a diameter. A body at distance x from the centre feels only the mass inside radius x, which grows as x cubed.","live":[],"does":[[420.5806458333333,"head_tunnel is shown on the screen, written out."],[423.98064583333326,"globe is shown on the screen, written out."],[423.98064583333326,"earth is shown on the screen, written out."],[427.2606458333333,"tunnel is shown on the screen, written out."],[429.4206458333333,"ball is shown on the screen, written out."],[432.9406458333333,"globe moves to a new place on the board."],[432.9406458333333,"derivation_3 is shown on the screen, written out."]]},{"start":436.3956458333333,"say":"So the force grows linearly, minus G M m over R cubed, times x. That is a spring law. The body oscillates harmonically about the centre, with angular frequency the square root of g over R.","live":["globe","head_tunnel","earth","tunnel","ball"],"does":[[437.38064583333323,"spring is shown on the screen, written out."],[438.10064583333326,"derivation_3 is shown on the screen, written out."],[444.66064583333326,"ball is redrawn as the numbers it depends on change."],[444.66064583333326,"spring is redrawn as the numbers it depends on change."],[444.66064583333326,"b_x ticks to -0.9."],[447.34064583333327,"derivation_3 is shown on the screen, written out."]]},{"start":451.0781458333333,"say":"Putting in numbers, the period is about eighty-four minutes. Drop a stone in at one end and it reaches the other end some forty-two minutes later, at rest, then falls back. Curiously, this is also the period of a satellite skimming the surface.","live":["globe","head_tunnel","earth","tunnel","ball","spring"],"does":[[452.8206458333333,"derivation_3 is shown on the screen, written out."],[455.0606458333333,"ball is redrawn as the numbers it depends on change."],[455.0606458333333,"spring is redrawn as the numbers it depends on change."],[455.0606458333333,"b_x ticks to 0.9."],[465.99564583333324,"derivation_3 is hidden from the screen — left the board."],[465.99564583333324,"globe is hidden from the screen — left the board."],[465.99564583333324,"earth is hidden from the screen — globe left the board."],[465.99564583333324,"tunnel is hidden from the screen — globe left the board."],[465.99564583333324,"ball is hidden from the screen — globe left the board."],[465.99564583333324,"spring is hidden from the screen — globe left the board."],[465.99564583333324,"head_tunnel is hidden from the screen — left the board."]]},{"start":467.1956458333333,"say":"Energy answers the next question: how fast must a body leave the surface never to return? Take zero potential energy at infinity. The minimum launch arrives there with nothing to spare, so its total energy is zero.","live":[],"does":[[467.1956458333333,"head_escape is shown on the screen, written out."],[475.4606458333333,"derivation_4 is shown on the screen, written out."]]},{"start":481.6406458333333,"say":"Solving, the escape speed is the square root of two G M over R, or the square root of two g R: about eleven point two kilometres per second for the Earth. It is exactly root two times the speed of a circular orbit grazing the surface.","live":["head_escape"],"does":[[482.0206458333332,"derivation_4 is shown on the screen, written out."],[493.6206458333332,"escape_note is shown on the screen, written out."],[497.4556458333333,"derivation_4 is hidden from the screen — left the board."],[497.4556458333333,"escape_note is hidden from the screen — left the board."],[497.4556458333333,"head_escape is hidden from the screen — left the board."]]},{"start":498.65564583333327,"say":"The same circular orbit reasoning weighs stars. Two stars of equal mass circle their common centre of mass, each at two hundred and twenty kilometres per second, with a period of fourteen point four days.","live":[],"does":[[498.65564583333327,"problem is shown on the screen, written out."],[501.4206458333332,"pair is shown on the screen, written out."],[501.4206458333332,"track is shown on the screen, written out."],[502.98064583333314,"star_a is shown on the screen, written out."],[502.98064583333314,"star_b is shown on the screen, written out."],[504.8206458333332,"centre is shown on the screen, written out."],[509.0206458333332,"star_a is redrawn as the numbers it depends on change."],[509.0206458333332,"star_b is redrawn as the numbers it depends on change."],[509.0206458333332,"angle_b ticks to 200.0."]]},{"start":511.79314583333326,"say":"Each star sits at radius r, but the other is two r away. So gravity, G M squared over two r squared, supplies the centripetal force M v squared over r. The radius comes from the motion itself: r is v T over two pi.","live":["pair","problem","track","star_a","star_b","centre"],"does":[[519.5806458333332,"pair moves to a new place on the board."],[519.5806458333332,"derivation_5 is shown on the screen, written out."],[523.3406458333332,"star_a is redrawn as the numbers it depends on change."],[523.3406458333332,"star_b is redrawn as the numbers it depends on change."],[523.3406458333332,"angle_b ticks to 380.0."],[524.4606458333332,"derivation_5 is shown on the screen, written out."]]},{"start":528.4131458333333,"say":"Solving, M equals four r v squared over G, about one point three times ten to the thirty-two kilograms, some sixty-four solar masses each. This is how astronomers weigh stars they will never visit.","live":null,"does":[[528.8206458333332,"derivation_5 is shown on the screen, written out."],[537.4606458333332,"mass_answer is shown on the screen, written out."],[542.478,"derivation_5 is hidden from the screen — left the board."],[542.478,"mass_answer is hidden from the screen — left the board."],[542.478,"pair is hidden from the screen — left the board."],[542.478,"track is hidden from the screen — pair left the board."],[542.478,"star_a is hidden from the screen — pair left the board."],[542.478,"star_b is hidden from the screen — pair left the board."],[542.478,"centre is hidden from the screen — pair left the board."],[542.478,"problem is hidden from the screen — left the board."]]}]},{"title":"The Two-Body Problem and the Conic Orbits","start":543.5196666666667,"end":825.3754166666668,"objects":{"body1":"a Point [blue] labelled \"m_1\" drawn in pairfig (location=(-1.0, -0.3))","body2":"a Point [red] labelled \"m_2\" drawn in pairfig (location=(1.6, 0.6))","cm":"a Point [gray] labelled \"upright(\"CM\")\" drawn in pairfig (location=(-0.133, 0.0))","conics":"an Axes (x_range=(-1.2, 2.4), y_range=(-2.5, 2.5), aspect=(3.6, 5.0))","derivation":"a Derivation [text] that says \"$arrow(r) &= r thin hat(u)_r \\ arrow(v) &= dot(r) thin hat(u)_r + r dot(theta) thin hat(u)_theta \\ a_r &= dot.double(r) - r dot(theta)^2 \\ a_theta &= 2 dot(r) dot(theta) + r dot.double(theta) = frac(1, r) frac(dif, dif t) (r^2 dot(theta))$\"","derivation_2":"a Derivation [text] that says \"$mu &= frac(m_1 m_2, m_1 + m_2), quad arrow(r) = arrow(r)_2 - arrow(r)_1 \\ mu thin dot.double(arrow(r)) &= -frac(k, r^2) thin hat(u)_r, quad k = G m_1 m_2 \\ L &= mu r^2 dot(theta) = upright(\"const\"), quad frac(dif A, dif t) = frac(L, 2 mu) …$\"","derivation_3":"a Derivation [text] that says \"$u &= frac(1, r), quad dot(r) = -frac(L, mu) frac(dif u, dif theta) \\ dot.double(r) &= -frac(L^2, mu^2) u^2 frac(dif^2 u, dif theta^2) \\ frac(dif^2 u, dif theta^2) + u &= frac(mu k, L^2) \\ u(theta) &= frac(mu k, L^2) (1 - epsilon cos theta)$\"","derivation_4":"a Derivation [text] that says \"$T &= frac(2 mu pi a b, L) \\ T^2 &= frac(4 pi^2 mu^2 a^2 b^2, L^2), quad b^2 = a D = frac(a L^2, mu k) \\ T^2 &= frac(4 pi^2 a^3, G (m_1 + m_2))$\"","eps_line":"a Math [text] that says \"$epsilon = sqrt(1 + frac(2 E L^2, mu k^2))$\"","focus":"a Point [yellow] labelled \"F\" drawn in conics","head_binet":"a Heading that says \"Binet's Equation and the Orbit\"","head_class":"a Heading that says \"Energy Decides the Shape\"","head_k3":"a Heading that says \"Kepler's Third Law, Exactly\"","head_polar":"a Heading that says \"Kinematics in Polar Coordinates\"","head_two":"a Heading that says \"Two Bodies, One Equation\"","kinds":"a Table [text] that says \"$epsilon$ $E$ Orbit $0$ $E_(min)$ circle $0 < epsilon < 1$ $E < 0$ ellipse $1$ $0$ parabola $epsilon > 1$ $E > 0$ hyperbola\" (rows=(('$epsilon$', '$E$', 'Orbit'), ('$0$', '$E_(min)$', 'circle'),…, header=True)","orbit_eq":"a Math [text] that says \"$r(theta) = frac(D, 1 - epsilon cos theta), quad D = frac(L^2, mu k)$\"","origin":"a Point [gray] labelled \"O\" drawn in plane","pairfig":"a Figure (x_range=(-1.6, 2.2), y_range=(-1.0, 1.2), aspect=(3.8, 2.2))","parametric_curve":"a ParametricCurve [blue] drawn in conics (function=<function>, t_range=(0.0, 6.283185307179586))","parametric_curve_2":"a ParametricCurve [gray] drawn in conics (function=<function>, t_range=(0.0, 6.283185307179586))","parametric_curve_3":"a ParametricCurve [green] drawn in conics (function=<function>, t_range=(0.85, 5.433185307179587))","parametric_curve_4":"a ParametricCurve [red] drawn in conics (function=<function>, t_range=(1.2, 5.083185307179586))","plane":"a Figure (x_range=(-0.3, 2.6), y_range=(-0.3, 2.2), aspect=(2.9, 2.5))","polar_angle":"an Angle [cyan] labelled \"theta\" drawn in plane (sides=((2.4, 0.0), (1.2237, 1.0307)))","radial":"a Line [text] drawn in plane (end=(1.2237, 1.0307))","sep":"a Vector [yellow] labelled \"arrow(r)\" drawn in pairfig (start=(-1.0, -0.3), end=(1.6, 0.6))","spot":"a Point [yellow] labelled \"P\" drawn in plane (location=(1.2237, 1.0307))","three":"a Text [text] that says \"Three or more bodies: no general closed-form solution. Integrate numerically in small steps $Delta t$.\"","turning":"a Math [text] that says \"$dot(hat(u))_r = dot(theta) thin hat(u)_theta, quad dot(hat(u))_theta = -dot(theta) thin hat(u)_r$\"","u_r":"a Vector [red] labelled \"hat(u)_r\" drawn in plane (start=(1.2237, 1.0307), end=(1.6443, 1.385))","u_t":"a Vector [green] labelled \"hat(u)_theta\" drawn in plane (start=(1.2237, 1.0307), end=(0.8694, 1.4514))","x_ref":"a Line [gray] drawn in plane (end=(2.4, 0.0))"},"beats":[{"start":543.5196666666667,"say":"To find every possible orbit we need a better description of motion under a central force. Use polar coordinates centred on the source: a distance r and an angle theta, with unit vectors u r pointing outward and u theta pointing along increasing angle.","live":[],"does":[[543.5196666666667,"head_polar is shown on the screen, written out."],[549.5196666666667,"plane is shown on the screen, written out."],[549.5196666666667,"origin is shown on the screen, written out."],[549.5196666666667,"x_ref is shown on the screen, written out."],[552.0796666666666,"plane moves to a new place on the board."],[552.0796666666666,"spot is shown on the screen, written out."],[552.0796666666666,"radial is shown on the screen, written out."],[552.0796666666666,"derivation is shown on the screen, written out."],[554.4796666666667,"polar_angle is shown on the screen, written out."],[556.2396666666667,"u_r is shown on the screen, written out."],[558.5596666666667,"u_t is shown on the screen, written out."]]},{"start":560.4796666666667,"say":"The unit vectors turn as the point moves. The derivative of u r is theta dot u theta, and the derivative of u theta is minus theta dot u r. With these, the velocity has a radial part r dot, and a transverse part r theta dot.","live":["plane","head_polar","origin","x_ref","spot","radial","polar_angle","u_r","u_t"],"does":[[563.7196666666667,"turning is shown on the screen, written out."],[572.0396666666667,"derivation is shown on the screen, written out."]]},{"start":577.5246666666667,"say":"Differentiate once more and collect terms. The radial acceleration is r double dot minus r theta dot squared, the second piece being the familiar centripetal term. The transverse acceleration is one over r times the time derivative of r squared theta dot.","live":["turning","plane","head_polar","origin","x_ref","spot","radial","polar_angle","u_r","u_t"],"does":[[580.7996666666667,"derivation is shown on the screen, written out."],[586.9996666666667,"derivation (the \"r dot(theta)^2\" part) is emphasized."],[588.6396666666667,"derivation is shown on the screen, written out."],[588.6396666666667,"derivation (the \"r dot(theta)^2\" part) is no longer emphasized."],[594.0346666666667,"derivation is hidden from the screen — left the board."],[594.0346666666667,"head_polar is hidden from the screen — left the board."],[594.0346666666667,"plane is hidden from the screen — left the board."],[594.0346666666667,"origin is hidden from the screen — plane left the board."],[594.0346666666667,"x_ref is hidden from the screen — plane left the board."],[594.0346666666667,"spot is hidden from the screen — plane left the board."],[594.0346666666667,"radial is hidden from the screen — plane left the board."],[594.0346666666667,"polar_angle is hidden from the screen — plane left the board."],[594.0346666666667,"u_r is hidden from the screen — plane left the board."],[594.0346666666667,"u_t is hidden from the screen — plane left the board."],[594.0346666666667,"turning is hidden from the screen — left the board."]]},{"start":595.2346666666667,"say":"Now two bodies, m one and m two, attract each other. Their centre of mass moves uniformly, so we watch only the separation vector r, from one body to the other.","live":[],"does":[[595.2346666666667,"head_two is shown on the screen, written out."],[596.3996666666667,"pairfig is shown on the screen, written out."],[597.0396666666667,"body1 is shown on the screen, written out."],[597.7596666666667,"body2 is shown on the screen, written out."],[600.0396666666667,"cm is shown on the screen, written out."],[603.0396666666667,"sep is shown on the screen, written out."]]},{"start":606.6896666666667,"say":"Subtracting the two equations of motion gives a single one, for a fictitious particle of reduced mass mu, attracted to a fixed centre by the force minus k over r squared, with k equal to G m one m two.","live":["pairfig","head_two","body1","body2","cm","sep"],"does":[[607.0796666666668,"pairfig moves to a new place on the board."],[607.0796666666668,"derivation_2 is shown on the screen, written out."],[610.5596666666667,"derivation_2 is shown on the screen, written out."]]},{"start":621.3096666666667,"say":"The force is central, so it exerts no torque, and the angular momentum L equals mu r squared theta dot, constant. The transverse acceleration vanishes, and the area swept per unit time, L over two mu, is constant. That is the second law of Kepler, now derived.","live":null,"does":[[624.8796666666667,"derivation_2 is shown on the screen, written out."]]},{"start":640.6171666666667,"say":"The energy is conserved too. Kinetic energy splits into radial motion and a centrifugal term, L squared over two mu r squared, which acts like a repulsive barrier added to the gravitational potential minus k over r.","live":null,"does":[[641.0796666666668,"derivation_2 is shown on the screen, written out."],[645.3996666666667,"derivation_2 (the \"frac(L^2, 2 mu r^2)\" part) is emphasized."],[650.6396666666667,"derivation_2 (the \"frac(L^2, 2 mu r^2)\" part) is no longer emphasized."],[654.5546666666667,"derivation_2 is hidden from the screen — left the board."],[654.5546666666667,"head_two is hidden from the screen — left the board."],[654.5546666666667,"pairfig is hidden from the screen — left the board."],[654.5546666666667,"body1 is hidden from the screen — pairfig left the board."],[654.5546666666667,"body2 is hidden from the screen — pairfig left the board."],[654.5546666666667,"cm is hidden from the screen — pairfig left the board."],[654.5546666666667,"sep is hidden from the screen — pairfig left the board."]]},{"start":655.7546666666667,"say":"The idea of Binet is to describe the orbit by r as a function of theta, and to use u equals one over r. The chain rule and constant L turn time derivatives into angle derivatives: r dot becomes minus L over mu, times d u by d theta.","live":[],"does":[[655.7546666666667,"head_binet is shown on the screen, written out."],[658.2796666666667,"conics is shown on the screen, written out."],[658.2796666666667,"focus is shown on the screen, written out."],[663.4396666666667,"conics moves to a new place on the board."],[663.4396666666667,"derivation_3 is shown on the screen, written out."]]},{"start":672.8921666666668,"say":"Differentiating again, r double dot is minus L squared over mu squared, u squared, times the second derivative of u. Insert this into the radial equation, and every power of u lines up.","live":["conics","head_binet","focus"],"does":[[674.2396666666667,"derivation_3 is shown on the screen, written out."]]},{"start":686.0346666666667,"say":"The result is astonishingly simple: the second derivative of u, plus u, equals the constant mu k over L squared. It is the equation of a harmonic oscillator in the angle, shifted by a constant.","live":null,"does":[[686.8396666666667,"derivation_3 is shown on the screen, written out."]]},{"start":700.0471666666667,"say":"Its general solution is the constant plus a cosine. Choosing the orientation, u equals mu k over L squared, times one minus epsilon cosine theta. Invert it, and r of theta is D over one minus epsilon cosine theta, with D equal to L squared over mu k.","live":null,"does":[[704.0796666666666,"derivation_3 is shown on the screen, written out."],[711.1196666666666,"orbit_eq is shown on the screen, written out."],[711.1196666666666,"parametric_curve is shown on the screen, drawn."],[716.7996666666666,"A box is drawn around orbit_eq."]]},{"start":719.2596666666667,"say":"This is the polar equation of a conic with its focus at the centre of force. The first law of Kepler appears as a theorem, and epsilon, the eccentricity, decides the shape.","live":["orbit_eq","conics","head_binet","focus","parametric_curve"],"does":[[722.1196666666666,"focus is indicated — a transient flash."],[730.0471666666667,"derivation_3 is hidden from the screen — left the board."],[730.0471666666667,"head_binet is hidden from the screen — left the board."],[730.0471666666667,"orbit_eq is hidden from the screen — left the board."]]},{"start":731.2471666666668,"say":"Evaluate the energy at closest approach, and the eccentricity turns out to be the square root of one plus two E L squared over mu k squared. The sign of the energy now classifies every orbit.","live":["conics","focus","parametric_curve"],"does":[[731.2471666666668,"head_class is shown on the screen, written out."],[731.2471666666668,"eps_line is shown on the screen, written out."],[742.1196666666666,"kinds is shown on the screen, written out."]]},{"start":744.5296666666667,"say":"At the minimum allowed energy, epsilon is zero: a circle. For negative energy, epsilon lies below one: a closed ellipse, a bound orbit like those of the planets.","live":["conics","focus","parametric_curve","eps_line","head_class"],"does":[[748.3196666666665,"parametric_curve_2 is shown on the screen, drawn."],[748.3196666666665,"kinds is shown on the screen, written out."],[753.0396666666666,"kinds is shown on the screen, written out."]]},{"start":756.6146666666667,"say":"At exactly zero energy, epsilon equals one: a parabola, the borderline case of a body that just escapes. For positive energy, epsilon exceeds one: a hyperbola, an open path that swings once past the centre and leaves forever, like a comet or a probe on a flyby.","live":["conics","focus","parametric_curve","eps_line","head_class","parametric_curve_2"],"does":[[760.5996666666665,"parametric_curve_3 is shown on the screen, drawn."],[760.5996666666665,"kinds is shown on the screen, written out."],[768.2396666666666,"parametric_curve_4 is shown on the screen, drawn."],[768.2396666666666,"kinds is shown on the screen, written out."],[775.4346666666668,"eps_line is hidden from the screen — left the board."],[775.4346666666668,"head_class is hidden from the screen — left the board."],[775.4346666666668,"kinds is hidden from the screen — left the board."]]},{"start":776.6346666666667,"say":"The ellipse also yields the third law, now exactly. Its area is pi a b, swept at the constant rate L over two mu, so the period is pi a b divided by that rate.","live":["conics","focus","parametric_curve","parametric_curve_2","parametric_curve_3","parametric_curve_4"],"does":[[776.6346666666667,"head_k3 is shown on the screen, written out."],[776.6346666666667,"parametric_curve_2 is hidden from the screen."],[776.6346666666667,"parametric_curve_3 is hidden from the screen."],[776.6346666666667,"parametric_curve_4 is hidden from the screen."],[776.6346666666667,"derivation_4 is shown on the screen, written out."]]},{"start":789.9946666666667,"say":"Square it, and use the geometric fact that b squared equals a times D. The angular momentum cancels, and T squared equals four pi squared a cubed over G times the total mass. The constant of the third law depends, slightly, on the mass of the planet itself.","live":["conics","focus","parametric_curve","head_k3"],"does":[[790.3996666666666,"derivation_4 is shown on the screen, written out."],[796.2396666666666,"derivation_4 is shown on the screen, written out."],[801.3596666666666,"A box is drawn around derivation_4."]]},{"start":808.1321666666668,"say":"Add a third body and this beautiful solvability is lost. Poincaré showed that the three-body problem has no general solution in closed form, so the Sun, Earth and Moon are followed numerically, advancing positions and velocities in small time steps.","live":null,"does":[[808.9196666666666,"three is shown on the screen, written out."],[824.33375,"conics is hidden from the screen — left the board."],[824.33375,"focus is hidden from the screen — conics left the board."],[824.33375,"parametric_curve is hidden from the screen — conics left the board."],[824.33375,"derivation_4 is hidden from the screen — left the board."],[824.33375,"head_k3 is hidden from the screen — left the board."],[824.33375,"three is hidden from the screen — left the board."]]}]},{"title":"Tides, Surface Gravity and Orbital Dynamics","start":825.3754166666668,"end":1099.9944375,"objects":{"c_vec":"a Vector [green] labelled \"omega^2 rho\" drawn in sphere (start=(0.766, 0.643), end=(1.12, 0.643))","capsule":"a Point [red] drawn in orbit_fig (location=((0.5 + (1.5 * cos(t_cap))), (1.4142135623730951 * sin(t_cap))))","derivation":"a Derivation [text] that says \"$g(h) &= frac(G M, (R + h)^2) approx g_0 (1 - frac(2 h, R)) \\ frac(Delta g, g) &approx -frac(2 Delta h, R) \\ g(lambda) &approx g_0 - omega^2 R cos^2 lambda \\ omega^2 R &approx 0.034 thin upright(\"m\") thin upright(\"s\")^(-2)$\"","derivation_2":"a Derivation [text] that says \"$Delta a &= frac(G M, (d - r)^2) - frac(G M, d^2) \\ &approx frac(2 G M r, d^3) \\ frac(Delta a_S, Delta a_L) &= frac(M_S, M_L) (frac(d_L, d_S))^3 approx 0.46$\"","derivation_3":"a Derivation [text] that says \"$frac(G M_S, (d - x)^2) - frac(G M_E, x^2) &= omega^2 (d - x) \\ omega^2 &= frac(G M_S, d^3) \\ x &approx d (frac(M_E, 3 M_S))^(1 / 3) approx 1.5 times 10^6 thin upright(\"km\")$\"","derivation_4":"a Derivation [text] that says \"$v_A thin 2 R &= v_B thin R \\ frac(1, 2) v_A^2 - frac(G M, 2 R) &= frac(1, 2) v_B^2 - frac(G M, R) \\ v_A &= sqrt(frac(G M, 3 R)) < v_0 = sqrt(frac(G M, 2 R)) \\ v_B &= 2 sqrt(frac(G M, 3 R))$\"","earth_g":"a Circle [blue] drawn in sphere","earth_l":"a Point [blue] labelled \"upright(\"Earth\")\" drawn in line_fig (location=(3.0, 0.0))","earth_t":"a Circle [blue] drawn in seas (radius=0.8)","equator":"a Line [gray] drawn in sphere (end=(1.3, 0.0))","far_pull":"a Vector [red] drawn in seas (start=(-0.8, 0.0), end=(-1.3, 0.0))","fx1":"a Text [text] that says \"Basin resonance: about $0.5$ m at the Po delta, up to $15$ m at Mont Saint-Michel.\"","fx2":"a Text [text] that says \"Solid Earth tides: the crust rises and falls by up to about $30$ cm.\"","fx3":"a Text [text] that says \"At CERN the $27$ km ring changes length by about $1$ mm, shifting the beam energy.\"","g_vec":"a Vector [red] labelled \"arrow(g)_0\" drawn in sphere (start=(0.766, 0.643), end=(0.345, 0.289))","head_fx":"a Heading that says \"What the Coasts and the Rock Do\"","head_g":"a Heading that says \"Gravity at the Surface of the Earth\"","head_l1":"a Heading that says \"The Lagrange Point $L_1$\"","head_tide":"a Heading that says \"Tides\"","join":"a Line [gray] drawn in line_fig (end=(3.0, 0.0))","kick":"a Vector [yellow] labelled \"Delta v\" drawn in orbit_fig (start=(2.0, 0.0), end=(2.0, -0.3))","l1_pt":"a Point [red] labelled \"L_1\" drawn in line_fig (location=(2.55, 0.0))","lat":"an Angle [cyan] labelled \"lambda\" drawn in sphere (sides=((1.3, 0.0), (0.766, 0.643)), radius=0.35)","launch":"a Math [text] that says \"$Delta v = v_0 - v_A approx 0.13 sqrt(frac(G M, R)), quad upright(\"backward\")$\"","line_fig":"a Figure (x_range=(-0.4, 3.6), y_range=(-0.9, 0.9), aspect=(4.0, 1.8))","low_pull":"a Vector [red] drawn in seas (start=(0.0, -0.8), end=(0.0, -0.55))","moon":"a Point [gray] labelled \"upright(\"Moon\")\" drawn in seas (location=(3.2, 0.0))","near_pull":"a Vector [red] drawn in seas (start=(0.8, 0.0), end=(1.3, 0.0))","orbit_fig":"a Figure (x_range=(-2.4, 2.4), y_range=(-2.4, 2.4), aspect=(1.0, 1.0))","planet":"a Circle [blue] drawn in orbit_fig (filled=True)","point_a":"a Point [yellow] labelled \"A\" drawn in orbit_fig (location=(2.0, 0.0))","point_b":"a Point [yellow] labelled \"B\" drawn in orbit_fig (location=(-1.0, 0.0))","problem":"a Text [text] that says \"A ship orbits at radius $2R$ and releases a capsule at $A$. Find the launch that makes it land at $B$, diametrically opposite, on the surface of radius $R$.\"","pull_earth":"a Vector [blue] drawn in line_fig (start=(2.55, -0.25), end=(2.95, -0.25))","pull_sun":"a Vector [yellow] drawn in line_fig (start=(2.55, 0.25), end=(1.85, 0.25))","rho_line":"a Line [green] labelled \"rho\" drawn in sphere (start=(0.0, 0.643), end=(0.766, 0.643), dashed=True)","seas":"a Figure (x_range=(-3.6, 3.6), y_range=(-1.3, 1.3), aspect=(7.2, 2.6))","ship_path":"a Circle [gray] drawn in orbit_fig (radius=2.0)","site":"a Point [yellow] drawn in sphere (location=(0.766, 0.643))","sphere":"a Figure (x_range=(-1.4, 1.8), y_range=(-1.4, 1.4), aspect=(3.2, 2.8))","spin_axis":"a Line [gray] drawn in sphere (start=(0.0, -1.3), end=(0.0, 1.3), dashed=True)","sun":"a Point [yellow] labelled \"upright(\"Sun\")\" drawn in seas (location=(-3.2, 0.0))","sun_l":"a Point [yellow] labelled \"upright(\"Sun\")\" drawn in line_fig","t_cap":"a VariableNumber","to_q":"a Line [gray] drawn in sphere (end=(0.766, 0.643))","top_pull":"a Vector [red] drawn in seas (start=(0.0, 0.8), end=(0.0, 0.55))","transfer":"a ParametricCurve [red] drawn in orbit_fig (function=<function>, t_range=(0.0, 3.141592653589793))","v_cap":"a Vector [red] labelled \"v_A\" drawn in orbit_fig (start=(2.1, 0.0), end=(2.1, 0.98))","v_low":"a Vector [red] labelled \"v_B\" drawn in orbit_fig (start=(-1.0, 0.0), end=(-1.0, -1.96))","v_ship":"a Vector [green] labelled \"v_0\" drawn in orbit_fig (start=(1.9, 0.0), end=(1.9, 1.2))"},"beats":[{"start":825.3754166666668,"say":"Back on Earth, the measured g is not one fixed number. With height h, the distance to the centre grows, and expanding to first order, g falls by the fraction two h over R.","live":[],"does":[[825.3754166666668,"head_g is shown on the screen, written out."],[825.9354166666667,"sphere is shown on the screen, written out."],[825.9354166666667,"earth_g is shown on the screen, written out."],[829.5754166666668,"sphere moves to a new place on the board."],[829.5754166666668,"derivation is shown on the screen, written out."],[835.2954166666667,"derivation is shown on the screen, written out."]]},{"start":837.8954166666667,"say":"Climb one kilometre and g drops by about three parts in ten thousand. Gravimeters detect far smaller changes, enough to map dense rock underground.","live":["sphere","head_g","earth_g"],"does":[[840.8554166666668,"derivation is indicated — a transient flash."]]},{"start":848.5304166666667,"say":"Latitude matters too, because the Earth rotates. A point at latitude lambda circles the axis at distance R cosine lambda, and in the rotating frame it feels a centrifugal acceleration directed away from the axis.","live":null,"does":[[848.8154166666668,"equator is shown on the screen, written out."],[850.9754166666668,"spin_axis is shown on the screen, written out."],[852.2954166666667,"site is shown on the screen, written out."],[852.2954166666667,"to_q is shown on the screen, written out."],[852.2954166666667,"g_vec is shown on the screen, written out."],[853.7754166666667,"lat is shown on the screen, written out."],[855.1354166666667,"rho_line is shown on the screen, written out."],[859.1354166666667,"c_vec is shown on the screen, written out."]]},{"start":863.0854166666668,"say":"Its component along the vertical reduces the effective g by omega squared R cosine squared lambda. At the equator that is about three hundredths of a metre per second squared, a third of a percent; the flattening of the Earth adds a comparable effect.","live":["sphere","head_g","earth_g","spin_axis","site","to_q","g_vec","equator","lat","rho_line","c_vec"],"does":[[863.8954166666667,"derivation is shown on the screen, written out."],[870.2954166666667,"derivation is shown on the screen, written out."],[878.5979166666667,"derivation is hidden from the screen — left the board."],[878.5979166666667,"head_g is hidden from the screen — left the board."],[878.5979166666667,"sphere is hidden from the screen — left the board."],[878.5979166666667,"earth_g is hidden from the screen — sphere left the board."],[878.5979166666667,"spin_axis is hidden from the screen — sphere left the board."],[878.5979166666667,"site is hidden from the screen — sphere left the board."],[878.5979166666667,"to_q is hidden from the screen — sphere left the board."],[878.5979166666667,"g_vec is hidden from the screen — sphere left the board."],[878.5979166666667,"equator is hidden from the screen — sphere left the board."],[878.5979166666667,"lat is hidden from the screen — sphere left the board."],[878.5979166666667,"rho_line is hidden from the screen — sphere left the board."],[878.5979166666667,"c_vec is hidden from the screen — sphere left the board."]]},{"start":879.7979166666668,"say":"Tides arise not from the pull of the Moon itself, but from its variation across the Earth. The whole Earth falls toward the Moon with the acceleration at its centre. The near side is pulled a little more, the far side a little less.","live":[],"does":[[879.7979166666668,"head_tide is shown on the screen, written out."],[883.3354166666668,"seas is shown on the screen, written out."],[883.3354166666668,"earth_t is shown on the screen, written out."],[883.3354166666668,"moon is shown on the screen, written out."],[889.6554166666667,"near_pull is shown on the screen, written out."],[891.7354166666668,"far_pull is shown on the screen, written out."]]},{"start":893.9254166666667,"say":"Subtract the central acceleration and expand for r much smaller than d. The difference is two G M r over d cubed: tidal effects fall as the inverse cube of distance, like the field of a dipole, because they are a difference of two nearly equal pulls.","live":["seas","head_tide","earth_t","moon","near_pull","far_pull"],"does":[[894.1754166666667,"seas moves to a new place on the board."],[894.1754166666667,"derivation_2 is shown on the screen, written out."],[899.2954166666667,"derivation_2 is shown on the screen, written out."]]},{"start":911.4279166666668,"say":"On the sides, the pull converges slightly inward. The result is two bulges, one facing the Moon and one opposite, and as the Earth turns beneath them each coast sees two high tides a day, about twelve hours apart.","live":null,"does":[[911.9754166666667,"top_pull is shown on the screen, written out."],[911.9754166666667,"low_pull is shown on the screen, written out."]]},{"start":926.0804166666668,"say":"Now compare the Sun. Its mass is twenty-seven million times that of the Moon, but it is about three hundred and ninety times farther away, and the cube of that ratio wins. The solar tide is about forty-six percent of the lunar one: roughly half.","live":["seas","head_tide","earth_t","moon","near_pull","far_pull","top_pull","low_pull"],"does":[[926.6554166666667,"derivation_2 is shown on the screen, written out."],[927.2154166666667,"sun is shown on the screen, written out."],[939.3754166666668,"derivation_2 (the \"0.46\" part) is emphasized."],[941.9679166666667,"derivation_2 (the \"0.46\" part) is no longer emphasized."]]},{"start":942.5679166666667,"say":"At new and full moon, the syzygies, Sun, Earth and Moon line up, the two tides add, and we get spring tides. At the quarter moons they partly cancel.","live":["seas","head_tide","earth_t","moon","near_pull","far_pull","top_pull","low_pull","sun"],"does":[[946.6954166666667,"sun is indicated — a transient flash."],[946.9163094416322,"moon is indicated — a transient flash."],[952.8329166666667,"derivation_2 is hidden from the screen — left the board."],[952.8329166666667,"head_tide is hidden from the screen — left the board."]]},{"start":954.0329166666668,"say":"What we observe on a coast depends on geography. Each basin has its own natural period, so resonance amplifies or suppresses the forcing: barely half a metre at the Po delta, fifteen metres at Mont Saint-Michel.","live":["seas","earth_t","moon","near_pull","far_pull","top_pull","low_pull","sun"],"does":[[954.0329166666668,"head_fx is shown on the screen, written out."],[954.0329166666668,"fx1 is shown on the screen, written out."]]},{"start":968.3904166666667,"say":"The rock itself is not rigid. The solid Earth rises and falls twice a day, and at CERN the twenty-seven kilometre ring of the collider stretches by about a millimetre, enough to shift the beam energy, so physicists correct for the Moon.","live":["seas","earth_t","moon","near_pull","far_pull","top_pull","low_pull","sun","fx1","head_fx"],"does":[[968.7354166666668,"fx2 is shown on the screen, written out."],[974.3354166666668,"fx3 is shown on the screen, written out."],[983.1104166666668,"fx1 is hidden from the screen — left the board."],[983.1104166666668,"fx2 is hidden from the screen — left the board."],[983.1104166666668,"fx3 is hidden from the screen — left the board."],[983.1104166666668,"head_fx is hidden from the screen — left the board."],[983.1104166666668,"seas is hidden from the screen — left the board."],[983.1104166666668,"earth_t is hidden from the screen — seas left the board."],[983.1104166666668,"moon is hidden from the screen — seas left the board."],[983.1104166666668,"near_pull is hidden from the screen — seas left the board."],[983.1104166666668,"far_pull is hidden from the screen — seas left the board."],[983.1104166666668,"top_pull is hidden from the screen — seas left the board."],[983.1104166666668,"low_pull is hidden from the screen — seas left the board."],[983.1104166666668,"sun is hidden from the screen — seas left the board."]]},{"start":984.3104166666667,"say":"One last equilibrium. Between the Sun and the Earth lies a point where a small body can orbit the Sun in step with the Earth. The pull of the Earth partly cancels that of the Sun, so the body needs less centripetal force than a free orbit at that radius would.","live":[],"does":[[984.3104166666667,"head_l1 is shown on the screen, written out."],[985.2954166666667,"line_fig is shown on the screen, written out."],[985.2954166666667,"join is shown on the screen, written out."],[985.2954166666667,"sun_l is shown on the screen, written out."],[985.2954166666667,"earth_l is shown on the screen, written out."],[988.3354166666668,"l1_pt is shown on the screen, written out."],[993.6954166666667,"pull_sun is shown on the screen, written out."],[993.9024226151136,"pull_earth is shown on the screen, written out."]]},{"start":1000.4304166666667,"say":"Require the net pull to equal omega squared times the distance from the Sun, with the angular velocity of the Earth itself. Expanding to first order in x over d, the distance from the Earth is d times the cube root of M E over three M S.","live":["line_fig","head_l1","join","sun_l","earth_l","l1_pt","pull_sun","pull_earth"],"does":[[1000.6954166666667,"line_fig moves to a new place on the board."],[1000.6954166666667,"derivation_3 is shown on the screen, written out."],[1005.5754166666667,"derivation_3 is shown on the screen, written out."],[1007.9754166666667,"derivation_3 is shown on the screen, written out."]]},{"start":1016.5404166666667,"say":"That is about one and a half million kilometres, one percent of the way to the Sun. Solar observatories are parked there, with an uninterrupted view of the Sun.","live":null,"does":[[1023.0954166666667,"l1_pt is indicated — a transient flash."],[1025.7254166666667,"derivation_3 is hidden from the screen — left the board."],[1025.7254166666667,"head_l1 is hidden from the screen — left the board."],[1025.7254166666667,"line_fig is hidden from the screen — left the board."],[1025.7254166666667,"join is hidden from the screen — line_fig left the board."],[1025.7254166666667,"sun_l is hidden from the screen — line_fig left the board."],[1025.7254166666667,"earth_l is hidden from the screen — line_fig left the board."],[1025.7254166666667,"l1_pt is hidden from the screen — line_fig left the board."],[1025.7254166666667,"pull_sun is hidden from the screen — line_fig left the board."],[1025.7254166666667,"pull_earth is hidden from the screen — line_fig left the board."]]},{"start":1026.9254166666667,"say":"Finally, a manoeuvre that uses everything at once. A ship circles a planet of radius R at radius two R. It releases a capsule at point A, which must land at B, diametrically opposite, on the surface.","live":[],"does":[[1026.9254166666667,"problem is shown on the screen, written out."],[1028.1754166666667,"orbit_fig is shown on the screen, written out."],[1028.1754166666667,"planet is shown on the screen, written out."],[1031.2954166666668,"ship_path is shown on the screen, written out."],[1036.0954166666668,"point_a is shown on the screen, written out."],[1038.0154166666666,"point_b is shown on the screen, written out."]]},{"start":1041.3179166666666,"say":"The free path of the capsule is half an ellipse, with A as apogee and B as perigee. Angular momentum fixes the speeds: v A times two R equals v B times R, so the capsule is twice as fast at B.","live":["orbit_fig","problem","planet","ship_path","point_a","point_b"],"does":[[1043.1754166666667,"transfer is shown on the screen, drawn."],[1044.9354166666667,"capsule is shown on the screen, written out."],[1047.2554166666669,"derivation_4 is shown on the screen, written out."]]},{"start":1056.2854166666666,"say":"Energy conservation between A and B adds a second equation. Substituting v B equals two v A, the capsule must leave A with speed the square root of G M over three R.","live":["orbit_fig","problem","planet","ship_path","point_a","point_b","transfer","capsule"],"does":[[1056.5754166666668,"derivation_4 is shown on the screen, written out."],[1060.4154166666667,"derivation_4 is shown on the screen, written out."]]},{"start":1068.8179166666666,"say":"But the ship moves at the square root of G M over two R, which is faster. So the capsule must be thrown backward, against the direction of the orbit, with a relative speed of about point one three root G M over R. Braking, not pushing, brings it down.","live":null,"does":[[1069.4154166666667,"v_ship is shown on the screen, written out."],[1072.8554166666668,"v_cap is shown on the screen, written out."],[1075.5754166666668,"kick is shown on the screen, written out."],[1078.2954166666668,"launch is shown on the screen, written out."]]},{"start":1085.6204166666666,"say":"Released this way, it falls along the ellipse and meets the surface at B with speed twice the square root of G M over three R. Every step used only the two conservation laws that the orbits of Kepler taught us to trust.","live":["launch","orbit_fig","problem","planet","ship_path","point_a","point_b","transfer","capsule","v_ship","v_cap","kick"],"does":[[1087.3354166666668,"capsule is redrawn as the numbers it depends on change."],[1087.3354166666668,"t_cap ticks to 3.141592653589793."],[1088.5754166666668,"derivation_4 is shown on the screen, written out."],[1090.4554166666667,"v_low is shown on the screen, written out."],[1098.9527708333335,"derivation_4 is hidden from the screen — left the board."],[1098.9527708333335,"launch is hidden from the screen — left the board."],[1098.9527708333335,"orbit_fig is hidden from the screen — left the board."],[1098.9527708333335,"planet is hidden from the screen — orbit_fig left the board."],[1098.9527708333335,"ship_path is hidden from the screen — orbit_fig left the board."],[1098.9527708333335,"point_a is hidden from the screen — orbit_fig left the board."],[1098.9527708333335,"point_b is hidden from the screen — orbit_fig left the board."],[1098.9527708333335,"transfer is hidden from the screen — orbit_fig left the board."],[1098.9527708333335,"capsule is hidden from the screen — orbit_fig left the board."],[1098.9527708333335,"v_ship is hidden from the screen — orbit_fig left the board."],[1098.9527708333335,"v_cap is hidden from the screen — orbit_fig left the board."],[1098.9527708333335,"kick is hidden from the screen — orbit_fig left the board."],[1098.9527708333335,"v_low is hidden from the screen — orbit_fig left the board."],[1098.9527708333335,"problem is hidden from the screen — left the board."]]}]}]},"durationSeconds":1100,"chapters":[{"title":"From Kepler to Newton","startSeconds":0,"narration":"Two lines of reasoning meet in this lecture. Kepler found, from years of careful observation, three rules that the planets obey. Newton showed that one force, falling as the inverse square of distance, explains all three. We will build that force, test it, and then derive every orbit it allows. Start with the observations. Here is one planetary orbit, and here is the Sun. The first law says the orbit is an ellipse, and the Sun sits not at its centre but at one focus. The second law concerns speed. Join the Sun to the planet. In one interval of time near the Sun, the line sweeps this green sector. Now let the planet travel round to the far side. There, in the same interval, it sweeps this thin sector, and the two areas are exactly equal. Far from the Sun the planet moves slowly; near it, quickly. The third law compares different planets. The square of the period, divided by the cube of the semi-major axis, is the same number for all of them. These three laws describe the motion; they do not explain it. Newton's argument is cleanest for a nearly circular orbit of radius r. The planet moves uniformly, so the only acceleration is centripetal, pointing at the Sun. So the force is m omega squared r. Writing omega as two pi over the period, the force becomes m times four pi squared over T squared, times r. Now use the third law: T squared equals k r cubed. Substitute, and the period disappears in a telling way. The force on the planet is proportional to its mass and falls as one over r squared. The third law of dynamics finishes the job. The planet pulls the Sun with an equal and opposite force, so the same expression must also be proportional to the mass M of the Sun. The constant four pi squared over k must therefore be G times M. This is the law of universal gravitation. Every pair of masses attracts along the line joining them, with a force G m M over r squared, and G is one universal constant. Look closely at one step we took for granted. On the left of the second law of dynamics sits the inertial mass, the resistance to changes of velocity. In the law of gravitation sits the gravitational mass, the charge that sources and feels the attraction. Dividing, the acceleration carries the ratio m g over m i. Only if these two masses are equal, for every material, does every body fall with the same acceleration, G M over r squared. That equality is the principle of equivalence. It can be tested with a plumb line. On the rotating Earth the bob feels gravity, proportional to m g, and a centrifugal pull, proportional to m i. If the ratio differed between materials, plumb lines of lead and of wood would hang at different angles. They do not, to better than one part in ten trillion in modern torsion experiments. With the two masses equal, a circular orbit follows at once. Gravity supplies the centripetal acceleration, v squared over r, so the orbital speed is the square root of G M over r. Closer orbits are faster. Measuring G is hard, because gravity between laboratory masses is tiny. Cavendish used a torsion balance. A light rod carrying two small masses m hangs from a thin fibre, and a mirror on the fibre reflects a light beam onto a distant scale, magnifying the smallest twist. Two large masses M are brought close, one on each side. Each attracts its neighbour with a force G m M over r squared, and the rod twists until the restoring torque of the fibre balances the torque of the two forces. The torque of the pair is two b times F, with b the lever arm. The fibre answers with kappa theta, where kappa is its calibrated torsion constant. Equating the two and solving gives G, about six point six seven times ten to the minus eleven, in newton metre squared per kilogram squared."},{"title":"Shells, Potentials and Classic Examples","startSeconds":261.8206458333333,"narration":"Newton's law is written for point masses, yet planets are large. The shell theorem tells us how an extended sphere attracts. Take a thin uniform shell of mass M and radius R, and a point P at distance r from its centre. Slice the shell into rings around the line to P. The ring at angle theta has radius R sine theta, and all of it lies at the same distance s from P. Its share of the mass is M over two, times sine theta d theta. The ring contributes a potential minus G d M over s. The law of cosines links s to theta, and differentiating it gives s d s equal to R r sine theta d theta, which turns the integral over angles into an integral over s. Everything cancels except a plain integral of d s, running from the nearest point of the shell to the farthest, so the potential is minus G M over two R r, times the length of that range. Outside, the range runs from r minus R to r plus R, a length two R. The potential is minus G M over r: exactly that of a point mass at the centre. Now move P inside. The range runs from R minus r to R plus r, a length two r, and the potential becomes minus G M over R, the same everywhere. A constant potential means zero field. Inside a uniform shell there is no gravity at all. Any spherically symmetric body is a nest of such shells. At radius r, the shells outside pull nothing, and the shells inside act as a point mass. So g of r is G times the mass enclosed, over r squared. Consider a body with a hollow core of radius R one, a shell of density rho two out to R two, and an outer layer of density rho three out to R three. The enclosed mass is the integral of four pi r squared rho over the radius. Follow the field outward. In the hollow core it is exactly zero. Through the first shell it grows as r cubed minus R one cubed, over r squared. Through the second it keeps rising, but more gently, because rho three is smaller. Beyond the surface the whole mass acts from the centre, and the field falls as one over r squared. The potential energy is minus m times the integral of g from r out to infinity: continuous everywhere, and flat inside the cavity, where the field vanishes. Here is a striking consequence. Treat the Earth as a uniform sphere and drill a straight frictionless tunnel along a diameter. A body at distance x from the centre feels only the mass inside radius x, which grows as x cubed. So the force grows linearly, minus G M m over R cubed, times x. That is a spring law. The body oscillates harmonically about the centre, with angular frequency the square root of g over R. Putting in numbers, the period is about eighty-four minutes. Drop a stone in at one end and it reaches the other end some forty-two minutes later, at rest, then falls back. Curiously, this is also the period of a satellite skimming the surface. Energy answers the next question: how fast must a body leave the surface never to return? Take zero potential energy at infinity. The minimum launch arrives there with nothing to spare, so its total energy is zero. Solving, the escape speed is the square root of two G M over R, or the square root of two g R: about eleven point two kilometres per second for the Earth. It is exactly root two times the speed of a circular orbit grazing the surface. The same circular orbit reasoning weighs stars. Two stars of equal mass circle their common centre of mass, each at two hundred and twenty kilometres per second, with a period of fourteen point four days. Each star sits at radius r, but the other is two r away. So gravity, G M squared over two r squared, supplies the centripetal force M v squared over r. The radius comes from the motion itself: r is v T over two pi. Solving, M equals four r v squared over G, about one point three times ten to the thirty-two kilograms, some sixty-four solar masses each. This is how astronomers weigh stars they will never visit."},{"title":"The Two-Body Problem and the Conic Orbits","startSeconds":543.5196666666667,"narration":"To find every possible orbit we need a better description of motion under a central force. Use polar coordinates centred on the source: a distance r and an angle theta, with unit vectors u r pointing outward and u theta pointing along increasing angle. The unit vectors turn as the point moves. The derivative of u r is theta dot u theta, and the derivative of u theta is minus theta dot u r. With these, the velocity has a radial part r dot, and a transverse part r theta dot. Differentiate once more and collect terms. The radial acceleration is r double dot minus r theta dot squared, the second piece being the familiar centripetal term. The transverse acceleration is one over r times the time derivative of r squared theta dot. Now two bodies, m one and m two, attract each other. Their centre of mass moves uniformly, so we watch only the separation vector r, from one body to the other. Subtracting the two equations of motion gives a single one, for a fictitious particle of reduced mass mu, attracted to a fixed centre by the force minus k over r squared, with k equal to G m one m two. The force is central, so it exerts no torque, and the angular momentum L equals mu r squared theta dot, constant. The transverse acceleration vanishes, and the area swept per unit time, L over two mu, is constant. That is the second law of Kepler, now derived. The energy is conserved too. Kinetic energy splits into radial motion and a centrifugal term, L squared over two mu r squared, which acts like a repulsive barrier added to the gravitational potential minus k over r. The idea of Binet is to describe the orbit by r as a function of theta, and to use u equals one over r. The chain rule and constant L turn time derivatives into angle derivatives: r dot becomes minus L over mu, times d u by d theta. Differentiating again, r double dot is minus L squared over mu squared, u squared, times the second derivative of u. Insert this into the radial equation, and every power of u lines up. The result is astonishingly simple: the second derivative of u, plus u, equals the constant mu k over L squared. It is the equation of a harmonic oscillator in the angle, shifted by a constant. Its general solution is the constant plus a cosine. Choosing the orientation, u equals mu k over L squared, times one minus epsilon cosine theta. Invert it, and r of theta is D over one minus epsilon cosine theta, with D equal to L squared over mu k. This is the polar equation of a conic with its focus at the centre of force. The first law of Kepler appears as a theorem, and epsilon, the eccentricity, decides the shape. Evaluate the energy at closest approach, and the eccentricity turns out to be the square root of one plus two E L squared over mu k squared. The sign of the energy now classifies every orbit. At the minimum allowed energy, epsilon is zero: a circle. For negative energy, epsilon lies below one: a closed ellipse, a bound orbit like those of the planets. At exactly zero energy, epsilon equals one: a parabola, the borderline case of a body that just escapes. For positive energy, epsilon exceeds one: a hyperbola, an open path that swings once past the centre and leaves forever, like a comet or a probe on a flyby. The ellipse also yields the third law, now exactly. Its area is pi a b, swept at the constant rate L over two mu, so the period is pi a b divided by that rate. Square it, and use the geometric fact that b squared equals a times D. The angular momentum cancels, and T squared equals four pi squared a cubed over G times the total mass. The constant of the third law depends, slightly, on the mass of the planet itself. Add a third body and this beautiful solvability is lost. Poincaré showed that the three-body problem has no general solution in closed form, so the Sun, Earth and Moon are followed numerically, advancing positions and velocities in small time steps."},{"title":"Tides, Surface Gravity and Orbital Dynamics","startSeconds":825.3754166666668,"narration":"Back on Earth, the measured g is not one fixed number. With height h, the distance to the centre grows, and expanding to first order, g falls by the fraction two h over R. Climb one kilometre and g drops by about three parts in ten thousand. Gravimeters detect far smaller changes, enough to map dense rock underground. Latitude matters too, because the Earth rotates. A point at latitude lambda circles the axis at distance R cosine lambda, and in the rotating frame it feels a centrifugal acceleration directed away from the axis. Its component along the vertical reduces the effective g by omega squared R cosine squared lambda. At the equator that is about three hundredths of a metre per second squared, a third of a percent; the flattening of the Earth adds a comparable effect. Tides arise not from the pull of the Moon itself, but from its variation across the Earth. The whole Earth falls toward the Moon with the acceleration at its centre. The near side is pulled a little more, the far side a little less. Subtract the central acceleration and expand for r much smaller than d. The difference is two G M r over d cubed: tidal effects fall as the inverse cube of distance, like the field of a dipole, because they are a difference of two nearly equal pulls. On the sides, the pull converges slightly inward. The result is two bulges, one facing the Moon and one opposite, and as the Earth turns beneath them each coast sees two high tides a day, about twelve hours apart. Now compare the Sun. Its mass is twenty-seven million times that of the Moon, but it is about three hundred and ninety times farther away, and the cube of that ratio wins. The solar tide is about forty-six percent of the lunar one: roughly half. At new and full moon, the syzygies, Sun, Earth and Moon line up, the two tides add, and we get spring tides. At the quarter moons they partly cancel. What we observe on a coast depends on geography. Each basin has its own natural period, so resonance amplifies or suppresses the forcing: barely half a metre at the Po delta, fifteen metres at Mont Saint-Michel. The rock itself is not rigid. The solid Earth rises and falls twice a day, and at CERN the twenty-seven kilometre ring of the collider stretches by about a millimetre, enough to shift the beam energy, so physicists correct for the Moon. One last equilibrium. Between the Sun and the Earth lies a point where a small body can orbit the Sun in step with the Earth. The pull of the Earth partly cancels that of the Sun, so the body needs less centripetal force than a free orbit at that radius would. Require the net pull to equal omega squared times the distance from the Sun, with the angular velocity of the Earth itself. Expanding to first order in x over d, the distance from the Earth is d times the cube root of M E over three M S. That is about one and a half million kilometres, one percent of the way to the Sun. Solar observatories are parked there, with an uninterrupted view of the Sun. Finally, a manoeuvre that uses everything at once. A ship circles a planet of radius R at radius two R. It releases a capsule at point A, which must land at B, diametrically opposite, on the surface. The free path of the capsule is half an ellipse, with A as apogee and B as perigee. Angular momentum fixes the speeds: v A times two R equals v B times R, so the capsule is twice as fast at B. Energy conservation between A and B adds a second equation. Substituting v B equals two v A, the capsule must leave A with speed the square root of G M over three R. But the ship moves at the square root of G M over two R, which is faster. So the capsule must be thrown backward, against the direction of the orbit, with a relative speed of about point one three root G M over R. Braking, not pushing, brings it down. Released this way, it falls along the ellipse and meets the surface at B with speed twice the square root of G M over three R. Every step used only the two conservation laws that the orbits of Kepler taught us to trust."}]}}