{"version":1,"lectureId":"01M14TZRRY2J3BBR5D2P0ZJHD3","attempt":0,"publication":{"slug":"deriving-keplerian-orbits-from-conservation-laws","title":"Deriving Keplerian Orbits from Conservation Laws","subject":"physics","summary":"A derivation-first account of Keplerian motion for physics students who know calculus and angular momentum. The lecture obtains equal areas from zero torque for any central force, introduces inverse-square gravity to derive the polar conic equation, classifies trajectories by eccentricity, derives the period and semi-major-axis relation, checks it against planetary data, and uses conserved energy to distinguish bound motion, escape, and the origin of escape velocity.","metaDescription":"Derive equal areas, conic orbits, Kepler's period law, orbital energy classes, and escape speed from conservation laws.","transcript":"We are going to derive orbital behavior rather than begin with a list of laws. The first result comes from angular momentum alone. It applies to every central force, before gravity or an inverse square has entered. A central force points along the line joining a moving body P to one fixed point O. The blue path could have many shapes. The defining feature is only that the force is radial at every point. Take moments about O. The torque is r cross F. A central force can be written as some scalar function of distance times the radial unit vector, so r and F are parallel. The cross product of parallel vectors is zero. Therefore the torque vanishes, and the rate of change of angular momentum vanishes with it. Angular momentum is therefore constant in magnitude and direction. Since L is perpendicular to r at every instant, every position vector remains in the single plane perpendicular to L. That fixed orbital plane is already a major simplification. The body may travel around O and the radial force turns with it, but the motion cannot leave this two-dimensional plane. Notice what has not appeared. We have not used the strength of the force, its sign, or any inverse-square dependence. Zero torque is the entire argument so far. Now let the body move from P to a nearby point Q. The two radius lines and the small displacement enclose a thin triangle. To first order, the triangle has area one half the magnitude of r cross d r. The cross product supplies base times perpendicular height, exactly the area measurement this little wedge needs. Divide by the elapsed time. The displacement d r divided by d t becomes velocity, so the rate at which area is swept out is one half the magnitude of r cross v. But angular momentum is r cross m v. Its magnitude is m times the cross product already standing here. The areal rate is therefore L divided by two m. Since L is constant, the areal rate is constant. Equal intervals of time must sweep equal areas, wherever the body happens to be on its path. Watch the area element move to different parts of the orbit. Its shape changes because both the radius and the displacement change, but the angular-momentum calculation keeps the area per unit time fixed. This is Kepler's equal-areas law, but its logical source is broader than Keplerian gravity. Any central force conserves angular momentum, fixes an orbital plane, and sweeps area at a constant rate. Gravity enters only when we ask for the shape of the path. Equal areas needed only a central force. Now we specialize to gravity. Work in the fixed orbital plane and write the acceleration as minus mu over r squared in the radial direction. Use polar unit vectors. The position is r times r hat. Differentiating once gives a radial velocity r dot and a transverse velocity r theta dot. Differentiate again. Radial acceleration is r double dot minus r theta dot squared. Tangential acceleration is r theta double dot plus two r dot theta dot. Gravity has no tangential component, so that second expression is zero. Multiplying through by r turns it into the time derivative of r squared theta dot. Therefore r squared theta dot is constant. Call it h, the angular momentum per unit mass. This is the same conservation law that produced equal areas, now written in polar coordinates. The inverse-square force changes the speed and bends the velocity, but the force remains radial and h remains fixed throughout the motion. The radial equation still contains time derivatives and the unknown r. A useful substitution is u of theta equal to one over r. We will describe the orbit by angle rather than by time. Since h equals r squared theta dot, replacing r by one over u gives theta dot equal to h u squared. Now differentiate r with respect to time by the chain rule. Since r is one over u, the factors of u cancel and r dot becomes minus h u prime. Differentiate once more. Another factor theta dot appears, so r double dot becomes minus h squared u squared u double prime. This substitution is not a guess at the orbit. It is a change of variable chosen because conservation of angular momentum makes every time derivative collapse into an angle derivative. Now use the radial equation. Radial acceleration equals minus mu over r squared. Substitute the expressions we just found for r, theta dot, and r double dot. Every term contains u squared. After substitution, the left side is minus h squared u squared times u double prime plus u. The right side is minus mu u squared. Cancel the common factor and divide by h squared. The nonlinear-looking orbital problem has become the linear equation u double prime plus u equals mu over h squared. A constant particular solution is mu over h squared. The homogeneous solution is a sine and cosine. Combine their amplitude into e and their phase into theta zero. Invert u to recover r. Define p as h squared over mu. The orbit is p divided by one plus e cosine theta minus theta zero. Three constants now carry the initial conditions. Theta zero chooses the direction of periapsis. The scale p comes from angular momentum. The dimensionless number e decides the shape. The conic sections have not been assumed. They emerged from the inverse-square force, conservation of angular momentum, and one change from r of time to one over r as a function of angle. The orbit equation already has the focus-polar form of a conic. To see that geometrically rather than merely name it, multiply through by the denominator. Since x equals r cosine theta, replace that product. The equation becomes r plus e x equals p. For positive e, rearrange. The quantity p over e minus x is the perpendicular distance from P to the vertical line x equals p over e. Draw that line as the directrix. The distance from P to the focus is e times its distance to the directrix. Equivalently, their ratio is the constant e. That is the focus-directrix definition of a conic, and e is its eccentricity. The force centre therefore occupies a focus of the conic. It is not generally the geometric centre of the curve. That distinction will matter for both speed and distance around an ellipse. Start with the closed cases. If e is zero, the angular term disappears and r equals the constant p. Every point stays the same distance from the focus, so the orbit is a circle. For e between zero and one, the denominator one plus e cosine theta remains positive for every direction. The radius stays finite all the way around, and the path closes as an ellipse. The focus is displaced from the geometric centre. The body comes close at periapsis and reaches its greatest distance at apoapsis. Equal areas then require it to move faster near periapsis and slower near apoapsis. Both circle and ellipse are bound geometric shapes. Their denominator never reaches zero, so the orbit never runs to infinite radius. At e equal to one, the denominator approaches zero only as theta approaches the backward direction pi. The radius grows without bound in that limiting direction, producing a parabola. A parabolic orbit is the boundary between closed motion and escape. It reaches arbitrarily large distance but has only one limiting escape direction. For e greater than one, the denominator reaches zero before theta reaches pi, when cosine theta equals minus one over e. The physical trajectory is one branch of a hyperbola. The hyperbola approaches an asymptote as the radius diverges. The body arrives from far away, swings past the focus, and departs again without closing its path. One integration constant classifies every inverse-square trajectory. Zero gives a circle. Values between zero and one give ellipses. One gives a parabola, and values greater than one give hyperbolas. Geometry has told us every possible shape. It has not yet told us which shape a particular launch selects. Energy will answer that later. First, for the closed ellipse, we can derive the relation between its size and its period. For a bound ellipse, one complete period sweeps the entire area of the ellipse. We already know the areal rate, so the period is total area divided by that constant rate. Let a be the semi-major axis and b the semi-minor axis. Their product times pi is the area of the ellipse. The areal rate is h over two, where h is the angular momentum per unit mass. This rate is constant everywhere on the orbit. Divide the full area pi a b by h over two. The period is two pi a b over h. This already explains why the changing orbital speed does not complicate the period calculation. Equal areas have packaged the entire uneven motion into the one constant h. The formula still contains b and h. The conic geometry supplies the relation p equals b squared over a. The orbit derivation supplied p equals h squared over mu. Combine the two expressions for p. Then h squared equals mu b squared over a. Now square the period formula. Its numerator is four pi squared a squared b squared, and its denominator is h squared. Substitute for h squared. The b squared cancels completely, and one power of a moves up from the denominator. The result is four pi squared over mu times a cubed. Therefore T squared over a cubed is four pi squared over mu. For every object orbiting the same central mass, that ratio is the same. This is Kepler's period law as a consequence, not an empirical rule placed at the beginning. The inverse-square force fixed the conic, and angular momentum converted its area into a period. Now compare the prediction with the actual planets. Measure a in astronomical units and T in years. For objects orbiting the Sun, the common ratio should be approximately one. Read the table from the inner Solar System outward. Mercury has semi-major axis zero point three eight seven astronomical units and period zero point two four one years. Venus has a equal to zero point seven two three and period zero point six one five years. Its ratio is again essentially one. Earth defines the convenient units: one astronomical unit and one year. Its ratio is exactly one in this unit system. Mars is farther out, at one point five two four astronomical units, and takes one point eight eight one years. The period grows faster than the orbital size itself, exactly as the three-halves power predicts. Jupiter is more than five times Earth's distance from the Sun, but its period is almost twelve years. Even across that much larger scale, the ratio remains zero point nine nine nine. The small deviations shown here come from rounded data. The agreement across the planets is the quantitative signature of one inverse-square gravitational parameter mu belonging to the Sun. Geometry classified the possible conics, but it did not tell us which conic a particular launch selects. For that we need a second conserved quantity: mechanical energy per unit mass. Begin with kinetic energy per unit mass, one half v squared. Its time derivative is acceleration dotted with velocity. In inverse-square gravity only the radial velocity contributes. The derivative of the gravitational potential per unit mass, minus mu over r, is the equal and opposite quantity: plus mu r dot over r squared. Add the two rates and they cancel. Therefore one half v squared minus mu over r is constant. Call this constant epsilon, the specific orbital energy. The example begins at negative energy. In these scaled units the launch speed is zero point eight. The gravitational well has enough depth that this motion cannot reach infinity. At infinite distance the potential approaches zero. Kinetic energy cannot be negative, so a trajectory with negative total energy cannot get all the way there. It must turn back. Now connect energy with the eccentricity already present in the orbit. Differentiating the conic equation gives the radial speed mu e over h times sine theta. The transverse speed is r theta dot, or h over r. Using the orbit equation, it becomes mu over h times one plus e cosine theta. These two velocity components are perpendicular. Square them and add. The result is mu squared over h squared times one plus e squared plus two e cosine theta. Insert this speed and the orbit's reciprocal radius into epsilon. Every term containing theta cancels. Energy cannot depend on where the body happens to be on the orbit. Solve for eccentricity. E squared is one plus two epsilon h squared over mu squared. Because h squared and mu squared are positive, the sign of energy decides whether e lies below, at, or above one. For an ellipse, combine this with h squared equal to mu a times one minus e squared. The specific energy becomes minus mu over two a. A larger bound orbit lies closer to zero energy. Negative energy gives e below one. The trajectory is a bound circle or ellipse, and the body must return rather than reach infinite distance. Increase the launch speed continuously. The velocity arrow grows and the energy marker moves rightward through zero. At zero energy, e equals one. That is the parabolic boundary: the body can reach arbitrarily large distance, but its speed tends to zero as it gets there. Positive energy gives e greater than one. The orbit is hyperbolic and unbound. Even at infinite distance, positive kinetic energy remains. The geometric classification and the physical classification are now the same statement. Ellipses are negative-energy orbits, the parabola is the zero-energy threshold, and hyperbolas carry positive energy. Escape velocity is simply the launch speed at that exact boundary. Lower the example from positive energy until it settles at the marginal speed, while the launch radius stays fixed. At radius r, specific energy is one half v squared minus mu over r. The least speed that can escape is the case epsilon equal to zero. Move the potential term to the other side and multiply by two. Escape speed squared is two mu over r. Take the positive square root. The escape speed is the square root of two mu over r. This is not a speed that turns gravity off. Gravity continues to act at every finite distance. It is the speed whose initial kinetic energy is just enough to raise the total energy to zero. At Earth's surface, ignoring the atmosphere and Earth's rotation, the result is about eleven point two kilometres per second. A faster launch has positive energy. Far away the potential tends to zero, so the remaining energy is kinetic and the craft retains a nonzero speed. Escape velocity is precisely the boundary where that leftover speed vanishes. The lecture now forms one connected chain. A central force gives zero torque, a fixed orbital plane, and equal areas. The inverse-square strength turns the radial equation into the polar conic equation. Eccentricity then distinguishes circle, ellipse, parabola, and hyperbola. For an ellipse, total area divided by constant areal rate produces T squared proportional to a cubed, in agreement with the planets. Finally, conserved energy chooses which conic a launch follows. Its zero level is the boundary between return and escape, and that boundary gives the escape-speed formula.","watch":{"version":1,"scenes":[{"title":"Equal Areas from Angular Momentum","start":0,"end":177.55370833333333,"objects":{"area_work":"a Derivation [text] that says \"$dif A &= frac(1, 2) abs(arrow(r) times dif arrow(r)) \\ frac(dif A, dif t) &= frac(1, 2) abs(arrow(r) times arrow(v)) \\ &= frac(L, 2 m)$\"","body":"a Point [yellow] labelled \"P\" drawn in orbit_plane (location=((2.8 * cos(phase)), (1.65 * sin(phase))))","card":"a Title that says \"Classical Mechanics — Deriving Keplerian Orbits from Conservation Laws\"","centre":"a Point [red] labelled \"O\" drawn in orbit_plane","displacement":"a Vector [green] labelled \"dif arrow(r)\" drawn in orbit_plane (start=((2.8 * cos(phase)), (1.65 * sin(phase))), end=((2.8 * cos((phase + 0.28))), (1.65 * sin((phase + 0.28)))))","equal_areas":"a Math [text] that says \"$frac(dif A, dif t) = frac(L, 2 m) = upright(\"constant\")$\"","force":"a Vector [red] labelled \"arrow(F)\" drawn in orbit_plane (start=((2.8 * cos(phase)), (1.65 * sin(phase))), end=((0.68 * (2.8 * cos(phase))), (0.68 * (1.65 * sin(phase)))))","gravity_not_used":"a Text [text] that says \"No inverse-square law has entered. Equal areas follows from zero torque.\"","heading_area":"a Heading that says \"Angular Momentum Becomes Swept Area\"","heading_central":"a Heading that says \"What Every Central Force Implies\"","l_constant":"a Math [text] that says \"$arrow(L) = upright(\"constant\")$\"","next_body":"a Point [green] labelled \"Q\" drawn in orbit_plane (location=((2.8 * cos((phase + 0.28))), (1.65 * sin((phase + 0.28)))))","next_radius":"a Line [green] drawn in orbit_plane (end=((2.8 * cos((phase + 0.28))), (1.65 * sin((phase + 0.28)))))","orbit_plane":"a Figure (x_range=(-3.4, 3.4), y_range=(-2.3, 2.3), aspect=(6.8, 4.6))","phase":"a VariableNumber (initial_value=0.55)","plane_fact":"a Math [text] that says \"$arrow(L) dot arrow(r) = 0$\"","position":"a Vector [yellow] labelled \"arrow(r)\" drawn in orbit_plane (end=((2.8 * cos(phase)), (1.65 * sin(phase))))","torque_work":"a Derivation [text] that says \"$arrow(tau) &= arrow(r) times arrow(F) \\ &= arrow(r) times F(r) hat(r) \\ &= arrow(0) \\ frac(dif arrow(L), dif t) &= arrow(tau) = arrow(0)$\"","trajectory":"a ParametricCurve [blue] labelled \"C\" drawn in orbit_plane (function=<function>, t_range=(0.0, 6.283185307179586))","wedge":"a Polygon [green] drawn in orbit_plane (vertices=((0.0, 0.0), ((2.8 * cos(phase)), (1.65 * sin(phase))), ((2.8 *…, fill_opacity=0.28)"},"beats":[{"start":0,"say":"We are going to derive orbital behavior rather than begin with a list of laws. The first result comes from angular momentum alone. It applies to every central force, before gravity or an inverse square has entered.","live":[],"does":[[0,"card is shown on the screen, written out."],[1.5,"card: enter:write-left-to-right."],[13.0145,"card is hidden from the screen — left the board."]]},{"start":14.2145,"say":"A central force points along the line joining a moving body P to one fixed point O. The blue path could have many shapes. The defining feature is only that the force is radial at every point.","live":null,"does":[[14.2145,"heading_central is shown on the screen, written out."],[14.737,"orbit_plane is shown on the screen, written out."],[15.132,"force is shown on the screen, written out."],[17.965,"body is shown on the screen, written out."],[17.965,"position is shown on the screen, written out."],[19.102,"centre is shown on the screen, written out."],[20.623,"trajectory is shown on the screen, drawn."]]},{"start":27.539499999999997,"say":"Take moments about O. The torque is r cross F. A central force can be written as some scalar function of distance times the radial unit vector, so r and F are parallel.","live":["orbit_plane","heading_central","trajectory","centre","body","position","force"],"does":[[29.558999999999997,"orbit_plane moves to a new place on the board."],[29.558999999999997,"torque_work is shown on the screen, written out."],[35.875,"torque_work is shown on the screen, written out."],[38.778,"position is indicated — a transient flash."],[38.778,"force is indicated — a transient flash."]]},{"start":40.399499999999996,"say":"The cross product of parallel vectors is zero. Therefore the torque vanishes, and the rate of change of angular momentum vanishes with it.","live":null,"does":[[42.977,"torque_work is shown on the screen, written out."],[45.183,"torque_work (the \"arrow(0)\" part) is indicated — a transient flash."],[46.367,"torque_work is shown on the screen, written out."]]},{"start":50.31099999999999,"say":"Angular momentum is therefore constant in magnitude and direction. Since L is perpendicular to r at every instant, every position vector remains in the single plane perpendicular to L.","live":null,"does":[[52.43599999999999,"l_constant is shown on the screen, written out."],[56.092999999999996,"plane_fact is shown on the screen, written out."],[57.520999999999994,"body is redrawn as the numbers it depends on change."],[57.520999999999994,"position is redrawn as the numbers it depends on change."],[57.520999999999994,"force is redrawn as the numbers it depends on change."],[57.520999999999994,"next_body is redrawn as the numbers it depends on change."],[57.520999999999994,"next_radius is redrawn as the numbers it depends on change."],[57.520999999999994,"displacement is redrawn as the numbers it depends on change."],[57.520999999999994,"wedge is redrawn as the numbers it depends on change."],[57.520999999999994,"phase ticks to 1.65."]]},{"start":63.35699999999999,"say":"That fixed orbital plane is already a major simplification. The body may travel around O and the radial force turns with it, but the motion cannot leave this two-dimensional plane.","live":["l_constant","plane_fact","orbit_plane","heading_central","trajectory","centre","body","position","force"],"does":[[68.836,"body is redrawn as the numbers it depends on change."],[68.836,"position is redrawn as the numbers it depends on change."],[68.836,"force is redrawn as the numbers it depends on change."],[68.836,"next_body is redrawn as the numbers it depends on change."],[68.836,"next_radius is redrawn as the numbers it depends on change."],[68.836,"displacement is redrawn as the numbers it depends on change."],[68.836,"wedge is redrawn as the numbers it depends on change."],[68.836,"phase ticks to 2.45."],[70.218,"force is indicated — a transient flash."]]},{"start":75.79849999999999,"say":"Notice what has not appeared. We have not used the strength of the force, its sign, or any inverse-square dependence. Zero torque is the entire argument so far.","live":null,"does":[[82.428,"gravity_not_used is shown on the screen, written out."],[88.01249999999999,"l_constant moves to a new place on the board."],[88.01249999999999,"gravity_not_used is hidden from the screen — left the board."],[88.01249999999999,"heading_central is hidden from the screen — left the board."],[88.01249999999999,"plane_fact is hidden from the screen — left the board."],[88.01249999999999,"torque_work is hidden from the screen — left the board."]]},{"start":89.21249999999999,"say":"Now let the body move from P to a nearby point Q. The two radius lines and the small displacement enclose a thin triangle.","live":["l_constant","orbit_plane","trajectory","centre","body","position","force"],"does":[[89.21249999999999,"heading_area is shown on the screen, written out."],[92.103,"next_body is shown on the screen, written out."],[92.103,"next_radius is shown on the screen, written out."],[95.029,"displacement is shown on the screen, written out."],[96.55,"wedge is shown on the screen, written out."]]},{"start":98.19449999999999,"say":"To first order, the triangle has area one half the magnitude of r cross d r. The cross product supplies base times perpendicular height, exactly the area measurement this little wedge needs.","live":["l_constant","orbit_plane","trajectory","centre","body","position","force","heading_area","next_body","next_radius","displacement","wedge"],"does":[[100.31899999999999,"area_work is shown on the screen, written out."],[102.09599999999999,"position is indicated — a transient flash."],[102.862,"displacement is indicated — a transient flash."]]},{"start":111.728,"say":"Divide by the elapsed time. The displacement d r divided by d t becomes velocity, so the rate at which area is swept out is one half the magnitude of r cross v.","live":null,"does":[[112.076,"area_work is shown on the screen, written out."],[121.968,"area_work (the \"arrow(r) times arrow(v)\" part) is indicated — a transient flash."]]},{"start":123.91499999999999,"say":"But angular momentum is r cross m v. Its magnitude is m times the cross product already standing here. The areal rate is therefore L divided by two m.","live":null,"does":[[132.738,"area_work is shown on the screen, written out."],[133.272,"area_work (the \"frac(L, 2 m)\" part) is emphasized."],[135.2345,"area_work (the \"frac(L, 2 m)\" part) is no longer emphasized."]]},{"start":135.8345,"say":"Since L is constant, the areal rate is constant. Equal intervals of time must sweep equal areas, wherever the body happens to be on its path.","live":null,"does":[[137.00699999999998,"equal_areas is shown on the screen, written out."],[140.188,"A box is drawn around equal_areas."]]},{"start":146.268,"say":"Watch the area element move to different parts of the orbit. Its shape changes because both the radius and the displacement change, but the angular-momentum calculation keeps the area per unit time fixed.","live":["l_constant","orbit_plane","trajectory","centre","body","position","force","equal_areas","heading_area","next_body","next_radius","displacement","wedge"],"does":[[147.94000000000003,"body is redrawn as the numbers it depends on change."],[147.94000000000003,"position is redrawn as the numbers it depends on change."],[147.94000000000003,"force is redrawn as the numbers it depends on change."],[147.94000000000003,"next_body is redrawn as the numbers it depends on change."],[147.94000000000003,"next_radius is redrawn as the numbers it depends on change."],[147.94000000000003,"displacement is redrawn as the numbers it depends on change."],[147.94000000000003,"wedge is redrawn as the numbers it depends on change."],[147.94000000000003,"phase ticks to 3.2."],[150.59900000000002,"body is redrawn as the numbers it depends on change."],[150.59900000000002,"position is redrawn as the numbers it depends on change."],[150.59900000000002,"force is redrawn as the numbers it depends on change."],[150.59900000000002,"next_body is redrawn as the numbers it depends on change."],[150.59900000000002,"next_radius is redrawn as the numbers it depends on change."],[150.59900000000002,"displacement is redrawn as the numbers it depends on change."],[150.59900000000002,"wedge is redrawn as the numbers it depends on change."],[150.59900000000002,"phase ticks to 4.35."]]},{"start":159.5,"say":"This is Kepler's equal-areas law, but its logical source is broader than Keplerian gravity. Any central force conserves angular momentum, fixes an orbital plane, and sweeps area at a constant rate. Gravity enters only when we ask for the shape of the path.","live":null,"does":[[160.661,"equal_areas is indicated — a transient flash."],[176.51204166666668,"area_work is hidden from the screen — left the board."],[176.51204166666668,"equal_areas is hidden from the screen — left the board."],[176.51204166666668,"heading_area is hidden from the screen — left the board."],[176.51204166666668,"l_constant is hidden from the screen — left the board."],[176.51204166666668,"orbit_plane is hidden from the screen — left the board."],[176.51204166666668,"trajectory is hidden from the screen — orbit_plane left the board."],[176.51204166666668,"centre is hidden from the screen — orbit_plane left the board."],[176.51204166666668,"body is hidden from the screen — orbit_plane left the board."],[176.51204166666668,"position is hidden from the screen — orbit_plane left the board."],[176.51204166666668,"force is hidden from the screen — orbit_plane left the board."],[176.51204166666668,"next_body is hidden from the screen — orbit_plane left the board."],[176.51204166666668,"next_radius is hidden from the screen — orbit_plane left the board."],[176.51204166666668,"displacement is hidden from the screen — orbit_plane left the board."],[176.51204166666668,"wedge is hidden from the screen — orbit_plane left the board."]]}]},{"title":"The Inverse-Square Orbit Equation","start":177.55370833333333,"end":402.0461875,"objects":{"angular_work":"a Derivation [text] that says \"$a_theta &= r dot(dot(theta))+2 dot(r) dot(theta)=0 \\ frac(dif, dif t)(r^2 dot(theta)) &= 0 \\ h = r^2 dot(theta) &= upright(\"constant\")$\"","body":"a Point [yellow] labelled \"P\" drawn in plane (location=(((1.55 / (1.0 + (0.52 * cos(phase)))) * cos(phase)), ((1.55 / …)","change_work":"a Derivation [text] that says \"$u(theta) &= frac(1, r) \\ dot(theta) &= h u^2 \\ dot(r) &= -h u' \\ dot(dot(r)) &= -h^2 u^2 u''$\"","focus":"a Point [red] labelled \"O\" drawn in plane","gravity":"a Vector [red] labelled \"arrow(a)\" drawn in plane (start=(((1.55 / (1.0 + (0.52 * cos(phase)))) * cos(phase)), ((1.55 / …, end=((0.68 * ((1.55 / (1.0 + (0.52 * cos(phase)))) * cos(phase))), …)","heading_change":"a Heading that says \"Change the Dependent Variable\"","heading_motion":"a Heading that says \"Motion in the Orbital Plane\"","heading_orbit":"a Heading that says \"The Orbit Falls Out\"","kinematics":"a Derivation [text] that says \"$arrow(r) &= r hat(r) \\ arrow(v) &= dot(r) hat(r) + r dot(theta) hat(theta) \\ arrow(a) &= (dot(dot(r))-r dot(theta)^2) hat(r) +(r dot(dot(theta))+2 dot(r) dot(theta)) hat(theta)$\"","orbit":"a ParametricCurve [blue] labelled \"C\" drawn in plane (function=<function>, t_range=(-3.141592653589793, 3.141592653589793))","orbit_equation":"a Math [text] that says \"$r(theta) = frac(p, 1+e cos(theta-theta_0))$\"","orbit_work":"a Derivation [text] that says \"$dot(dot(r))-r dot(theta)^2 &= -frac(mu, r^2) \\ -h^2 u^2 (u''+u) &= -mu u^2 \\ u''+u &= frac(mu, h^2) \\ u(theta) &= frac(mu, h^2) (1+e cos(theta-theta_0))$\"","p_relation":"a Math [text] that says \"$p = frac(h^2, mu)$\"","phase":"a VariableNumber (initial_value=0.42)","plane":"a Figure (x_range=(-3.9, 2.7), y_range=(-2.7, 2.7), aspect=(6.6, 5.4))","position":"a Vector [yellow] labelled \"arrow(r)\" drawn in plane (end=(((1.55 / (1.0 + (0.52 * cos(phase)))) * cos(phase)), ((1.55 / …)","t2":"a Math [text] that says \"$h = r^2 dot(theta) &= upright(\"constant\")$\"","velocity":"a Vector [green] labelled \"arrow(v)\" drawn in plane (start=(((1.55 / (1.0 + (0.52 * cos(phase)))) * cos(phase)), ((1.55 / …, end=((((1.55 / (1.0 + (0.52 * cos(phase)))) * cos(phase)) + ((0.75 …)"},"beats":[{"start":177.55370833333333,"say":"Equal areas needed only a central force. Now we specialize to gravity. Work in the fixed orbital plane and write the acceleration as minus mu over r squared in the radial direction.","live":[],"does":[[177.55370833333333,"heading_motion is shown on the screen, written out."],[181.87270833333332,"orbit is shown on the screen, drawn."],[184.34570833333333,"plane is shown on the screen, written out."],[185.18170833333335,"gravity is shown on the screen, written out."],[188.36270833333333,"focus is shown on the screen, written out."],[188.36270833333333,"body is shown on the screen, written out."],[188.36270833333333,"position is shown on the screen, written out."]]},{"start":190.27470833333334,"say":"Use polar unit vectors. The position is r times r hat. Differentiating once gives a radial velocity r dot and a transverse velocity r theta dot.","live":["plane","heading_motion","orbit","focus","body","position","gravity"],"does":[[193.06070833333334,"plane moves to a new place on the board."],[193.06070833333334,"kinematics is shown on the screen, written out."],[196.16070833333333,"kinematics is shown on the screen, written out."],[197.97170833333334,"kinematics (the \"dot(r) hat(r)\" part) is emphasized."],[198.36670833333335,"velocity is shown on the screen, written out."],[200.08470833333334,"kinematics (the \"dot(r) hat(r)\" part) is no longer emphasized."],[200.08470833333334,"kinematics (the \"r dot(theta) hat(theta)\" part) is emphasized."],[202.53420833333334,"kinematics (the \"r dot(theta) hat(theta)\" part) is no longer emphasized."]]},{"start":203.13420833333333,"say":"Differentiate again. Radial acceleration is r double dot minus r theta dot squared. Tangential acceleration is r theta double dot plus two r dot theta dot.","live":["plane","heading_motion","orbit","focus","body","position","gravity","velocity"],"does":[[204.34170833333334,"kinematics is shown on the screen, written out."],[205.87370833333333,"kinematics (the \"dot(dot(r))-r dot(theta)^2\" part) is indicated — a transient flash."],[210.56470833333333,"kinematics (the \"r dot(dot(theta))+2 dot(r) dot(theta)\" part) is indicated — a transient flash."]]},{"start":216.38870833333334,"say":"Gravity has no tangential component, so that second expression is zero. Multiplying through by r turns it into the time derivative of r squared theta dot.","live":null,"does":[[220.82370833333334,"angular_work is shown on the screen, written out."],[224.91070833333333,"angular_work is shown on the screen, written out."]]},{"start":228.06470833333333,"say":"Therefore r squared theta dot is constant. Call it h, the angular momentum per unit mass. This is the same conservation law that produced equal areas, now written in polar coordinates.","live":null,"does":[[230.51470833333332,"angular_work is shown on the screen, written out."],[232.38370833333335,"angular_work (the \"h\" part) is indicated — a transient flash."],[240.71970833333333,"body is redrawn as the numbers it depends on change."],[240.71970833333333,"position is redrawn as the numbers it depends on change."],[240.71970833333333,"gravity is redrawn as the numbers it depends on change."],[240.71970833333333,"velocity is redrawn as the numbers it depends on change."],[240.71970833333333,"phase ticks to 1.25."]]},{"start":242.22570833333333,"say":"The inverse-square force changes the speed and bends the velocity, but the force remains radial and h remains fixed throughout the motion.","live":null,"does":[[243.86270833333333,"body is redrawn as the numbers it depends on change."],[243.86270833333333,"position is redrawn as the numbers it depends on change."],[243.86270833333333,"gravity is redrawn as the numbers it depends on change."],[243.86270833333333,"velocity is redrawn as the numbers it depends on change."],[243.86270833333333,"phase ticks to 2.1."],[247.53070833333334,"gravity is indicated — a transient flash."],[250.71220833333334,"angular_work is hidden from the screen — left the board."],[250.71220833333334,"heading_motion is hidden from the screen — left the board."],[250.71220833333334,"kinematics is hidden from the screen — left the board."]]},{"start":251.91220833333333,"say":"The radial equation still contains time derivatives and the unknown r. A useful substitution is u of theta equal to one over r. We will describe the orbit by angle rather than by time.","live":["plane","orbit","focus","body","position","gravity","velocity"],"does":[[251.91220833333333,"heading_change is shown on the screen, written out."],[257.60070833333333,"change_work is shown on the screen, written out."]]},{"start":266.01420833333333,"say":"Since h equals r squared theta dot, replacing r by one over u gives theta dot equal to h u squared.","live":["plane","orbit","focus","body","position","gravity","velocity","heading_change"],"does":[[271.41270833333334,"change_work is shown on the screen, written out."]]},{"start":274.79920833333335,"say":"Now differentiate r with respect to time by the chain rule. Since r is one over u, the factors of u cancel and r dot becomes minus h u prime.","live":null,"does":[[278.0037083333333,"change_work is shown on the screen, written out."],[284.2727083333333,"change_work (the \"-h u'\" part) is indicated — a transient flash."]]},{"start":286.5562083333333,"say":"Differentiate once more. Another factor theta dot appears, so r double dot becomes minus h squared u squared u double prime.","live":null,"does":[[287.7287083333333,"change_work is shown on the screen, written out."],[295.83270833333336,"change_work (the \"u''\" part) is indicated — a transient flash."]]},{"start":297.46570833333334,"say":"This substitution is not a guess at the orbit. It is a change of variable chosen because conservation of angular momentum makes every time derivative collapse into an angle derivative.","live":null,"does":[[307.38070833333336,"body is redrawn as the numbers it depends on change."],[307.38070833333336,"position is redrawn as the numbers it depends on change."],[307.38070833333336,"gravity is redrawn as the numbers it depends on change."],[307.38070833333336,"velocity is redrawn as the numbers it depends on change."],[307.38070833333336,"phase ticks to 2.75."],[308.7157083333333,"change_work is hidden from the screen — left the board."],[308.7157083333333,"heading_change is hidden from the screen — left the board."]]},{"start":309.9157083333333,"say":"Now use the radial equation. Radial acceleration equals minus mu over r squared. Substitute the expressions we just found for r, theta dot, and r double dot.","live":["plane","orbit","focus","body","position","gravity","velocity"],"does":[[309.9157083333333,"heading_orbit is shown on the screen, written out."],[310.9837083333333,"orbit_work is shown on the screen, written out."]]},{"start":323.33270833333336,"say":"Every term contains u squared. After substitution, the left side is minus h squared u squared times u double prime plus u. The right side is minus mu u squared.","live":["plane","orbit","focus","body","position","gravity","velocity","heading_orbit"],"does":[[325.1087083333333,"orbit_work (the \"u^2\" part) is indicated — a transient flash."],[327.14070833333335,"orbit_work is shown on the screen, written out."],[330.04370833333337,"orbit_work (the \"u^2#2\" part) is indicated — a transient flash."]]},{"start":337.1562083333333,"say":"Cancel the common factor and divide by h squared. The nonlinear-looking orbital problem has become the linear equation u double prime plus u equals mu over h squared.","live":null,"does":[[337.4117083333333,"orbit_work is shown on the screen, written out."],[344.02970833333336,"orbit_work (the \"u''+u\" part) is indicated — a transient flash."]]},{"start":350.2487083333333,"say":"A constant particular solution is mu over h squared. The homogeneous solution is a sine and cosine. Combine their amplitude into e and their phase into theta zero.","live":null,"does":[[350.7477083333333,"orbit_work is shown on the screen, written out."],[359.72270833333334,"orbit_work (the \"e\" part) is indicated — a transient flash."],[361.4757083333333,"orbit_work (the \"theta_0\" part) is indicated — a transient flash."]]},{"start":363.2537083333333,"say":"Invert u to recover r. Define p as h squared over mu. The orbit is p divided by one plus e cosine theta minus theta zero.","live":null,"does":[[365.9417083333333,"p_relation is shown on the screen, written out."],[369.3897083333333,"orbit_equation is shown on the screen, written out."],[374.3932083333333,"A box is drawn around orbit_equation."]]},{"start":374.9932083333333,"say":"Three constants now carry the initial conditions. Theta zero chooses the direction of periapsis. The scale p comes from angular momentum. The dimensionless number e decides the shape.","live":["plane","orbit","focus","body","position","gravity","velocity","p_relation","orbit_equation","heading_orbit"],"does":[[379.7537083333333,"orbit_equation (the \"theta_0\" part) is indicated — a transient flash."],[382.2957083333333,"orbit_equation (the \"p\" part) is indicated — a transient flash."],[387.42770833333327,"orbit_equation (the \"e\" part) is indicated — a transient flash."]]},{"start":388.8052083333333,"say":"The conic sections have not been assumed. They emerged from the inverse-square force, conservation of angular momentum, and one change from r of time to one over r as a function of angle.","live":null,"does":[[389.3277083333333,"orbit_equation is indicated — a transient flash."],[401.00452083333334,"heading_orbit is hidden from the screen — left the board."],[401.00452083333334,"orbit_equation is hidden from the screen — left the board."],[401.00452083333334,"orbit_work is hidden from the screen — left the board."],[401.00452083333334,"p_relation is hidden from the screen — left the board."],[401.00452083333334,"plane is hidden from the screen — left the board."],[401.00452083333334,"orbit is hidden from the screen — plane left the board."],[401.00452083333334,"focus is hidden from the screen — plane left the board."],[401.00452083333334,"body is hidden from the screen — plane left the board."],[401.00452083333334,"position is hidden from the screen — plane left the board."],[401.00452083333334,"gravity is hidden from the screen — plane left the board."],[401.00452083333334,"velocity is hidden from the screen — plane left the board."]]}]},{"title":"Eccentricity and the Conic Sections","start":402.0461875,"end":593.156625,"objects":{"circle_curve":"a ParametricCurve [blue] drawn in circle_figure (function=<function>, t_range=(-3.141592653589793, 3.141592653589793))","circle_figure":"a Figure (x_range=(-2.3, 2.3), y_range=(-2.3, 2.3), aspect=(1.0, 1.0))","circle_focus":"a Point [red] labelled \"F\" drawn in circle_figure","circle_formula":"a Math [text] that says \"$e = 0, quad r = p$\"","classification":"a Block [text] that says \"$e=0$: circle. $0<e<1$: ellipse. $e=1$: parabola. $e>1$: hyperbola.\"","conic_work":"a Derivation [text] that says \"$r (1 + e cos(theta)) &= p \\ r + e x &= p \\ r &= e (frac(p, e) - x) \\ frac(r, d_perp) &= e$\"","directrix":"a Line [gray] labelled \"d\" drawn in focus_figure (start=(3.0, -2.8), end=(3.0, 2.8), dashed=True)","ellipse_curve":"a ParametricCurve [red] drawn in ellipse_figure (function=<function>, t_range=(-3.141592653589793, 3.141592653589793))","ellipse_figure":"a Figure (x_range=(-3.7, 2.2), y_range=(-2.3, 2.3), aspect=(5.9, 4.6))","ellipse_focus":"a Point [yellow] labelled \"F\" drawn in ellipse_figure","ellipse_formula":"a Math [text] that says \"$0 < e < 1$\"","focus":"a Point [red] labelled \"F\" drawn in focus_figure","focus_curve":"a ParametricCurve [blue] drawn in focus_figure (function=<function>, t_range=(-3.141592653589793, 3.141592653589793))","focus_distance":"a Line [yellow] labelled \"r\" drawn in focus_figure (end=(0.8149495023197076, 1.0269653122367388))","focus_figure":"a Figure (x_range=(-4.8, 3.4), y_range=(-3.0, 3.0), aspect=(8.2, 6.0))","heading_closed":"a Heading that says \"The Closed Cases\"","heading_conic":"a Heading that says \"Why the Orbit Is a Conic\"","heading_open":"a Heading that says \"The Open Cases\"","heading_summary":"a Heading that says \"One Number, Four Shapes\"","hyperbola_curve":"a ParametricCurve [green] drawn in hyperbola_figure (function=<function>, t_range=(-2.05, 2.05))","hyperbola_figure":"a Figure (x_range=(-2.4, 2.0), y_range=(-3.5, 3.5), aspect=(4.4, 7.0))","hyperbola_focus":"a Point [red] labelled \"F\" drawn in hyperbola_figure","hyperbola_formula":"a Math [text] that says \"$e > 1$\"","line_distance":"a Line [green] labelled \"d_perp\" drawn in focus_figure (start=(0.8149495023197076, 1.0269653122367388), end=(3.0, 1.0269653122367388))","parabola_curve":"a ParametricCurve [yellow] drawn in parabola_figure (function=<function>, t_range=(-2.42, 2.42))","parabola_figure":"a Figure (x_range=(-4.6, 2.0), y_range=(-3.6, 3.6), aspect=(6.6, 7.2))","parabola_focus":"a Point [red] labelled \"F\" drawn in parabola_figure","parabola_formula":"a Math [text] that says \"$e = 1$\"","sample":"a Point [yellow] labelled \"P\" drawn in focus_figure (location=(0.8149495023197076, 1.0269653122367388))","tex":"a Tex [text] that says \"Circle\"","tex_2":"a Tex [text] that says \"Ellipse\"","tex_3":"a Tex [text] that says \"Parabola\"","tex_4":"a Tex [text] that says \"Hyperbola\""},"beats":[{"start":402.0461875,"say":"The orbit equation already has the focus-polar form of a conic. To see that geometrically rather than merely name it, multiply through by the denominator.","live":[],"does":[[402.0461875,"heading_conic is shown on the screen, written out."],[402.0461875,"focus_figure is shown on the screen, written out."],[402.30118749999997,"focus_curve is shown on the screen, drawn."],[404.06618749999996,"focus is shown on the screen, written out."],[409.4761875,"focus_figure moves to a new place on the board."],[409.4761875,"conic_work is shown on the screen, written out."]]},{"start":412.27068749999995,"say":"Since x equals r cosine theta, replace that product. The equation becomes r plus e x equals p.","live":["focus_figure","heading_conic","focus_curve","focus"],"does":[[415.3121875,"conic_work is shown on the screen, written out."],[417.3911875,"sample is shown on the screen, written out."],[417.3911875,"focus_distance is shown on the screen, written out."]]},{"start":421.77568749999995,"say":"For positive e, rearrange. The quantity p over e minus x is the perpendicular distance from P to the vertical line x equals p over e. Draw that line as the directrix.","live":["focus_figure","heading_conic","focus_curve","focus","sample","focus_distance"],"does":[[423.28518749999995,"conic_work is shown on the screen, written out."],[428.2191875,"line_distance is shown on the screen, written out."],[433.93118749999996,"directrix is shown on the screen, written out."]]},{"start":435.68668749999995,"say":"The distance from P to the focus is e times its distance to the directrix. Equivalently, their ratio is the constant e. That is the focus-directrix definition of a conic, and e is its eccentricity.","live":["focus_figure","heading_conic","focus_curve","focus","sample","focus_distance","directrix","line_distance"],"does":[[437.2481875,"focus_distance is indicated — a transient flash."],[439.4541875,"line_distance is indicated — a transient flash."],[441.5791875,"conic_work is shown on the screen, written out."]]},{"start":449.42368749999997,"say":"The force centre therefore occupies a focus of the conic. It is not generally the geometric centre of the curve. That distinction will matter for both speed and distance around an ellipse.","live":null,"does":[[451.4901875,"focus is indicated — a transient flash."],[460.56868749999995,"conic_work is hidden from the screen — left the board."],[460.56868749999995,"focus_figure is hidden from the screen — left the board."],[460.56868749999995,"focus_curve is hidden from the screen — focus_figure left the board."],[460.56868749999995,"focus is hidden from the screen — focus_figure left the board."],[460.56868749999995,"sample is hidden from the screen — focus_figure left the board."],[460.56868749999995,"focus_distance is hidden from the screen — focus_figure left the board."],[460.56868749999995,"directrix is hidden from the screen — focus_figure left the board."],[460.56868749999995,"line_distance is hidden from the screen — focus_figure left the board."],[460.56868749999995,"heading_conic is hidden from the screen — left the board."]]},{"start":461.76868749999994,"say":"Start with the closed cases. If e is zero, the angular term disappears and r equals the constant p. Every point stays the same distance from the focus, so the orbit is a circle.","live":[],"does":[[461.76868749999994,"heading_closed is shown on the screen, written out."],[465.06618749999996,"circle_formula is shown on the screen, written out."],[468.45618749999994,"circle_figure is shown on the screen, written out."],[472.55418749999995,"circle_focus is shown on the screen, written out."],[474.1221875,"circle_curve is shown on the screen, drawn."]]},{"start":475.60418749999997,"say":"For e between zero and one, the denominator one plus e cosine theta remains positive for every direction. The radius stays finite all the way around, and the path closes as an ellipse.","live":["circle_formula","circle_figure","heading_closed","circle_curve","circle_focus"],"does":[[476.50918749999994,"ellipse_formula is shown on the screen, written out."],[484.18418749999995,"ellipse_figure is shown on the screen, written out."],[487.24918749999995,"ellipse_curve is shown on the screen, drawn."],[487.24918749999995,"ellipse_focus is shown on the screen, written out."]]},{"start":488.7666875,"say":"The focus is displaced from the geometric centre. The body comes close at periapsis and reaches its greatest distance at apoapsis. Equal areas then require it to move faster near periapsis and slower near apoapsis.","live":["circle_formula","circle_figure","ellipse_formula","ellipse_figure","heading_closed","circle_curve","circle_focus","ellipse_curve","ellipse_focus"],"does":[[489.27718749999997,"ellipse_focus is indicated — a transient flash."],[493.43418749999995,"ellipse_curve is indicated — a transient flash."]]},{"start":503.9596875,"say":"Both circle and ellipse are bound geometric shapes. Their denominator never reaches zero, so the orbit never runs to infinite radius.","live":null,"does":[[504.6221875,"circle_formula is indicated — a transient flash."],[505.2721875,"ellipse_formula is indicated — a transient flash."],[512.7601874999999,"circle_figure is hidden from the screen — left the board."],[512.7601874999999,"circle_curve is hidden from the screen — circle_figure left the board."],[512.7601874999999,"circle_focus is hidden from the screen — circle_figure left the board."],[512.7601874999999,"circle_formula is hidden from the screen — left the board."],[512.7601874999999,"ellipse_figure is hidden from the screen — left the board."],[512.7601874999999,"ellipse_curve is hidden from the screen — ellipse_figure left the board."],[512.7601874999999,"ellipse_focus is hidden from the screen — ellipse_figure left the board."],[512.7601874999999,"ellipse_formula is hidden from the screen — left the board."],[512.7601874999999,"heading_closed is hidden from the screen — left the board."]]},{"start":513.9601875,"say":"At e equal to one, the denominator approaches zero only as theta approaches the backward direction pi. The radius grows without bound in that limiting direction, producing a parabola.","live":[],"does":[[513.9601875,"heading_open is shown on the screen, written out."],[515.3421874999999,"parabola_formula is shown on the screen, written out."],[521.5881875,"parabola_figure is shown on the screen, written out."],[525.1641874999999,"parabola_curve is shown on the screen, drawn."],[525.1641874999999,"parabola_focus is shown on the screen, written out."]]},{"start":526.6696875,"say":"A parabolic orbit is the boundary between closed motion and escape. It reaches arbitrarily large distance but has only one limiting escape direction.","live":["parabola_formula","parabola_figure","heading_open","parabola_curve","parabola_focus"],"does":[[528.3531875,"parabola_curve is indicated — a transient flash."]]},{"start":536.7551874999999,"say":"For e greater than one, the denominator reaches zero before theta reaches pi, when cosine theta equals minus one over e. The physical trajectory is one branch of a hyperbola.","live":null,"does":[[537.4051875,"hyperbola_formula is shown on the screen, written out."],[538.7871875,"hyperbola_figure is shown on the screen, written out."],[547.9701875,"hyperbola_curve is shown on the screen, drawn."],[547.9701875,"hyperbola_focus is shown on the screen, written out."]]},{"start":549.6731875,"say":"The hyperbola approaches an asymptote as the radius diverges. The body arrives from far away, swings past the focus, and departs again without closing its path.","live":["parabola_formula","parabola_figure","hyperbola_formula","hyperbola_figure","heading_open","parabola_curve","parabola_focus","hyperbola_curve","hyperbola_focus"],"does":[[551.4961874999999,"hyperbola_curve is indicated — a transient flash."],[557.7421875,"hyperbola_focus is indicated — a transient flash."],[561.1326875,"heading_open is hidden from the screen — left the board."],[561.1326875,"hyperbola_figure is hidden from the screen — left the board."],[561.1326875,"hyperbola_curve is hidden from the screen — hyperbola_figure left the board."],[561.1326875,"hyperbola_focus is hidden from the screen — hyperbola_figure left the board."],[561.1326875,"hyperbola_formula is hidden from the screen — left the board."],[561.1326875,"parabola_figure is hidden from the screen — left the board."],[561.1326875,"parabola_curve is hidden from the screen — parabola_figure left the board."],[561.1326875,"parabola_focus is hidden from the screen — parabola_figure left the board."],[561.1326875,"parabola_formula is hidden from the screen — left the board."]]},{"start":561.7326875,"say":"One integration constant classifies every inverse-square trajectory. Zero gives a circle. Values between zero and one give ellipses. One gives a parabola, and values greater than one give hyperbolas.","live":[],"does":[[561.7326875,"heading_summary is shown on the screen, written out."],[562.0811874999999,"classification is shown on the screen, written out."],[567.7811875,"classification (the \"circle\" part) is indicated — a transient flash."],[571.2641874999999,"classification (the \"ellipse\" part) is indicated — a transient flash."],[573.1801875,"classification (the \"parabola\" part) is indicated — a transient flash."],[575.5941875,"classification (the \"hyperbola\" part) is indicated — a transient flash."]]},{"start":577.3091875,"say":"Geometry has told us every possible shape. It has not yet told us which shape a particular launch selects. Energy will answer that later. First, for the closed ellipse, we can derive the relation between its size and its period.","live":["classification","heading_summary"],"does":[[592.1149583333333,"classification is hidden from the screen — left the board."],[592.1149583333333,"heading_summary is hidden from the screen — left the board."]]}]},{"title":"The Period Law and the Planets","start":593.156625,"end":793.8776875,"objects":{"body":"a Point [yellow] labelled \"P\" drawn in ellipse_figure (location=(2.105056041594506, 1.0879847087529306))","ellipse":"a ParametricCurve [blue] drawn in ellipse_figure (function=<function>, t_range=(0.0, 6.283185307179586))","ellipse_figure":"a Figure (x_range=(-3.6, 3.6), y_range=(-2.6, 2.6), aspect=(7.2, 5.2))","focus":"a Point [red] labelled \"F\" drawn in ellipse_figure (location=(-2.262189205172724, 0.0))","heading_area":"a Heading that says \"Area Divided by Areal Rate\"","heading_planets":"a Heading that says \"The Solar System Check\"","heading_scale":"a Heading that says \"Eliminate the Hidden Quantities\"","interpretation":"a Text [text] that says \"For bodies orbiting the same central mass, the ratio $T^2/a^3$ is common.\"","kepler_three":"a Math [text] that says \"$frac(T^2, a^3) = frac(4 pi^2, mu)$\"","period_formula":"a Math [text] that says \"$T = frac(2 pi a b, h)$\"","period_work":"a Derivation [text] that says \"$upright(\"Area of ellipse\") &= pi a b \\ frac(dif A, dif t) &= frac(h, 2) \\ T &= frac(pi a b, h/2) \\ &= frac(2 pi a b, h)$\"","planet_table":"a Table [text] that says \"Planet $a$ AU $T$ yr $T^2/a^3$ Mercury 0.387 0.241 1.002 Venus 0.723 0.615 1.001 Earth 1.000 1.000 1.000 Mars 1.524 1.881 0.999 Jupiter 5.203 11.862 0.999\" (rows=(('Planet', '$a$ AU', '$T$ yr', '$T^2/a^3$'), ('Mercury', '0.38…, header=True)","radius":"a Line [gray] drawn in ellipse_figure (start=(-2.262189205172724, 0.0), end=(2.105056041594506, 1.0879847087529306))","scale_work":"a Derivation [text] that says \"$p &= frac(b^2, a) \\ h^2 &= mu p = frac(mu b^2, a) \\ T^2 &= frac(4 pi^2 a^2 b^2, h^2) \\ &= frac(4 pi^2, mu) a^3$\"","semi_major":"a Line [yellow] labelled \"a\" drawn in ellipse_figure (end=(2.8, 0.0))","semi_minor":"a Line [green] labelled \"b\" drawn in ellipse_figure (end=(0.0, 1.65))","solar_form":"a Math [text] that says \"$frac(T^2, a^3) approx 1 thin upright(\"yr\")^2 / upright(\"AU\")^3$\""},"beats":[{"start":593.156625,"say":"For a bound ellipse, one complete period sweeps the entire area of the ellipse. We already know the areal rate, so the period is total area divided by that constant rate.","live":[],"does":[[593.156625,"heading_area is shown on the screen, written out."],[593.516625,"focus is shown on the screen, written out."],[593.795625,"ellipse_figure is shown on the screen, written out."],[593.795625,"ellipse is shown on the screen, drawn."],[595.199625,"body is shown on the screen, written out."],[595.199625,"radius is shown on the screen, written out."]]},{"start":604.646625,"say":"Let a be the semi-major axis and b the semi-minor axis. Their product times pi is the area of the ellipse.","live":["ellipse_figure","heading_area","ellipse","focus","body","radius"],"does":[[605.656625,"semi_major is shown on the screen, written out."],[607.688625,"semi_minor is shown on the screen, written out."],[611.217625,"ellipse_figure moves to a new place on the board."],[611.217625,"period_work is shown on the screen, written out."]]},{"start":613.373625,"say":"The areal rate is h over two, where h is the angular momentum per unit mass. This rate is constant everywhere on the orbit.","live":["ellipse_figure","heading_area","ellipse","focus","body","radius","semi_major","semi_minor"],"does":[[614.255625,"period_work is shown on the screen, written out."],[621.395625,"radius is indicated — a transient flash."]]},{"start":622.808625,"say":"Divide the full area pi a b by h over two. The period is two pi a b over h.","live":null,"does":[[623.156625,"period_work is shown on the screen, written out."],[627.243625,"period_work is shown on the screen, written out."],[627.8356249999999,"period_formula is shown on the screen, written out."],[629.6291249999999,"A box is drawn around period_formula."]]},{"start":630.229125,"say":"This already explains why the changing orbital speed does not complicate the period calculation. Equal areas have packaged the entire uneven motion into the one constant h.","live":["period_formula","ellipse_figure","heading_area","ellipse","focus","body","radius","semi_major","semi_minor"],"does":[[640.3356249999999,"period_formula is indicated — a transient flash."],[641.554625,"period_formula moves to a new place on the board."],[641.554625,"heading_area is hidden from the screen — left the board."],[641.554625,"period_work is hidden from the screen — left the board."],[641.554625,"The box around period_formula is lifted."]]},{"start":642.7546249999999,"say":"The formula still contains b and h. The conic geometry supplies the relation p equals b squared over a.","live":["period_formula","ellipse_figure","ellipse","focus","body","radius","semi_major","semi_minor"],"does":[[642.7546249999999,"heading_scale is shown on the screen, written out."],[644.3456249999999,"semi_minor is indicated — a transient flash."],[647.351625,"scale_work is shown on the screen, written out."],[649.5806249999999,"semi_major is indicated — a transient flash."]]},{"start":650.895125,"say":"The orbit derivation supplied p equals h squared over mu. Combine the two expressions for p. Then h squared equals mu b squared over a.","live":["period_formula","ellipse_figure","ellipse","focus","body","radius","semi_major","semi_minor","heading_scale"],"does":[[653.257625,"scale_work (the \"h^2\" part) is emphasized."],[655.521625,"scale_work is shown on the screen, written out."],[661.570625,"scale_work (the \"h^2\" part) is no longer emphasized."]]},{"start":662.170625,"say":"Now square the period formula. Its numerator is four pi squared a squared b squared, and its denominator is h squared.","live":null,"does":[[662.7976249999999,"scale_work is shown on the screen, written out."],[663.319625,"period_formula is indicated — a transient flash."]]},{"start":671.953625,"say":"Substitute for h squared. The b squared cancels completely, and one power of a moves up from the denominator. The result is four pi squared over mu times a cubed.","live":null,"does":[[672.301625,"scale_work is shown on the screen, written out."],[683.064625,"scale_work (the \"a^3\" part) is indicated — a transient flash."]]},{"start":684.8021249999999,"say":"Therefore T squared over a cubed is four pi squared over mu. For every object orbiting the same central mass, that ratio is the same.","live":null,"does":[[685.208625,"kepler_three is shown on the screen, written out."],[691.408625,"A box is drawn around kepler_three."]]},{"start":695.444625,"say":"This is Kepler's period law as a consequence, not an empirical rule placed at the beginning. The inverse-square force fixed the conic, and angular momentum converted its area into a period.","live":["period_formula","ellipse_figure","ellipse","focus","body","radius","semi_major","semi_minor","kepler_three","heading_scale"],"does":[[708.111625,"kepler_three moves to a new place on the board."],[708.111625,"ellipse_figure is hidden from the screen — left the board."],[708.111625,"ellipse is hidden from the screen — ellipse_figure left the board."],[708.111625,"focus is hidden from the screen — ellipse_figure left the board."],[708.111625,"body is hidden from the screen — ellipse_figure left the board."],[708.111625,"radius is hidden from the screen — ellipse_figure left the board."],[708.111625,"semi_major is hidden from the screen — ellipse_figure left the board."],[708.111625,"semi_minor is hidden from the screen — ellipse_figure left the board."],[708.111625,"heading_scale is hidden from the screen — left the board."],[708.111625,"period_formula is hidden from the screen — left the board."],[708.111625,"scale_work is hidden from the screen — left the board."]]},{"start":709.3116249999999,"say":"Now compare the prediction with the actual planets. Measure a in astronomical units and T in years. For objects orbiting the Sun, the common ratio should be approximately one.","live":["kepler_three"],"does":[[709.3116249999999,"heading_planets is shown on the screen, written out."],[718.958625,"interpretation is shown on the screen, written out."],[720.735625,"solar_form is shown on the screen, written out."]]},{"start":722.217625,"say":"Read the table from the inner Solar System outward. Mercury has semi-major axis zero point three eight seven astronomical units and period zero point two four one years.","live":["kepler_three","solar_form","interpretation","heading_planets"],"does":[[723.030625,"planet_table is shown on the screen, written out."],[725.805625,"planet_table is shown on the screen, written out."],[731.679625,"planet_table is indicated — a transient flash."]]},{"start":734.346125,"say":"Venus has a equal to zero point seven two three and period zero point six one five years. Its ratio is again essentially one.","live":null,"does":[[734.694625,"planet_table is shown on the screen, written out."],[739.500625,"planet_table (the \"1.001\" part) is emphasized."],[743.808125,"planet_table (the \"1.001\" part) is no longer emphasized."]]},{"start":744.4081249999999,"say":"Earth defines the convenient units: one astronomical unit and one year. Its ratio is exactly one in this unit system.","live":null,"does":[[744.756625,"planet_table is shown on the screen, written out."],[750.398625,"planet_table is indicated — a transient flash."]]},{"start":753.472125,"say":"Mars is farther out, at one point five two four astronomical units, and takes one point eight eight one years. The period grows faster than the orbital size itself, exactly as the three-halves power predicts.","live":null,"does":[[753.878625,"planet_table is shown on the screen, written out."],[766.4746250000001,"kepler_three is indicated — a transient flash."]]},{"start":768.003125,"say":"Jupiter is more than five times Earth's distance from the Sun, but its period is almost twelve years. Even across that much larger scale, the ratio remains zero point nine nine nine.","live":null,"does":[[768.351625,"planet_table is shown on the screen, written out."],[777.918625,"planet_table (the \"0.999\" part) is emphasized."],[779.7296249999999,"planet_table (the \"0.999\" part) is no longer emphasized."]]},{"start":780.329625,"say":"The small deviations shown here come from rounded data. The agreement across the planets is the quantitative signature of one inverse-square gravitational parameter mu belonging to the Sun.","live":null,"does":[[784.485625,"planet_table (the \"column=4\" part) is emphasized."],[792.5860208333334,"planet_table (the \"column=4\" part) is no longer emphasized."],[792.8360208333334,"heading_planets is hidden from the screen — left the board."],[792.8360208333334,"interpretation is hidden from the screen — left the board."],[792.8360208333334,"kepler_three is hidden from the screen — left the board."],[792.8360208333334,"planet_table is hidden from the screen — left the board."],[792.8360208333334,"solar_form is hidden from the screen — left the board."]]}]},{"title":"Energy, Bound Orbits, and Escape","start":793.8776875,"end":1127.5007916666666,"objects":{"bound":"a Text [text] that says \"$epsilon<0$: $e<1$, a bound circle or ellipse.\"","central_body":"a Circle [blue] labelled \"M\" drawn in launch (radius=0.42, filled=True)","conic_energy_work":"a Derivation [text] that says \"$dot(r) &= frac(mu e, h) sin(theta) \\ r dot(theta) &= frac(mu, h) (1+e cos(theta)) \\ v^2 &= frac(mu^2, h^2) (1+e^2+2e cos(theta)) \\ epsilon &= frac(mu^2, 2h^2) (e^2-1)$\"","craft":"a Point [yellow] labelled \"m\" drawn in launch (location=(1.45, 0.0))","earth_escape":"a Math [text] that says \"$r=R_E arrow.r v_(upright(\"esc\")) approx 11.2 thin upright(\"km/s\")$\"","ellipse_energy":"a Math [text] that says \"$epsilon = -frac(mu, 2a), quad e<1$\"","energy":"a VariableNumber (initial_value=-0.68, format_spec='.2f')","energy_eccentricity":"a Math [text] that says \"$e^2 = 1+frac(2 epsilon h^2, mu^2)$\"","energy_line":"a NumberLine labelled \"epsilon\" (x_range=(-1.0, 1.0), include_numbers=True, ticks_every=0.5)","energy_point":"a Point [yellow] labelled \"-0.68\" drawn in energy_line (location=(<VariableNumber energy = 0.62>, 0.0))","energy_work":"a Derivation [text] that says \"$frac(dif, dif t) frac(v^2, 2) &= -frac(mu, r^2) dot(r) \\ frac(dif, dif t) (-frac(mu, r)) &= +frac(mu, r^2) dot(r) \\ frac(dif, dif t) (frac(v^2, 2)-frac(mu, r)) &= 0 \\ epsilon = frac(v^2, 2)-frac(mu, r) &= upright(\"constant\")$\"","escape_work":"a Derivation [text] that says \"$epsilon &= frac(v^2, 2)-frac(mu, r) \\ 0 &= frac(v_(upright(\"esc\"))^2, 2)-frac(mu, r) \\ v_(upright(\"esc\"))^2 &= frac(2mu, r) \\ v_(upright(\"esc\")) &= sqrt(frac(2mu, r))$\"","heading_classes":"a Heading that says \"Bound, Marginal, or Unbound\"","heading_energy":"a Heading that says \"A Second Conservation Law\"","heading_escape":"a Heading that says \"Where Escape Velocity Comes From\"","heading_relation":"a Heading that says \"Energy Chooses the Conic\"","heading_summary":"a Heading that says \"The Derivation as One Chain\"","launch":"a Figure (x_range=(-0.8, 2.4), y_range=(-0.7, 2.4), aspect=(3.2, 3.1))","launch_radius":"a Line [gray] labelled \"r\" drawn in launch (end=(1.45, 0.0))","marginal":"a Text [text] that says \"$epsilon=0$: $e=1$, the parabolic escape boundary.\"","point":"a Point [yellow] drawn in energy_line (location=(-0.68, 0.0))","point_2":"a Point [yellow] drawn in energy_line","speed":"a VariableNumber (initial_value=0.8, format_spec='.2f')","summary":"a Block [text] that says \"Central force $arrow.r$ fixed plane and equal areas. Inverse square $arrow.r$ $r=p slash (1+e cos theta)$. Ellipse area $arrow.r$ $T^2=(4 pi^2 slash mu)a^3$. Energy sign $arrow.r$ conic class and escape threshold.\"","unbound":"a Text [text] that says \"$epsilon>0$: $e>1$, an unbound hyperbola.\"","velocity":"a Vector [yellow] labelled \"v = 0.80\" drawn in launch (start=(1.45, 0.0), end=(1.45, <VariableNumber speed = 1.4142135623730951>))"},"beats":[{"start":793.8776875,"say":"Geometry classified the possible conics, but it did not tell us which conic a particular launch selects. For that we need a second conserved quantity: mechanical energy per unit mass.","live":[],"does":[[793.8776875,"heading_energy is shown on the screen, written out."],[798.8236875,"launch is shown on the screen, written out."],[798.8236875,"central_body is shown on the screen, written out."],[798.8236875,"craft is shown on the screen, written out."],[798.8236875,"launch_radius is shown on the screen, written out."],[803.9436875,"launch moves to a new place on the board."],[803.9436875,"velocity is shown on the screen, written out."],[803.9436875,"energy_line is shown on the screen, written out."],[803.9436875,"energy_point is shown on the screen, written out."]]},{"start":806.2386875,"say":"Begin with kinetic energy per unit mass, one half v squared. Its time derivative is acceleration dotted with velocity. In inverse-square gravity only the radial velocity contributes.","live":["launch","energy_line","heading_energy","central_body","craft","launch_radius","velocity","energy_point"],"does":[[807.1786875,"energy_work is shown on the screen, written out."],[817.7436875,"energy_work (the \"-frac(mu, r^2) dot(r)\" part) is emphasized."],[819.7176875,"energy_work (the \"-frac(mu, r^2) dot(r)\" part) is no longer emphasized."]]},{"start":820.3176874999999,"say":"The derivative of the gravitational potential per unit mass, minus mu over r, is the equal and opposite quantity: plus mu r dot over r squared.","live":null,"does":[[822.2446874999999,"energy_work is shown on the screen, written out."],[826.9346875,"energy_work (the \"+frac(mu, r^2) dot(r)\" part) is emphasized."],[831.2886874999999,"energy_work (the \"+frac(mu, r^2) dot(r)\" part) is no longer emphasized."]]},{"start":831.8886875,"say":"Add the two rates and they cancel. Therefore one half v squared minus mu over r is constant. Call this constant epsilon, the specific orbital energy.","live":null,"does":[[833.6066875,"energy_work is shown on the screen, written out."],[838.2856875,"energy_work is shown on the screen, written out."],[840.6886875,"energy_work (the \"epsilon\" part) is emphasized."],[843.5046874999999,"energy_work (the \"epsilon\" part) is no longer emphasized."]]},{"start":844.1046875,"say":"The example begins at negative energy. In these scaled units the launch speed is zero point eight. The gravitational well has enough depth that this motion cannot reach infinity.","live":null,"does":[[845.7126875,"energy_point is indicated — a transient flash."],[849.1146875,"velocity is indicated — a transient flash."]]},{"start":856.5761875,"say":"At infinite distance the potential approaches zero. Kinetic energy cannot be negative, so a trajectory with negative total energy cannot get all the way there. It must turn back.","live":null,"does":[[861.9396875,"point is shown on the screen, grown."],[863.9396875,"point is hidden from the screen."],[868.5686875,"energy_work is hidden from the screen — left the board."],[868.5686875,"heading_energy is hidden from the screen — left the board."]]},{"start":869.7686874999999,"say":"Now connect energy with the eccentricity already present in the orbit. Differentiating the conic equation gives the radial speed mu e over h times sine theta.","live":["launch","energy_line","central_body","craft","launch_radius","velocity","energy_point"],"does":[[869.7686874999999,"heading_relation is shown on the screen, written out."],[877.1416875,"conic_energy_work is shown on the screen, written out."]]},{"start":881.4911875,"say":"The transverse speed is r theta dot, or h over r. Using the orbit equation, it becomes mu over h times one plus e cosine theta.","live":["launch","energy_line","central_body","craft","launch_radius","velocity","energy_point","heading_relation"],"does":[[881.9906874999999,"conic_energy_work is shown on the screen, written out."],[886.7386875,"conic_energy_work (the \"1+e cos(theta)\" part) is emphasized."],[892.3581875,"conic_energy_work (the \"1+e cos(theta)\" part) is no longer emphasized."]]},{"start":892.9581875,"say":"These two velocity components are perpendicular. Square them and add. The result is mu squared over h squared times one plus e squared plus two e cosine theta.","live":null,"does":[[896.1506875,"conic_energy_work is shown on the screen, written out."],[897.9156875,"conic_energy_work (the \"1+e^2+2e cos(theta)\" part) is emphasized."],[904.5446875,"conic_energy_work (the \"1+e^2+2e cos(theta)\" part) is no longer emphasized."]]},{"start":905.1446874999999,"say":"Insert this speed and the orbit's reciprocal radius into epsilon. Every term containing theta cancels. Energy cannot depend on where the body happens to be on the orbit.","live":null,"does":[[905.4926875,"conic_energy_work is shown on the screen, written out."],[912.0646875,"conic_energy_work (the \"e^2-1\" part) is emphasized."],[917.4511875,"conic_energy_work (the \"e^2-1\" part) is no longer emphasized."]]},{"start":918.0511875,"say":"Solve for eccentricity. E squared is one plus two epsilon h squared over mu squared. Because h squared and mu squared are positive, the sign of energy decides whether e lies below, at, or above one.","live":null,"does":[[918.3526875,"energy_eccentricity is shown on the screen, written out."],[929.4286875,"A box is drawn around energy_eccentricity."]]},{"start":935.1136875,"say":"For an ellipse, combine this with h squared equal to mu a times one minus e squared. The specific energy becomes minus mu over two a. A larger bound orbit lies closer to zero energy.","live":["launch","energy_line","central_body","craft","launch_radius","velocity","energy_point","energy_eccentricity","heading_relation"],"does":[[940.0476874999999,"ellipse_energy is shown on the screen, written out."],[947.8376874999999,"energy_point is indicated — a transient flash."],[949.0571874999999,"conic_energy_work is hidden from the screen — left the board."],[949.0571874999999,"ellipse_energy is hidden from the screen — left the board."],[949.0571874999999,"energy_eccentricity is hidden from the screen — left the board."],[949.0571874999999,"heading_relation is hidden from the screen — left the board."]]},{"start":949.6571875,"say":"Negative energy gives e below one. The trajectory is a bound circle or ellipse, and the body must return rather than reach infinite distance.","live":["launch","energy_line","central_body","craft","launch_radius","velocity","energy_point"],"does":[[949.6571875,"heading_classes is shown on the screen, written out."],[949.9586875,"bound is shown on the screen, written out."],[951.3176874999999,"energy_point is indicated — a transient flash."]]},{"start":960.1256874999999,"say":"Increase the launch speed continuously. The velocity arrow grows and the energy marker moves rightward through zero.","live":["launch","energy_line","central_body","craft","launch_radius","velocity","energy_point","bound","heading_classes"],"does":[[960.5316875,"velocity is redrawn as the numbers it depends on change."],[960.5316875,"energy_point is redrawn as the numbers it depends on change."],[960.5316875,"speed ticks to 1.8."],[960.5316875,"energy ticks to 0.62."]]},{"start":968.7251875,"say":"At zero energy, e equals one. That is the parabolic boundary: the body can reach arbitrarily large distance, but its speed tends to zero as it gets there.","live":null,"does":[[969.3056875,"marginal is shown on the screen, written out."],[969.3056875,"point_2 is shown on the screen, grown."],[971.3056875,"point_2 is hidden from the screen."]]},{"start":981.1211874999999,"say":"Positive energy gives e greater than one. The orbit is hyperbolic and unbound. Even at infinite distance, positive kinetic energy remains.","live":["launch","energy_line","central_body","craft","launch_radius","velocity","energy_point","bound","marginal","heading_classes"],"does":[[981.4696875,"unbound is shown on the screen, written out."],[981.4696875,"energy_point is indicated — a transient flash."]]},{"start":991.8571875,"say":"The geometric classification and the physical classification are now the same statement. Ellipses are negative-energy orbits, the parabola is the zero-energy threshold, and hyperbolas carry positive energy.","live":["launch","energy_line","central_body","craft","launch_radius","velocity","energy_point","bound","marginal","unbound","heading_classes"],"does":[[1005.6036875,"launch moves to a new place on the board."],[1005.6036875,"bound is hidden from the screen — left the board."],[1005.6036875,"energy_line is hidden from the screen — left the board."],[1005.6036875,"energy_point is hidden from the screen — energy_line left the board."],[1005.6036875,"heading_classes is hidden from the screen — left the board."],[1005.6036875,"marginal is hidden from the screen — left the board."],[1005.6036875,"unbound is hidden from the screen — left the board."]]},{"start":1006.8036874999999,"say":"Escape velocity is simply the launch speed at that exact boundary. Lower the example from positive energy until it settles at the marginal speed, while the launch radius stays fixed.","live":["launch","central_body","craft","launch_radius","velocity"],"does":[[1006.8036874999999,"heading_escape is shown on the screen, written out."],[1011.5636874999999,"velocity is redrawn as the numbers it depends on change."],[1011.5636874999999,"speed ticks to 1.4142135623730951."]]},{"start":1019.0476874999999,"say":"At radius r, specific energy is one half v squared minus mu over r. The least speed that can escape is the case epsilon equal to zero.","live":["launch","central_body","craft","launch_radius","velocity","heading_escape"],"does":[[1020.8816875,"escape_work is shown on the screen, written out."],[1028.7536874999998,"escape_work is shown on the screen, written out."]]},{"start":1030.2821875,"say":"Move the potential term to the other side and multiply by two. Escape speed squared is two mu over r.","live":null,"does":[[1035.5996874999998,"escape_work is shown on the screen, written out."],[1036.5166875,"escape_work (the \"frac(2mu, r)\" part) is indicated — a transient flash."]]},{"start":1038.8991875,"say":"Take the positive square root. The escape speed is the square root of two mu over r.","live":null,"does":[[1040.1816875,"escape_work is shown on the screen, written out."],[1041.7146874999999,"A box is drawn around escape_work."]]},{"start":1045.8551874999998,"say":"This is not a speed that turns gravity off. Gravity continues to act at every finite distance. It is the speed whose initial kinetic energy is just enough to raise the total energy to zero.","live":null,"does":[[1057.5926875,"escape_work is indicated — a transient flash."]]},{"start":1059.0981875,"say":"At Earth's surface, ignoring the atmosphere and Earth's rotation, the result is about eleven point two kilometres per second.","live":null,"does":[[1059.7016875,"earth_escape is shown on the screen, written out."],[1064.2186875,"earth_escape is indicated — a transient flash."]]},{"start":1067.2216875,"say":"A faster launch has positive energy. Far away the potential tends to zero, so the remaining energy is kinetic and the craft retains a nonzero speed. Escape velocity is precisely the boundary where that leftover speed vanishes.","live":["launch","central_body","craft","launch_radius","velocity","earth_escape","heading_escape"],"does":[[1082.9186875,"earth_escape is hidden from the screen — left the board."],[1082.9186875,"escape_work is hidden from the screen — left the board."],[1082.9186875,"heading_escape is hidden from the screen — left the board."],[1082.9186875,"launch is hidden from the screen — left the board."],[1082.9186875,"central_body is hidden from the screen — launch left the board."],[1082.9186875,"craft is hidden from the screen — launch left the board."],[1082.9186875,"launch_radius is hidden from the screen — launch left the board."],[1082.9186875,"velocity is hidden from the screen — launch left the board."]]},{"start":1084.1186874999999,"say":"The lecture now forms one connected chain. A central force gives zero torque, a fixed orbital plane, and equal areas.","live":[],"does":[[1084.1186874999999,"heading_summary is shown on the screen, written out."],[1086.5916875,"summary is shown on the screen, written out."],[1087.8226875,"summary (the \"Central force\" part) is emphasized."]]},{"start":1093.6471875,"say":"The inverse-square strength turns the radial equation into the polar conic equation. Eccentricity then distinguishes circle, ellipse, parabola, and hyperbola.","live":["summary","heading_summary"],"does":[[1094.2276875,"summary (the \"Central force\" part) is no longer emphasized."],[1094.2276875,"summary (the \"Inverse square\" part) is emphasized."]]},{"start":1105.6601875,"say":"For an ellipse, total area divided by constant areal rate produces T squared proportional to a cubed, in agreement with the planets.","live":null,"does":[[1106.3686874999999,"summary (the \"Ellipse area\" part) is emphasized."],[1106.3686874999999,"summary (the \"Inverse square\" part) is no longer emphasized."]]},{"start":1114.9326875,"say":"Finally, conserved energy chooses which conic a launch follows. Its zero level is the boundary between return and escape, and that boundary gives the escape-speed formula.","live":null,"does":[[1116.6396875,"summary (the \"Ellipse area\" part) is no longer emphasized."],[1116.6396875,"summary (the \"Energy sign\" part) is emphasized."],[1126.2091249999999,"summary (the \"Energy sign\" part) is no longer emphasized."],[1126.4591249999999,"heading_summary is hidden from the screen — left the board."],[1126.4591249999999,"summary is hidden from the screen — left the board."]]}]}]},"durationSeconds":1128,"chapters":[{"title":"Equal Areas from Angular Momentum","startSeconds":0,"narration":"We are going to derive orbital behavior rather than begin with a list of laws. The first result comes from angular momentum alone. It applies to every central force, before gravity or an inverse square has entered. A central force points along the line joining a moving body P to one fixed point O. The blue path could have many shapes. The defining feature is only that the force is radial at every point. Take moments about O. The torque is r cross F. A central force can be written as some scalar function of distance times the radial unit vector, so r and F are parallel. The cross product of parallel vectors is zero. Therefore the torque vanishes, and the rate of change of angular momentum vanishes with it. Angular momentum is therefore constant in magnitude and direction. Since L is perpendicular to r at every instant, every position vector remains in the single plane perpendicular to L. That fixed orbital plane is already a major simplification. The body may travel around O and the radial force turns with it, but the motion cannot leave this two-dimensional plane. Notice what has not appeared. We have not used the strength of the force, its sign, or any inverse-square dependence. Zero torque is the entire argument so far. Now let the body move from P to a nearby point Q. The two radius lines and the small displacement enclose a thin triangle. To first order, the triangle has area one half the magnitude of r cross d r. The cross product supplies base times perpendicular height, exactly the area measurement this little wedge needs. Divide by the elapsed time. The displacement d r divided by d t becomes velocity, so the rate at which area is swept out is one half the magnitude of r cross v. But angular momentum is r cross m v. Its magnitude is m times the cross product already standing here. The areal rate is therefore L divided by two m. Since L is constant, the areal rate is constant. Equal intervals of time must sweep equal areas, wherever the body happens to be on its path. Watch the area element move to different parts of the orbit. Its shape changes because both the radius and the displacement change, but the angular-momentum calculation keeps the area per unit time fixed. This is Kepler's equal-areas law, but its logical source is broader than Keplerian gravity. Any central force conserves angular momentum, fixes an orbital plane, and sweeps area at a constant rate. Gravity enters only when we ask for the shape of the path."},{"title":"The Inverse-Square Orbit Equation","startSeconds":177.55370833333333,"narration":"Equal areas needed only a central force. Now we specialize to gravity. Work in the fixed orbital plane and write the acceleration as minus mu over r squared in the radial direction. Use polar unit vectors. The position is r times r hat. Differentiating once gives a radial velocity r dot and a transverse velocity r theta dot. Differentiate again. Radial acceleration is r double dot minus r theta dot squared. Tangential acceleration is r theta double dot plus two r dot theta dot. Gravity has no tangential component, so that second expression is zero. Multiplying through by r turns it into the time derivative of r squared theta dot. Therefore r squared theta dot is constant. Call it h, the angular momentum per unit mass. This is the same conservation law that produced equal areas, now written in polar coordinates. The inverse-square force changes the speed and bends the velocity, but the force remains radial and h remains fixed throughout the motion. The radial equation still contains time derivatives and the unknown r. A useful substitution is u of theta equal to one over r. We will describe the orbit by angle rather than by time. Since h equals r squared theta dot, replacing r by one over u gives theta dot equal to h u squared. Now differentiate r with respect to time by the chain rule. Since r is one over u, the factors of u cancel and r dot becomes minus h u prime. Differentiate once more. Another factor theta dot appears, so r double dot becomes minus h squared u squared u double prime. This substitution is not a guess at the orbit. It is a change of variable chosen because conservation of angular momentum makes every time derivative collapse into an angle derivative. Now use the radial equation. Radial acceleration equals minus mu over r squared. Substitute the expressions we just found for r, theta dot, and r double dot. Every term contains u squared. After substitution, the left side is minus h squared u squared times u double prime plus u. The right side is minus mu u squared. Cancel the common factor and divide by h squared. The nonlinear-looking orbital problem has become the linear equation u double prime plus u equals mu over h squared. A constant particular solution is mu over h squared. The homogeneous solution is a sine and cosine. Combine their amplitude into e and their phase into theta zero. Invert u to recover r. Define p as h squared over mu. The orbit is p divided by one plus e cosine theta minus theta zero. Three constants now carry the initial conditions. Theta zero chooses the direction of periapsis. The scale p comes from angular momentum. The dimensionless number e decides the shape. The conic sections have not been assumed. They emerged from the inverse-square force, conservation of angular momentum, and one change from r of time to one over r as a function of angle."},{"title":"Eccentricity and the Conic Sections","startSeconds":402.0461875,"narration":"The orbit equation already has the focus-polar form of a conic. To see that geometrically rather than merely name it, multiply through by the denominator. Since x equals r cosine theta, replace that product. The equation becomes r plus e x equals p. For positive e, rearrange. The quantity p over e minus x is the perpendicular distance from P to the vertical line x equals p over e. Draw that line as the directrix. The distance from P to the focus is e times its distance to the directrix. Equivalently, their ratio is the constant e. That is the focus-directrix definition of a conic, and e is its eccentricity. The force centre therefore occupies a focus of the conic. It is not generally the geometric centre of the curve. That distinction will matter for both speed and distance around an ellipse. Start with the closed cases. If e is zero, the angular term disappears and r equals the constant p. Every point stays the same distance from the focus, so the orbit is a circle. For e between zero and one, the denominator one plus e cosine theta remains positive for every direction. The radius stays finite all the way around, and the path closes as an ellipse. The focus is displaced from the geometric centre. The body comes close at periapsis and reaches its greatest distance at apoapsis. Equal areas then require it to move faster near periapsis and slower near apoapsis. Both circle and ellipse are bound geometric shapes. Their denominator never reaches zero, so the orbit never runs to infinite radius. At e equal to one, the denominator approaches zero only as theta approaches the backward direction pi. The radius grows without bound in that limiting direction, producing a parabola. A parabolic orbit is the boundary between closed motion and escape. It reaches arbitrarily large distance but has only one limiting escape direction. For e greater than one, the denominator reaches zero before theta reaches pi, when cosine theta equals minus one over e. The physical trajectory is one branch of a hyperbola. The hyperbola approaches an asymptote as the radius diverges. The body arrives from far away, swings past the focus, and departs again without closing its path. One integration constant classifies every inverse-square trajectory. Zero gives a circle. Values between zero and one give ellipses. One gives a parabola, and values greater than one give hyperbolas. Geometry has told us every possible shape. It has not yet told us which shape a particular launch selects. Energy will answer that later. First, for the closed ellipse, we can derive the relation between its size and its period."},{"title":"The Period Law and the Planets","startSeconds":593.156625,"narration":"For a bound ellipse, one complete period sweeps the entire area of the ellipse. We already know the areal rate, so the period is total area divided by that constant rate. Let a be the semi-major axis and b the semi-minor axis. Their product times pi is the area of the ellipse. The areal rate is h over two, where h is the angular momentum per unit mass. This rate is constant everywhere on the orbit. Divide the full area pi a b by h over two. The period is two pi a b over h. This already explains why the changing orbital speed does not complicate the period calculation. Equal areas have packaged the entire uneven motion into the one constant h. The formula still contains b and h. The conic geometry supplies the relation p equals b squared over a. The orbit derivation supplied p equals h squared over mu. Combine the two expressions for p. Then h squared equals mu b squared over a. Now square the period formula. Its numerator is four pi squared a squared b squared, and its denominator is h squared. Substitute for h squared. The b squared cancels completely, and one power of a moves up from the denominator. The result is four pi squared over mu times a cubed. Therefore T squared over a cubed is four pi squared over mu. For every object orbiting the same central mass, that ratio is the same. This is Kepler's period law as a consequence, not an empirical rule placed at the beginning. The inverse-square force fixed the conic, and angular momentum converted its area into a period. Now compare the prediction with the actual planets. Measure a in astronomical units and T in years. For objects orbiting the Sun, the common ratio should be approximately one. Read the table from the inner Solar System outward. Mercury has semi-major axis zero point three eight seven astronomical units and period zero point two four one years. Venus has a equal to zero point seven two three and period zero point six one five years. Its ratio is again essentially one. Earth defines the convenient units: one astronomical unit and one year. Its ratio is exactly one in this unit system. Mars is farther out, at one point five two four astronomical units, and takes one point eight eight one years. The period grows faster than the orbital size itself, exactly as the three-halves power predicts. Jupiter is more than five times Earth's distance from the Sun, but its period is almost twelve years. Even across that much larger scale, the ratio remains zero point nine nine nine. The small deviations shown here come from rounded data. The agreement across the planets is the quantitative signature of one inverse-square gravitational parameter mu belonging to the Sun."},{"title":"Energy, Bound Orbits, and Escape","startSeconds":793.8776875,"narration":"Geometry classified the possible conics, but it did not tell us which conic a particular launch selects. For that we need a second conserved quantity: mechanical energy per unit mass. Begin with kinetic energy per unit mass, one half v squared. Its time derivative is acceleration dotted with velocity. In inverse-square gravity only the radial velocity contributes. The derivative of the gravitational potential per unit mass, minus mu over r, is the equal and opposite quantity: plus mu r dot over r squared. Add the two rates and they cancel. Therefore one half v squared minus mu over r is constant. Call this constant epsilon, the specific orbital energy. The example begins at negative energy. In these scaled units the launch speed is zero point eight. The gravitational well has enough depth that this motion cannot reach infinity. At infinite distance the potential approaches zero. Kinetic energy cannot be negative, so a trajectory with negative total energy cannot get all the way there. It must turn back. Now connect energy with the eccentricity already present in the orbit. Differentiating the conic equation gives the radial speed mu e over h times sine theta. The transverse speed is r theta dot, or h over r. Using the orbit equation, it becomes mu over h times one plus e cosine theta. These two velocity components are perpendicular. Square them and add. The result is mu squared over h squared times one plus e squared plus two e cosine theta. Insert this speed and the orbit's reciprocal radius into epsilon. Every term containing theta cancels. Energy cannot depend on where the body happens to be on the orbit. Solve for eccentricity. E squared is one plus two epsilon h squared over mu squared. Because h squared and mu squared are positive, the sign of energy decides whether e lies below, at, or above one. For an ellipse, combine this with h squared equal to mu a times one minus e squared. The specific energy becomes minus mu over two a. A larger bound orbit lies closer to zero energy. Negative energy gives e below one. The trajectory is a bound circle or ellipse, and the body must return rather than reach infinite distance. Increase the launch speed continuously. The velocity arrow grows and the energy marker moves rightward through zero. At zero energy, e equals one. That is the parabolic boundary: the body can reach arbitrarily large distance, but its speed tends to zero as it gets there. Positive energy gives e greater than one. The orbit is hyperbolic and unbound. Even at infinite distance, positive kinetic energy remains. The geometric classification and the physical classification are now the same statement. Ellipses are negative-energy orbits, the parabola is the zero-energy threshold, and hyperbolas carry positive energy. Escape velocity is simply the launch speed at that exact boundary. Lower the example from positive energy until it settles at the marginal speed, while the launch radius stays fixed. At radius r, specific energy is one half v squared minus mu over r. The least speed that can escape is the case epsilon equal to zero. Move the potential term to the other side and multiply by two. Escape speed squared is two mu over r. Take the positive square root. The escape speed is the square root of two mu over r. This is not a speed that turns gravity off. Gravity continues to act at every finite distance. It is the speed whose initial kinetic energy is just enough to raise the total energy to zero. At Earth's surface, ignoring the atmosphere and Earth's rotation, the result is about eleven point two kilometres per second. A faster launch has positive energy. Far away the potential tends to zero, so the remaining energy is kinetic and the craft retains a nonzero speed. Escape velocity is precisely the boundary where that leftover speed vanishes. The lecture now forms one connected chain. A central force gives zero torque, a fixed orbital plane, and equal areas. The inverse-square strength turns the radial equation into the polar conic equation. Eccentricity then distinguishes circle, ellipse, parabola, and hyperbola. For an ellipse, total area divided by constant areal rate produces T squared proportional to a cubed, in agreement with the planets. Finally, conserved energy chooses which conic a launch follows. Its zero level is the boundary between return and escape, and that boundary gives the escape-speed formula."}]}}