WEBVTT 00:00:00.058 --> 00:00:04.529 We are going to derive orbital behavior rather than begin with a list of laws. 00:00:04.725 --> 00:00:07.954 The first result comes from angular momentum alone. 00:00:08.150 --> 00:00:13.271 It applies to every central force, before gravity or an inverse square has entered. 00:00:14.668 --> 00:00:18.125 A central force points along the line joining a moving body P 00:00:18.208 --> 00:00:20.206 to one fixed point O. 00:00:20.820 --> 00:00:23.271 The blue path could have many shapes. 00:00:23.468 --> 00:00:27.741 The defining feature is only that the force is radial at every point. 00:00:28.793 --> 00:00:30.674 Take moments about O. 00:00:30.778 --> 00:00:35.597 The torque is r cross F. A central force can be written as some scalar function 00:00:35.597 --> 00:00:40.323 of distance times the radial unit vector, so r and F are parallel. 00:00:41.119 --> 00:00:44.301 The cross product of parallel vectors is zero. 00:00:44.498 --> 00:00:46.272 Therefore the torque vanishes, 00:00:46.355 --> 00:00:49.873 and the rate of change of angular momentum vanishes with it. 00:00:50.577 --> 00:00:54.479 Angular momentum is therefore constant in magnitude and direction. 00:00:54.791 --> 00:00:59.958 Since L is perpendicular to r at every instant, every position vector remains 00:00:59.958 --> 00:01:02.617 in the single plane perpendicular to L. 00:01:03.414 --> 00:01:07.234 That fixed orbital plane is already a major simplification. 00:01:07.338 --> 00:01:11.028 The body may travel around O and the radial force turns with it, 00:01:11.111 --> 00:01:14.293 but the motion cannot leave this two-dimensional plane. 00:01:15.090 --> 00:01:17.134 Notice what has not appeared. 00:01:17.330 --> 00:01:20.557 We have not used the strength of the force, its sign, 00:01:20.640 --> 00:01:22.997 or any inverse-square dependence. 00:01:23.101 --> 00:01:26.074 Zero torque is the entire argument so far. 00:01:27.470 --> 00:01:30.917 Now let the body move from P to a nearby point Q. 00:01:31.000 --> 00:01:35.888 The two radius lines and the small displacement enclose a thin triangle. 00:01:36.685 --> 00:01:42.421 To first order, the triangle has area one half the magnitude of r cross d r. 00:01:42.525 --> 00:01:45.692 The cross product supplies base times perpendicular height, 00:01:45.775 --> 00:01:49.108 exactly the area measurement this little wedge needs. 00:01:49.905 --> 00:01:52.170 Divide by the elapsed time. 00:01:52.366 --> 00:01:55.941 The displacement d r divided by d t becomes velocity, 00:01:56.024 --> 00:02:01.017 so the rate at which area is swept out is one half the magnitude of r cross v. 00:02:01.813 --> 00:02:05.167 But angular momentum is r cross m v. 00:02:05.250 --> 00:02:09.477 Its magnitude is m times the cross product already standing here. 00:02:09.673 --> 00:02:13.006 The areal rate is therefore L divided by two m. 00:02:13.617 --> 00:02:17.310 Since L is constant, the areal rate is constant. 00:02:17.506 --> 00:02:20.477 Equal intervals of time must sweep equal areas, 00:02:20.560 --> 00:02:23.300 wherever the body happens to be on its path. 00:02:24.004 --> 00:02:27.523 Watch the area element move to different parts of the orbit. 00:02:27.719 --> 00:02:31.398 Its shape changes because both the radius and the displacement change, 00:02:31.481 --> 00:02:36.323 but the angular-momentum calculation keeps the area per unit time fixed. 00:02:37.538 --> 00:02:39.556 This is Kepler's equal-areas law, 00:02:39.639 --> 00:02:42.842 but its logical source is broader than Keplerian gravity. 00:02:42.925 --> 00:02:47.416 Any central force conserves angular momentum, fixes an orbital plane, 00:02:47.499 --> 00:02:50.124 and sweeps area at a constant rate. 00:02:50.320 --> 00:02:53.827 Gravity enters only when we ask for the shape of the path. 00:02:54.898 --> 00:02:57.894 Equal areas needed only a central force. 00:02:58.346 --> 00:03:00.715 Now we specialize to gravity. 00:03:00.912 --> 00:03:02.826 Work in the fixed orbital plane 00:03:02.909 --> 00:03:08.111 and write the acceleration as minus mu over r squared in the radial direction. 00:03:09.325 --> 00:03:11.451 Use polar unit vectors. 00:03:11.647 --> 00:03:14.632 The position is r times r hat. 00:03:14.735 --> 00:03:18.158 Differentiating once gives a radial velocity r dot 00:03:18.241 --> 00:03:21.238 and a transverse velocity r theta dot. 00:03:21.976 --> 00:03:23.649 Differentiate again. 00:03:23.787 --> 00:03:28.444 Radial acceleration is r double dot minus r theta dot squared. 00:03:28.687 --> 00:03:34.458 Tangential acceleration is r theta double dot plus two r dot theta dot. 00:03:35.301 --> 00:03:40.433 Gravity has no tangential component, so that second expression is zero. 00:03:40.746 --> 00:03:46.076 Multiplying through by r turns it into the time derivative of r squared theta dot. 00:03:46.779 --> 00:03:49.935 Therefore r squared theta dot is constant. 00:03:50.018 --> 00:03:53.816 Call it h, the angular momentum per unit mass. 00:03:54.012 --> 00:03:57.818 This is the same conservation law that produced equal areas, 00:03:57.901 --> 00:04:00.015 now written in polar coordinates. 00:04:00.812 --> 00:04:04.769 The inverse-square force changes the speed and bends the velocity, 00:04:04.852 --> 00:04:09.358 but the force remains radial and h remains fixed throughout the motion. 00:04:10.871 --> 00:04:15.608 The radial equation still contains time derivatives and the unknown r. 00:04:15.921 --> 00:04:20.833 A useful substitution is u of theta equal to one over r. 00:04:21.029 --> 00:04:24.873 We will describe the orbit by angle rather than by time. 00:04:25.785 --> 00:04:28.895 Since h equals r squared theta dot, 00:04:28.978 --> 00:04:33.658 replacing r by one over u gives theta dot equal to h u squared. 00:04:34.419 --> 00:04:38.693 Now differentiate r with respect to time by the chain rule. 00:04:38.854 --> 00:04:45.462 Since r is one over u, the factors of u cancel and r dot becomes minus h u prime. 00:04:46.223 --> 00:04:48.033 Differentiate once more. 00:04:48.116 --> 00:04:49.777 Another factor theta 00:04:49.880 --> 00:04:56.719 dot appears, so r double dot becomes minus h squared u squared u double prime. 00:04:57.516 --> 00:05:00.266 This substitution is not a guess at the orbit. 00:05:00.349 --> 00:05:05.084 It is a change of variable chosen because conservation of angular momentum 00:05:05.167 --> 00:05:09.347 makes every time derivative collapse into an angle derivative. 00:05:10.744 --> 00:05:12.820 Now use the radial equation. 00:05:12.903 --> 00:05:17.084 Radial acceleration equals minus mu over r squared. 00:05:17.280 --> 00:05:23.005 Substitute the expressions we just found for r, theta dot, and r double dot. 00:05:23.488 --> 00:05:26.182 Every term contains u squared. 00:05:26.495 --> 00:05:27.609 After substitution, 00:05:27.609 --> 00:05:32.960 the left side is minus h squared u squared times u double prime plus u. 00:05:33.043 --> 00:05:36.004 The right side is minus mu u squared. 00:05:36.487 --> 00:05:40.319 Cancel the common factor and divide by h squared. 00:05:40.423 --> 00:05:45.810 The nonlinear-looking orbital problem has become the linear equation u double prime 00:05:45.810 --> 00:05:49.038 plus u equals mu over h squared. 00:05:49.742 --> 00:05:53.760 A constant particular solution is mu over h squared. 00:05:53.956 --> 00:05:57.428 The homogeneous solution is a sine and cosine. 00:05:57.532 --> 00:06:02.339 Combine their amplitude into e and their phase into theta zero. 00:06:03.321 --> 00:06:05.351 Invert u to recover r. 00:06:05.434 --> 00:06:08.672 Define p as h squared over mu. 00:06:08.755 --> 00:06:13.818 The orbit is p divided by one plus e cosine theta minus theta zero. 00:06:14.486 --> 00:06:17.785 Three constants now carry the initial conditions. 00:06:17.935 --> 00:06:21.372 Theta zero chooses the direction of periapsis. 00:06:21.569 --> 00:06:24.646 The scale p comes from angular momentum. 00:06:24.750 --> 00:06:27.851 The dimensionless number e decides the shape. 00:06:28.554 --> 00:06:31.133 The conic sections have not been assumed. 00:06:31.236 --> 00:06:36.273 They emerged from the inverse-square force, conservation of angular momentum, 00:06:36.356 --> 00:06:40.955 and one change from r of time to one over r as a function of angle. 00:06:42.052 --> 00:06:46.035 The orbit equation already has the focus-polar form of a conic. 00:06:46.232 --> 00:06:49.260 To see that geometrically rather than merely name it, 00:06:49.343 --> 00:06:51.608 multiply through by the denominator. 00:06:52.312 --> 00:06:56.643 Since x equals r cosine theta, replace that product. 00:06:56.839 --> 00:07:00.869 The equation becomes r plus e x equals p. 00:07:01.572 --> 00:07:04.093 For positive e, rearrange. 00:07:04.289 --> 00:07:08.688 The quantity p over e minus x is the perpendicular distance from P 00:07:08.771 --> 00:07:12.008 to the vertical line x equals p over e. 00:07:12.091 --> 00:07:14.542 Draw that line as the directrix. 00:07:15.338 --> 00:07:20.261 The distance from P to the focus is e times its distance to the directrix. 00:07:20.458 --> 00:07:23.651 Equivalently, their ratio is the constant e. 00:07:23.755 --> 00:07:29.329 That is the focus-directrix definition of a conic, and e is its eccentricity. 00:07:30.125 --> 00:07:34.015 The force centre therefore occupies a focus of the conic. 00:07:34.212 --> 00:07:37.637 It is not generally the geometric centre of the curve. 00:07:37.741 --> 00:07:42.363 That distinction will matter for both speed and distance around an ellipse. 00:07:43.759 --> 00:07:45.977 Start with the closed cases. 00:07:46.174 --> 00:07:52.108 If e is zero, the angular term disappears and r equals the constant p. 00:07:52.304 --> 00:07:57.030 Every point stays the same distance from the focus, so the orbit is a circle. 00:07:57.827 --> 00:07:59.927 For e between zero and one, 00:08:00.010 --> 00:08:05.363 the denominator one plus e cosine theta remains positive for every direction. 00:08:05.466 --> 00:08:10.715 The radius stays finite all the way around, and the path closes as an ellipse. 00:08:11.418 --> 00:08:14.589 The focus is displaced from the geometric centre. 00:08:14.785 --> 00:08:20.417 The body comes close at periapsis and reaches its greatest distance at apoapsis. 00:08:20.614 --> 00:08:26.594 Equal areas then require it to move faster near periapsis and slower near apoapsis. 00:08:27.391 --> 00:08:30.805 Both circle and ellipse are bound geometric shapes. 00:08:31.001 --> 00:08:36.436 Their denominator never reaches zero, so the orbit never runs to infinite radius. 00:08:37.832 --> 00:08:43.277 At e equal to one, the denominator approaches zero only as theta approaches 00:08:43.277 --> 00:08:45.565 the backward direction pi. 00:08:45.761 --> 00:08:50.638 The radius grows without bound in that limiting direction, producing a parabola. 00:08:51.435 --> 00:08:55.499 A parabolic orbit is the boundary between closed motion and escape. 00:08:55.696 --> 00:09:00.863 It reaches arbitrarily large distance but has only one limiting escape direction. 00:09:01.915 --> 00:09:06.940 For e greater than one, the denominator reaches zero before theta reaches pi, 00:09:07.023 --> 00:09:10.438 when cosine theta equals minus one over e. 00:09:10.634 --> 00:09:13.979 The physical trajectory is one branch of a hyperbola. 00:09:14.682 --> 00:09:18.898 The hyperbola approaches an asymptote as the radius diverges. 00:09:19.094 --> 00:09:21.089 The body arrives from far away, 00:09:21.172 --> 00:09:25.481 swings past the focus, and departs again without closing its path. 00:09:26.277 --> 00:09:30.945 One integration constant classifies every inverse-square trajectory. 00:09:31.142 --> 00:09:32.908 Zero gives a circle. 00:09:33.104 --> 00:09:36.739 Values between zero and one give ellipses. 00:09:36.936 --> 00:09:41.522 One gives a parabola, and values greater than one give hyperbolas. 00:09:42.574 --> 00:09:45.698 Geometry has told us every possible shape. 00:09:45.895 --> 00:09:49.738 It has not yet told us which shape a particular launch selects. 00:09:49.935 --> 00:09:52.060 Energy will answer that later. 00:09:52.257 --> 00:09:53.973 First, for the closed ellipse, 00:09:54.056 --> 00:09:57.505 we can derive the relation between its size and its period. 00:09:58.648 --> 00:10:03.618 For a bound ellipse, one complete period sweeps the entire area of the ellipse. 00:10:03.931 --> 00:10:05.612 We already know the areal rate, 00:10:05.695 --> 00:10:09.318 so the period is total area divided by that constant rate. 00:10:10.231 --> 00:10:14.748 Let a be the semi-major axis and b the semi-minor axis. 00:10:14.945 --> 00:10:18.347 Their product times pi is the area of the ellipse. 00:10:19.144 --> 00:10:24.822 The areal rate is h over two, where h is the angular momentum per unit mass. 00:10:25.018 --> 00:10:27.980 This rate is constant everywhere on the orbit. 00:10:28.683 --> 00:10:32.492 Divide the full area pi a b by h over two. 00:10:32.689 --> 00:10:35.940 The period is two pi a b over h. 00:10:36.644 --> 00:10:38.246 This already explains why 00:10:38.246 --> 00:10:42.577 the changing orbital speed does not complicate the period calculation. 00:10:42.681 --> 00:10:47.964 Equal areas have packaged the entire uneven motion into the one constant h. 00:10:49.268 --> 00:10:51.739 The formula still contains b and h. 00:10:51.822 --> 00:10:56.978 The conic geometry supplies the relation p equals b squared over a. 00:10:57.658 --> 00:11:01.708 The orbit derivation supplied p equals h squared over mu. 00:11:01.791 --> 00:11:04.195 Combine the two expressions for p. 00:11:04.369 --> 00:11:08.305 Then h squared equals mu b squared over a. 00:11:09.079 --> 00:11:11.387 Now square the period formula. 00:11:11.470 --> 00:11:15.056 Its numerator is four pi squared a squared b squared, 00:11:15.139 --> 00:11:17.729 and its denominator is h squared. 00:11:18.526 --> 00:11:20.721 Substitute for h squared. 00:11:20.917 --> 00:11:26.514 The b squared cancels completely, and one power of a moves up from the denominator. 00:11:26.618 --> 00:11:30.717 The result is four pi squared over mu times a cubed. 00:11:31.328 --> 00:11:35.822 Therefore T squared over a cubed is four pi squared over mu. 00:11:36.018 --> 00:11:41.081 For every object orbiting the same central mass, that ratio is the same. 00:11:41.622 --> 00:11:44.384 This is Kepler's period law as a consequence, 00:11:44.467 --> 00:11:47.286 not an empirical rule placed at the beginning. 00:11:47.369 --> 00:11:49.979 The inverse-square force fixed the conic, 00:11:50.062 --> 00:11:53.859 and angular momentum converted its area into a period. 00:11:55.163 --> 00:11:58.217 Now compare the prediction with the actual planets. 00:11:58.321 --> 00:12:02.246 Measure a in astronomical units and T in years. 00:12:02.443 --> 00:12:07.470 For objects orbiting the Sun, the common ratio should be approximately one. 00:12:08.685 --> 00:12:11.646 Read the table from the inner Solar System outward. 00:12:11.843 --> 00:12:16.787 Mercury has semi-major axis zero point three eight seven astronomical units 00:12:16.870 --> 00:12:19.901 and period zero point two four one years. 00:12:20.814 --> 00:12:22.137 Venus has a equal 00:12:22.137 --> 00:12:27.153 to zero point seven two three and period zero point six one five years. 00:12:27.350 --> 00:12:30.079 Its ratio is again essentially one. 00:12:30.876 --> 00:12:33.396 Earth defines the convenient units: 00:12:33.592 --> 00:12:36.194 one astronomical unit and one year. 00:12:36.390 --> 00:12:39.862 Its ratio is exactly one in this unit system. 00:12:40.659 --> 00:12:45.081 Mars is farther out, at one point five two four astronomical units, 00:12:45.164 --> 00:12:47.939 and takes one point eight eight one years. 00:12:48.136 --> 00:12:51.944 The period grows faster than the orbital size itself, exactly 00:12:51.944 --> 00:12:54.081 as the three-halves power predicts. 00:12:55.296 --> 00:12:58.765 Jupiter is more than five times Earth's distance from the Sun, 00:12:58.848 --> 00:13:01.566 but its period is almost twelve years. 00:13:01.669 --> 00:13:06.790 Even across that much larger scale, the ratio remains zero point nine nine nine. 00:13:07.331 --> 00:13:10.745 The small deviations shown here come from rounded data. 00:13:10.942 --> 00:13:14.112 The agreement across the planets is the quantitative signature 00:13:14.112 --> 00:13:19.035 of one inverse-square gravitational parameter mu belonging to the Sun. 00:13:20.078 --> 00:13:22.687 Geometry classified the possible conics, 00:13:22.771 --> 00:13:26.383 but it did not tell us which conic a particular launch selects. 00:13:26.579 --> 00:13:31.642 For that we need a second conserved quantity: mechanical energy per unit mass. 00:13:32.555 --> 00:13:36.933 Begin with kinetic energy per unit mass, one half v squared. 00:13:37.129 --> 00:13:40.962 Its time derivative is acceleration dotted with velocity. 00:13:41.158 --> 00:13:45.733 In inverse-square gravity only the radial velocity contributes. 00:13:46.530 --> 00:13:51.891 The derivative of the gravitational potential per unit mass, minus mu over r, 00:13:51.975 --> 00:13:57.618 is the equal and opposite quantity: plus mu r dot over r squared. 00:13:58.520 --> 00:14:00.947 Add the two rates and they cancel. 00:14:01.399 --> 00:14:06.067 Therefore one half v squared minus mu over r is constant. 00:14:06.263 --> 00:14:10.618 Call this constant epsilon, the specific orbital energy. 00:14:11.403 --> 00:14:14.364 The example begins at negative energy. 00:14:14.677 --> 00:14:18.521 In these scaled units the launch speed is zero point eight. 00:14:18.973 --> 00:14:23.954 The gravitational well has enough depth that this motion cannot reach infinity. 00:14:24.751 --> 00:14:28.083 At infinite distance the potential approaches zero. 00:14:28.280 --> 00:14:30.704 Kinetic energy cannot be negative, 00:14:30.788 --> 00:14:34.791 so a trajectory with negative total energy cannot get all the way there. 00:14:34.875 --> 00:14:36.373 It must turn back. 00:14:37.886 --> 00:14:42.194 Now connect energy with the eccentricity already present in the orbit. 00:14:42.507 --> 00:14:45.952 Differentiating the conic equation gives the radial speed 00:14:46.036 --> 00:14:49.276 mu e over h times sine theta. 00:14:50.049 --> 00:14:54.473 The transverse speed is r theta dot, or h over r. 00:14:54.566 --> 00:15:00.128 Using the orbit equation, it becomes mu over h times one plus e cosine theta. 00:15:00.820 --> 00:15:05.441 These two velocity components are perpendicular. Square them and add. 00:15:05.615 --> 00:15:07.147 The result is mu squared 00:15:07.147 --> 00:15:12.547 over h squared times one plus e squared plus two e cosine theta. 00:15:13.250 --> 00:15:17.675 Insert this speed and the orbit's reciprocal radius into epsilon. 00:15:17.871 --> 00:15:20.786 Every term containing theta cancels. 00:15:20.983 --> 00:15:24.989 Energy cannot depend on where the body happens to be on the orbit. 00:15:25.693 --> 00:15:27.667 Solve for eccentricity. 00:15:27.771 --> 00:15:32.822 E squared is one plus two epsilon h squared over mu squared. 00:15:33.019 --> 00:15:36.302 Because h squared and mu squared are positive, 00:15:36.386 --> 00:15:41.750 the sign of energy decides whether e lies below, at, or above one. 00:15:42.663 --> 00:15:43.717 For an ellipse, 00:15:43.801 --> 00:15:49.104 combine this with h squared equal to mu a times one minus e squared. 00:15:49.188 --> 00:15:53.020 The specific energy becomes minus mu over two a. 00:15:53.193 --> 00:15:56.816 A larger bound orbit lies closer to zero energy. 00:15:57.590 --> 00:16:00.121 Negative energy gives e below one. 00:16:00.214 --> 00:16:02.350 The trajectory is a bound circle 00:16:02.350 --> 00:16:07.029 or ellipse, and the body must return rather than reach infinite distance. 00:16:07.826 --> 00:16:10.474 Increase the launch speed continuously. 00:16:10.670 --> 00:16:15.303 The velocity arrow grows and the energy marker moves rightward through zero. 00:16:16.007 --> 00:16:19.015 At zero energy, e equals one. 00:16:19.212 --> 00:16:24.224 That is the parabolic boundary: the body can reach arbitrarily large distance, 00:16:24.308 --> 00:16:27.154 but its speed tends to zero as it gets there. 00:16:27.950 --> 00:16:30.931 Positive energy gives e greater than one. 00:16:31.015 --> 00:16:34.023 The orbit is hyperbolic and unbound. 00:16:34.220 --> 00:16:38.644 Even at infinite distance, positive kinetic energy remains. 00:16:39.348 --> 00:16:41.112 The geometric classification 00:16:41.112 --> 00:16:44.817 and the physical classification are now the same statement. 00:16:45.129 --> 00:16:50.711 Ellipses are negative-energy orbits, the parabola is the zero-energy threshold, 00:16:50.795 --> 00:16:53.548 and hyperbolas carry positive energy. 00:16:54.944 --> 00:16:59.508 Escape velocity is simply the launch speed at that exact boundary. 00:16:59.704 --> 00:17:03.997 Lower the example from positive energy until it settles at the marginal speed, 00:17:04.081 --> 00:17:06.636 while the launch radius stays fixed. 00:17:07.433 --> 00:17:13.482 At radius r, specific energy is one half v squared minus mu over r. 00:17:13.795 --> 00:17:17.813 The least speed that can escape is the case epsilon equal to zero. 00:17:18.609 --> 00:17:22.627 Move the potential term to the other side and multiply by two. 00:17:22.824 --> 00:17:26.203 Escape speed squared is two mu over r. 00:17:27.000 --> 00:17:29.160 Take the positive square root. 00:17:29.357 --> 00:17:33.258 The escape speed is the square root of two mu over r. 00:17:34.055 --> 00:17:36.830 This is not a speed that turns gravity off. 00:17:37.027 --> 00:17:40.499 Gravity continues to act at every finite distance. 00:17:40.696 --> 00:17:44.434 It is the speed whose initial kinetic energy is just enough 00:17:44.434 --> 00:17:46.838 to raise the total energy to zero. 00:17:47.635 --> 00:17:51.440 At Earth's surface, ignoring the atmosphere and Earth's rotation, 00:17:51.524 --> 00:17:54.985 the result is about eleven point two kilometres per second. 00:17:55.781 --> 00:17:58.475 A faster launch has positive energy. 00:17:58.927 --> 00:18:01.339 Far away the potential tends to zero, 00:18:01.423 --> 00:18:06.254 so the remaining energy is kinetic and the craft retains a nonzero speed. 00:18:06.358 --> 00:18:11.432 Escape velocity is precisely the boundary where that leftover speed vanishes. 00:18:12.829 --> 00:18:15.755 The lecture now forms one connected chain. 00:18:15.859 --> 00:18:21.676 A central force gives zero torque, a fixed orbital plane, and equal areas. 00:18:22.380 --> 00:18:25.120 The inverse-square strength turns the radial equation 00:18:25.120 --> 00:18:27.350 into the polar conic equation. 00:18:27.547 --> 00:18:33.190 Eccentricity then distinguishes circle, ellipse, parabola, and hyperbola. 00:18:33.987 --> 00:18:36.843 For an ellipse, total area divided by constant 00:18:36.843 --> 00:18:42.428 areal rate produces T squared proportional to a cubed, in agreement with the planets. 00:18:43.341 --> 00:18:47.777 Finally, conserved energy chooses which conic a launch follows. 00:18:47.973 --> 00:18:51.616 Its zero level is the boundary between return and escape, 00:18:51.700 --> 00:18:54.557 and that boundary gives the escape-speed formula.