Deriving Keplerian Orbits from Conservation Laws

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.