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Angular momentum measures how a point mass's momentum passes around a chosen origin. Motion straight toward or away from that origin contributes zero; transverse momentum contributes more when it passes farther away. The picture shows the sign convention before the calculation.

Fix an origin $O$ and nonrotating axes in an inertial frame. For a nonrelativistic point mass of constant mass $m>0$, let $\mathbf{r}$ be its position from $O$ and $\mathbf{p}=m\mathbf{v}$ its mechanical momentum in that same reference frame. Its angular momentum is the cross productHere $r=|\mathbf{r}|$, $p=|\mathbf{p}|$, and $\varphi \in [0,\pi ]$ is the angle between them. The right-hand rule gives the direction: curl from $\mathbf{r}$ toward $\mathbf{p}$, and the thumb points along ${\mathbf{L}}_O$. Thus angular momentum is an axial vector, perpendicular to both position and momentum. Its International System of Units unit is $\mathrm{kg}{\mathrm{m}}^2/s$. A point mass need not follow a circle to have angular momentum.^[ref]

With $x$ rightward, $y$ upward, and $+z$ out of the page, the cross product gives $L_z=x{p}_y-y{p}_x$. TakeThen ${\mathbf{L}}_O=(0,0,6)\mathrm{kg}{\mathrm{m}}^2/s$. Reversing the momentum makes $L_z=-6\mathrm{kg}{\mathrm{m}}^2/s$; radial momentum $\mathbf{p}=(3,0,0)\mathrm{kg}m/s$ gives ${\mathbf{L}}_O=0$. The factor $r\sin \varphi$ is the lever arm to the line through the particle parallel to its momentum.

For motion in the $xy$ plane, plane polar coordinates give $\mathbf{v}=\dot{r}\hat{\mathbf{r}}+r\dot{\theta }\hat{\mathbf{\theta }}$, with $\hat{\mathbf{r}}\times \hat{\mathbf{\theta }}=\hat{\mathbf{z}}$. ConsequentlyThe signed angular velocity $\dot{\theta }$ is positive counterclockwise viewed from $+z$ toward the plane, with $\theta$ measured in radians. Radial velocity does not enter the cross product. Under conservation of angular momentum, this relation predicts faster angular motion at smaller $r$ and constant areal velocity $\dot{A}=L_z/(2m)$ for consistently oriented swept area. That is the geometric content of Kepler's second law.

For several point masses, add their angular momenta about the same origin: ${\mathbf{L}}_O={\sum }_i{\mathbf{r}}_i\times {\mathbf{p}}_i$. The origin matters: angular momentum under an origin shift explains why the same moving particle can have different values about two fixed points. The central axis of a momentum system and angular momentum decomposition about the centre of mass make this dependence useful.

The bound relative Kepler orbit is an ellipse with the attracting centre at one focus of a conic. In polar coordinates, its separation is the displayed function of true anomaly $f$, the angle measured from periapsis. For $a=1$ and $e=0.6$, it becomes $r=0.64/(1+0.6\cos f)$, giving $0.4$ at $f=0$ and $1.6$ at $f=\pi$. Setting orbital eccentricity to zero gives a circular orbit. This specifies the path; Kepler's second law specifies how it is traversed.

Kepler's laws describe three connected features of an elliptic orbit: its shape, how fast it is traversed, and how its size determines its orbital period. Kepler's first law gives an ellipse with the attracting centre at one focus of a conic; Kepler's second law gives equal swept areas in equal times; Kepler's third law relates the squared orbital period to the cubed semi-major axis. Together they turn a drawing of a Kepler orbit into a prediction of where the bodies will be and when they will return. ^[ref]

Here the mathematical model is an isolated two-body problem in Newtonian gravity, with point masses $M,m>0$ and a bound, noncolliding elliptic orbit. Let $\mathbf{r}={\mathbf{r}}_m-{\mathbf{r}}_M$ be the relative displacement, $r=|\mathbf{r}|$ the separation, and $\mu =G(M+m)$ the gravitational parameter, where $G$ is the gravitational constant. Newton's law of universal gravitation then givesThe relative Kepler orbit has semi-major axis $a$ and orbital eccentricity $0\le e<1$. Both bodies move around their shared astronomical barycenter, the system's centre of mass. In its rest frame their displacements are ${\mathbf{r}}_m=M\mathbf{r}/(M+m)$ and ${\mathbf{r}}_M=-m\mathbf{r}/(M+m)$. Their individual semi-major axes therefore add to $a$; the relative ellipse's attracting focus of a conic represents zero separation, rather than a body fixed in space.

One example will connect all three laws. After nondimensionalization, choose $a=1$, $e=0.6$ and $\mu =1$. The semi-minor axis is $b=a\sqrt{1-e^2}=0.8$, while the periapsis distance and apoapsis distance areThus $a=(0.4+1.6)/2=1$ measures the ellipse's size even though the separation changes fourfold. The specific angular momentum will be $0.8$, the two endpoint speeds $2$ and $0.5$, and the orbital period $2\pi$. These are values in the chosen scaled units, not measurements of a particular planet.

Johannes Kepler published Kepler's first law and Kepler's second law in Astronomia Nova in 1609 and Kepler's third law in Harmonices Mundi in 1619. These publication dates are documented by ^[ref]. Their appeal is that three short statements organize both geometry and timing; Newtonian gravity supplies a common dynamical explanation.

For an isolated two-body problem in Newtonian gravity, the relative elliptic orbit's orbital period $T$ obeysHere $a$ is the relative semi-major axis, and $M+m$ is the sum of the two masses. The separation $r$ varies during an elliptic orbit, so replacing $a$ by $r$ works only for a circular orbit, where $r=a$. Likewise, either body's individual semi-major axis about the centre of mass must first be converted to the relative one. When $m$ is negligible compared with $M$, the denominator can be approximated by $GM$.

For fixed gravitational parameter, the orbital period depends on the semi-major axis and is independent of orbital eccentricity. Our $a=1$, $\mu =1$ example gives $T=2\pi$ both for $e=0.6$ and for a circular orbit with $e=0$. The eccentric body's rapidly changing speed compensates for its different path. A compact way to see the cancellation uses Kepler's second law: divide the ellipse's area $\pi ab$ by its areal velocity $h/2$. With $b=a\sqrt{1-e^2}$ and specific angular momentum $h=\sqrt{\mu a(1-e^2)}$, the result is $2\pi \sqrt{a^3/\mu }$.

Compare two separate Kepler orbits with the same gravitational parameter, $a_1=1$ and $a_2=4$:With $\mu =1$, their orbital periods are $2\pi$ and $16\pi$; measured in units of $T_1$, these are $1$ and $8$. Quadrupling the size therefore takes eight times as long to complete a circuit. This power law makes comparisons possible without first choosing physical units.

The same relation can be inverted to weigh a system: $M+m=4{\pi }^2{a}^3/(G{T}^2)$. An observed relative semi-major axis and orbital period determine the total mass, while additional information is needed to separate the two masses. This is useful for binary stars, as discussed in NASA, Orbits and Kepler's Laws: measuring masses. Geometry and a clock together become a weighing instrument.

Why infer an area law from astronomy? In the authored account The great Martian catastrophe and how Kepler fixed it by Owen Gingerich, a 13 August 1593 Mars observation has ecliptic longitude $346^{\circ }7.5^{\prime }$, against predicted $342^{\circ }{0}^{\prime }$ and $351^{\circ }26^{\prime }$. These are Earth-based directions with a date, not measured Sun-based areas. The observer's armillary sphere used graduated circles and sighting arms to read declination; reversing it allowed a second reading. Comparing stars at different altitudes exposed refraction, and a planet-based check revealed an erroneous solar correction table. The apparent Mars distance signal had been contaminated by that correction. Johannes Kepler subsequently corrected the observer's Earth orbit, then tested Mars trajectories. The account describes typical angular errors of $2^{\prime }$ or less ($120$ arcseconds): small residuals needed care.^[ref]

The Tufts class notes Kepler's Astronomia Nova and the Orbit of Mars explain the key limitation: an area requires a trajectory and distances, so Kepler's second law cannot be directly tested from angular observations alone. They describe checking the combined ellipse and area law against 28 dated observations around seven oppositions during 1583-1595, using an Earth model to predict Earth-based directions; agreement improved greatly but some discrepancies remained.^[ref]

The inferential chain is useful today too: recover relative positions and times, compare swept areas, and ask what forces are compatible with their evolution. Agreement with equal areas supports zero torque about the chosen origin under the assumptions below; it alone does not identify an inverse-square law. Recovering the radial force also needs the trajectory and its timing, as in inferring an inverse-square force from Kepler laws. The two diagrams here are calculated examples, not the historical measurements.

With the Sun fixed, its gravitational force on the planet is a central force. In the fixed orbital plane, the transverse acceleration in polar coordinates therefore vanishes:During $dt$, the position vector sweeps an infinitesimal sector of area $dA=\frac{1}{2}{r}^{2}d\theta$. Hence the areal velocity isIntegrating over any two equal time intervals gives equal swept areas, proving Kepler's second law. Here $L$ is the signed component of angular momentum normal to the orbital plane. The argument uses conservation of angular momentum; an inverse-square dependence is not needed for this particular law, only a central force.

In a short time $dt$, the radius vector sweeps an oriented area $d\mathcal{A}=\frac{1}{2}{r}^{2}d\theta$. ThusThe constant $h$ is twice the oriented areal velocity, or the angular momentum per unit mass. Its constancy is Kepler's second law: equal areas are swept in equal times.

For a circular orbit of radius $a$, set $\ddot{r}=0$ in the radial equation. The balance isHence a circular orbit exists for any $a>0$ with this nonzero $|h|$. Its period gives Kepler's third law:To check circular-orbit stability for a power-law central force directly, keep $h$ fixed and set $r=a+\rho$. Expandingabout the balanced radius givesThe linear radial perturbation therefore obeysIt remains bounded and oscillates, proving stability under the stipulated small perturbations. Equivalently, the specific effective potential $h^2/(2r^2)-GM/r$ has a strict minimum at $a$.

For inferring an inverse-square force from Kepler laws, first use the area swept by an infinitesimal radius sector:Kepler's second law makes the areal velocity constant, so $l=r^2\dot{\theta }$ is constant. Then $r\ddot{\theta }+2\dot{r}\dot{\theta }=0$, meaning that the acceleration has no transverse component. Newton's second law therefore gives a central force directed along the Sun–planet line.

For the radial dependence, set $u=1/r$. Since $\dot{\theta }=l{u}^2$,The radial acceleration is the Binet equation expressionThe focus-based ellipse has $u=(1+\epsilon \cos \theta )/A$, hence $u_{\theta \theta }+u=1/A$. ThusFor a planet of mass $m$, the force is $F_r=-m{l}^2/(A{r}^2)$: it is attractive and obeys the inverse-square law. This deduction uses both the shape of the orbit and its constant areal velocity; the orbit shape alone would not determine its time-dependent acceleration.

A central force has the form $\mathbf{F}=f(r)\hat{\mathbf{r}}$, so it always points along the line from a fixed centre to the particle. Its torque about the centre vanishes:Thus angular momentum is conserved. Since $\mathbf{r}\cdot \mathbf{L}=0$, the trajectory lies in the fixed plane through the origin perpendicular to $\mathbf{L}$.

The area swept in time $dt$ is $dA=\frac{1}{2}|\mathbf{r}\times d\mathbf{r}|$, sowhich is constant. This proves Kepler's second law.

Let $\mathbf{r}$ be the orbiting point mass's position relative to the attracting body, and let its velocity be $\mathbf{v}=\dot{\mathbf{r}}$. Its angular momentum isConservation of angular momentum keeps both the magnitude and direction of $\mathbf{L}$ fixed. Since $\mathbf{L}$ is perpendicular to $\mathbf{r}$, a nonradial orbit lies in a fixed plane. During a short interval $dt$, the radius vector sweeps a thin triangle of areaThus the areal velocity is constant:Equal time intervals give equal swept areas: this is Kepler's second law. In plane polar coordinates, the same statement is $L=m{r}^2\dot{\varphi }$ and $\dot{A}=r^2\dot{\varphi }/2$. The attracting body is at the relevant focus of the ellipse, not its geometric centre. The area argument needs conserved angular momentum, not specifically an ellipse; a central force provides this conservation because its torque $\mathbf{r}\times \mathbf{F}$ vanishes.

For the two-body approximation with a dominant central mass, Kepler's laws of planetary motion are:

The first law includes a circular Kepler orbit as a special case. The third-law constant is the same for bodies orbiting the same dominant mass.

For a light test particle of mass $m$ outside a spherical source of mass $M$, Newton's law of universal gravitation gives $\mathbf{F}=-GMm\hat{\mathbf{r}}/r^2$. It is a central force, so its torque about the source vanishes and angular momentum $\mathbf{L}=m\mathbf{r}\times \mathbf{v}$ is conserved. The motion therefore lies in a fixed plane when $L\ne 0$. Its gravitational potential energy is $-GMm/r$, and conservation of mechanical energy givesHere $ϵ$ is specific orbital energy and $h$ is specific angular momentum. A Kepler orbit has polar equationThe angle is measured from periapsis. The conic section is an ellipse for $ϵ<0$ and $0\le e<1$, a parabola for $ϵ=0$ and $e=1$, and a hyperbola for $ϵ>0$ and $e>1$. The circle is the $e=0$ elliptical case. The special case $h=0$ is radial motion and must be handled as a degenerate limit. Bound ellipses have the source at one focus, which is Kepler's first law.

For a bound orbit of semi-major axis $a$, $ϵ=-GM/(2a)$. Combining this with the energy equation gives the vis-viva equation:The particle therefore moves faster near periapsis and slower near apoapsis. Conservation of angular momentum also gives constant areal velocity, $dA/dt=h/2$, expressing Kepler's second law. Using the ellipse's area and $h^2=GMa(1-e^2)$ gives the orbital periodwhich is Kepler's third law.

A circular orbit follows from centripetal acceleration $v_c^2/r=GM/r^2$, giving $v_c=\sqrt{GM/r}$ and $T=2\pi \sqrt{r^3/(GM)}$. The minimum escape velocity from radius $r$ follows from the marginally unbound condition $ϵ=0$:These ideal orbits neglect drag force, other gravitating bodies and changes in the central source. For two comparable masses, the relative two-body problem uses $G(M+m)$ instead of $GM$. An orbiting object is continuously in free fall; the inward gravitational force changes its direction even when its speed is constant. These inverse-square results assume the physical variation of gravity with radius, rather than extending a near-surface constant-$g$ approximation to an entire orbit.

Signed swept area counts the area traversed by a position vector, positively or negatively according to its direction of turning. For a continuously differentiable planar trajectory, its accumulated value relative to $t_0$ is the line integralA short segment contributes the oriented triangle area $\frac{1}{2}\mathbf{n}\cdot (\mathbf{r}\times d\mathbf{r})$; taking its time derivative gives areal velocity. Repeated turns count repeated area; a reversal subtracts it. Thus this quantity need not equal the area of the union of all previously swept regions. The origin matters: the relevant origin for Kepler's second law is the attracting focus of the ellipse, rather than its geometric centre.

Taylor's theorem expresses a sufficiently differentiable function as a finite Taylor polynomial plus a Taylor remainder. One convenient version assumes $f$ is $n+1$ times continuously differentiable on the segment between $a$ and $x$. Thenfor some intermediate point $\xi$, when $x\ne a$. Thus a bound $|f^{(n+1)}|\le M$ on that segment gives $|R_{n,a}(x)|\le M|x-a{|}^{n+1}/(n+1)!$. The intermediate point usually need not be computed: bounding the derivative throughout the segment is enough. This is why local approximation can yield reliable numerical bounds. A finite Taylor polynomial does not require an infinite Taylor series to represent the function.

Historically, the Royal Society holds Brook Taylor's letter of 26 July 1712 to John Machin. Its catalogue connects his work on Kepler's second law with the expansion later published in the 1715 Methodus incrementorum directa et inversa. This is a documented route from an orbital calculation to a general approximation method, rather than a claim that every earlier use of such expansions began with Taylor. See Royal Society archive, letter MS/82/2.