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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.

Conservation of angular momentum says that total angular momentum stays fixed when the net external torque about the chosen fixed inertial origin vanishes. Individual angular momenta and angular velocities can change. The radius change at constant angular momentum below makes this distinction visible: pulling two rotating point masses inward lowers the moment of inertia, raises the angular velocity, and supplies work that increases the kinetic energy.

Follow the same constant-mass point masses throughout, in an inertial frame, with all positions measured from a fixed origin $O$. If the internal pair forces are opposite and central, meaning directed along each pair's separation, thenThus zero net external torque throughout an interval implies constant total angular momentum throughout that interval. A vanishing external torque component, such as ${\tau }_z^{\mathrm{ext}}=0$, conserves the corresponding $L_z$ even when the other components change. This is the balance law for a fixed collection; angular-momentum flux must also be accounted for when matter crosses its boundary. ^[ref]

The central qualification is essential. Write ${\mathbf{F}}_{ij}$ for the internal force on member $i$ from member $j$. Under Newton's third law, ${\mathbf{F}}_{ji}=-{\mathbf{F}}_{ij}$, but the sum of their torques isThis vanishes when ${\mathbf{F}}_{ij}$ is parallel to ${\mathbf{r}}_i-{\mathbf{r}}_j$. Equal-and-opposite forces alone do not suffice: two opposite transverse forces with separated lines of action form a couple with nonzero torque. Cancellation of the internal pair torques, together with Newton's second law, proves the total balance above.

For one point mass under a central force about $O$, $\mathbf{F}=f(r)\hat{\mathbf{r}}$ gives ${\mathbf{\tau }}_O=\mathbf{r}\times \mathbf{F}=0$. The force may still change the speed and kinetic energy; its zero torque preserves angular momentum. With nonzero ${\mathbf{L}}_O$, $\mathbf{r}\cdot {\mathbf{L}}_O=0$ keeps the motion in a fixed plane. Choosing the $z$ axis perpendicular to that plane, $r^2\dot{\theta }=L_z/m$ governs its angular progress. This removes an unknown from the central-force radial turning point calculation and helps reduce the Kepler orbit to a radial problem.

There is a historical geometric version of this result. In Book I, Section II, Proposition I of the Principia, Newton proves that motion under a central force toward a fixed centre stays in a plane and sweeps equal areas in equal times, using successive central impulses and equal triangle areas. ^[ref] In the modern notation above, constant areal velocity is $\dot{A}=L_z/(2m)$: the old geometric proof and the angular momentum calculation express the same invariant.

Take an illustrative elliptic Kepler orbit with semi-major axis $a=1$, orbital eccentricity $e=0.6$ and gravitational parameter $\mu =1$, in consistent dimensionless units. Its semi-minor axis is $b=a\sqrt{1-e^2}=0.8$, mean motion is $n=\sqrt{\mu /a^3}=1$, and specific angular momentum magnitude is $h=\sqrt{\mu a(1-e^2)}=0.8$. About the attracting focus,The two sectors below each span $\Delta t=0.5$, giving $\Delta A=0.2$. Near periapsis the short radius sweeps a wide angle; near apoapsis the long radius sweeps a narrow angle. Constant areal velocity therefore does not require constant angular velocity or constant speed.

To compute the sectors without mistaking equal angles for equal times, parametrize the ellipse by eccentric anomaly $E$:This follows by integrating $\frac{1}{2}(xdy-ydx)$. Kepler's equation gives $E-e\sin E=M=n(t-t_{\mathrm{p}})$, with mean anomaly $M$ measured from periapsis time $t_{\mathrm{p}}$. Consequently $\Delta A=abn\Delta t/2=h\Delta t/2$. The displayed sectors use mean anomaly intervals $[-0.25,0.25]$ and $[\pi -0.25,\pi +0.25]$, rather than equal eccentric anomaly increments.

Constant areal velocity gives constant specific angular momentum $l=r^2\dot{\theta }$, hence zero transverse acceleration and a central force. For a focus-based ellipse $u=1/r=(1+\epsilon \cos \theta )/A$, one has $u_{\theta \theta }+u=1/A$. The Binet equation then gives the displayed radial acceleration, proving an attractive inverse-square law. An orbit's shape alone does not specify its acceleration without a time law.

The relative displacement of a Kepler orbit sweeps equal areas about the attracting focus of a conic in equal times. A short interval near periapsis covers a wide angle at small separation; the same interval near apoapsis covers a narrow angle at large separation. The swept areas are equal even though the travelled distances are different. In particular, the sectors have curved ellipse boundaries: triangles formed by joining their endpoints only approximate the swept areas for short intervals.

Let $h=|\mathbf{r}\times \dot{\mathbf{r}}|$ be the relative specific angular momentum. The areal velocity isA central force exerts zero torque about its centre, so conservation of angular momentum makes this rate constant. This explanation also applies to other central forces; the ellipse shape specifically comes from Newtonian gravity.

In the $a=1$, $e=0.6$, $\mu =1$ example, specific angular momentum is $h=\sqrt{\mu a(1-e^2)}=\sqrt{0.64}=0.8$, giving areal velocity $0.4$. At periapsis and apoapsis, the velocity is perpendicular to the relative displacement, so $h=rv$. The endpoint speeds are consequentlyThe fourfold speed difference is consistent with the fourfold separation difference. Away from these endpoints, $h/r$ gives only the transverse component of velocity; the full speed follows from the vis-viva equation.

To actually place equal-time markers, use Kepler's equation. Write mean anomaly as $\mathcal{M}=nt$, with $t=0$ at periapsis and mean motion $n=\sqrt{\mu /a^3}=1$. The different symbol $\mathcal{M}$ avoids confusing this angle with the mass $M$. Solve for eccentric anomaly $E$ in radians, then compute the relative position:This timing procedure is explained in NASA educational archive, How Orbital Motion is Calculated.

For the first interval $0\le t\le \pi /6$, the final eccentric anomaly is $E\approx 1.041495$. For the equal interval $\pi \le t\le 7\pi /6$ starting at apoapsis, it ends at $E\approx 3.471065$. Both swept areas are $\Delta A=(h/2)\Delta t=0.4\pi /6=\pi /15\approx 0.209440$, one twelfth of the entire ellipse's area $0.8\pi$. The diagram computes its markers from Kepler's equation, making visible how uniform time steps become uneven spatial steps.

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.

A central force has zero torque, so $h=r^2\dot{\theta }$ is constant and the angular momentum is $mh$. The areal velocity is $h/2$. Put $u=1/r$ and let primes denote angular derivatives. Then $\dot{r}=-h{u}^{\prime }$, $\ddot{r}=-h^2{u}^2{u}^{\prime \prime }$, and $r{\dot{\theta }}^2=h^2{u}^3$. The radial equation with inward force magnitude $f(u)$ is $m(\ddot{r}-r{\dot{\theta }}^2)=-f(u)$, giving the Binet equationChoose the angular origin along a diameter of a radius-$R$ circle through $O$. Its polar equation is $r=2R\cos \theta$, so $u=(2R{)}^{-1}\sec \theta$ on $-\pi /2<\theta <\pi /2$. Since $(\sec \theta {)}^{\prime \prime }+\sec \theta =2{\sec }^3\theta$, substitution gives $u^{\prime \prime }+u=8R^2{u}^3$ and henceThe force is attractive. For fixed force coefficient $c$ and mass, $R|mh|=\sqrt{mc/8}$, so $R$ is inversely proportional to the magnitude of the angular momentum. The time along the circle isThis circular orbit through an inverse-fifth-power singularity reaches the force center at the two endpoints of this angular interval. The integral is finite; continuation through the singular collision is not specified by the force law alone.

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.

The transfer ellipse has semi-major axis and semi-minor axisConservation of angular momentum gives constant areal velocity $h/2$. The satellite sweeps half the ellipse, whose area is $\pi ab/2$, soUsing part (i),This is half the orbital period from Kepler's third 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.

Kepler's laws of planetary motion state that, in the relative two-body description, bound trajectories are ellipses with the attracting body at a focus, the radius vector sweeps equal areas in equal times, and the square of the period is proportional to the cube of the semimajor axis for a fixed total mass. Their Newtonian derivation is especially informative. With relative coordinate $\mathbf{r}$ and ${\mu }_g=G(M+m)$, the inverse-square acceleration is $-{\mu }_g\mathbf{r}/r^3$. Its zero torque conserves $\mathbf{h}=\mathbf{r}\times \dot{\mathbf{r}}$, fixing a plane and the areal velocity $h/2$.

For $u=1/r$, the orbit equation is $u^{\prime \prime }+u={\mu }_g/h^2$ as a function of polar angle. ThereforeBound negative-energy motion has $0\le e<1$ and semilatus rectum $h^2/{\mu }_g=a(1-e^2)$. Dividing ellipse area by areal velocity givesFor a planet much lighter than its star, replace $M+m$ by $M$. Zero-energy and positive-energy encounters are parabolic and hyperbolic rather than periodic ellipses. Perturbations from other bodies and departures from point-mass Newtonian gravity can precess or otherwise modify the orbit; Kepler's laws are a two-body approximation, not an exact rule for every observed planetary path.

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 test particle outside a spherical central mass, the gravitational force is central, so angular momentum is conserved and the orbit is planar. With specific angular momentum $h=r^2\dot{\theta }$ and specific orbital energy $ϵ={\dot{r}}^2/2+h^2/(2r^2)-GM/r$, the radial effective potential combines attraction with the centrifugal barrier. Its nonradial solutions are conic sections:For nonzero $h$, negative energy gives an ellipse with $0\le e<1$, zero energy a parabolic Kepler orbit and positive energy a hyperbolic Kepler orbit. The circle is the special case $e=0$. Exactly radial motion is a separate zero-angular-momentum case and may hit the central body; a mathematical point-mass trajectory is not automatically a physically noncolliding orbit.

Conservation of $h$ gives constant areal velocity $h/2$, so equal areas are swept in equal times. For a bound ellipse of semimajor axis $a$, $ϵ=-GM/(2a)$ and $T^2=4{\pi }^2{a}^3/(GM)$. A circular orbit has speed $\sqrt{GM/r}$, while the minimum local escape speed is $\sqrt{2GM/r}$, obtained by setting the total energy to zero. These are the test-mass two-body results: a comparable second mass requires the relative-coordinate gravitational parameter $G(M+m)$, and nonspherical structure or drag can perturb the conic and its conserved quantities.

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.

Assume a constant positive mass $m$, a twice differentiable trajectory in the fixed plane, a fixed origin in an inertial frame, and Newton's second law for the net force. The product rule gives the short proofThe term $\mathbf{v}\times \mathbf{v}$ vanishes. This is the planar angular momentum balance written in area language.^[ref]

On a connected time interval, areal velocity is constant if and only if ${\tau }_n=0$ throughout. Since both position and acceleration lie in the fixed plane, the net torque has only that normal component. For $r\ne 0$, zero net torque is therefore equivalent to net force parallel to $\mathbf{r}$: its instantaneous direction is central, possibly inward, outward or zero. This diagnoses the net force along this motion, not its dependence on position or time elsewhere. Opposing transverse forces may cancel.

A central force about a fixed origin gives conservation of angular momentum, irrespective of its radial dependence: an inverse-square law is sufficient but unnecessary. With nonzero conserved angular momentum, $\mathbf{r}\cdot \mathbf{L}=0$ fixes the orbital plane. Zero angular momentum gives radial motion and zero areal velocity; it does not select a unique orbital plane. The converse needs the fixed-plane assumption: a constant projected areal velocity in three dimensions only constrains one torque component. Also, $r=0$ makes the instantaneous torque vanish regardless of force and makes plane polar coordinates singular. A moving origin or changing plane requires additional terms.

A path along a circle can have variable areal velocity. As an illustrative controlled trajectory, keep a point mass on a circle of radius $R$ with angleIts acceleration in polar coordinates and the required net force arewhere ${\mathbf{e}}_r$ and ${\mathbf{e}}_{\theta }$ are the radial and transverse unit vectors. A circular guide can supply the radial force, while a tangential drive supplies $mR\alpha$. HenceFor $R=m={\omega }_0=\alpha =1$ in illustrative units, the three equal time intervals $[0,0.5]$, $[0.5,1]$ and $[1,1.5]$ sweep areas $0.3125$, $0.4375$ and $0.5625$. Their difference comes from the transverse force's positive torque, even though the radius and path shape stay fixed. The area under each areal velocity graph segment equals the corresponding swept area.