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The centre of mass of an astronomical system. In an isolated two-body problem, both bodies follow orbits around their shared astronomical barycenter; the Kepler orbit of their relative displacement has a semi-major axis equal to the sum of their individual semi-major axes.

A binary planetesimal is a gravitationally bound pair of planetesimals. Its internal two-body problem and its centre of mass orbit respond differently to external forces, including gas drag.

The Cavendish experiment makes the attraction predicted by Newton's law of universal gravitation measurable between laboratory masses. Two small test masses hang at the ends of a horizontal rod in a Cavendish torsion balance; nearby larger source masses exert a tiny torque. A thin suspension fiber supplies an opposing torque as it twists. Measuring its twist and calibrating its torsion constant lets an experimenter infer the gravitational force and, in a modern analysis, the gravitational constant. ^[ref]

The beautiful feature is that the two horizontal forces can cancel as vectors while their torques add. The instrument converts weak attraction into a rotation, with sensitivity set by the fiber's torsion constant. Earth's much larger downward gravitational weight is supported by the suspension rather than supplying the intended torque about its vertical axis. A laboratory measurement can therefore probe the same Newtonian gravitational field that governs the two-body problem.

Henry Cavendish's original paper was Experiments to determine the Density of the Earth, published in Philosophical Transactions, volume 88 (1798), pp. 469–526, and read on 21 June 1798. On p. 469 he credits John Michell with the apparatus idea and explains that Michell died before using it; the instrument reached Cavendish through Francis John Hyde Wollaston. Cavendish's stated aim was Earth's mean mass density. Calling the experiment a determination of the gravitational constant is a modern interpretation, not Cavendish's own formulation of his result. ^[ref]

The connection between Earth's mean mass density and the gravitational constant follows in a spherical, nonrotating model of Earth. With Earth mass $M_E$, radius $R_E$, and surface gravitational acceleration $g$, the spherical shell theorem and Newton's law of universal gravitation giveThus a local comparison of gravitational attractions, together with knowledge of Earth's size and surface gravitational acceleration, reaches a property of the entire planet. These are modern equations explaining the connection, not equations quoted from Cavendish's paper.

For a worked example, use an explicitly simplified two-near-pair model. Each test point mass is $m$, each source point mass is $M$, and each lever arm is $L$, so the rod's full length is $2L$. Neglect the rod's mass. At zero twist put the test masses at $(-L,0)$ and $(L,0)$, and the sources at $(-L,-r)$ and $(L,r)$. Here $r$ is the near-pair center separation. Ignore each source's attraction on the test mass at the opposite end. Evaluate the two near-pair forces at zero twist and treat their torque as constant for a small twist, requiring $L|\theta |\ll r$. This is a teaching model, not a precision model of Cavendish's actual apparatus.

By Newton's law of universal gravitation, each retained force has magnitude $F=GmM/r^2$. The left force points down and the right force points up. Using torque about the suspension axis, both contributions are counterclockwise when viewed from above:The net retained horizontal force is zero, yet the net torque is positive. Moving the sources to $(-L,r)$ and $(L,-r)$ reverses both forces and hence the torque. In a real apparatus, the opposite source also attracts each test mass; that correction is explicitly included in the university derivation. ^[ref]

For illustrative parameters, take $m=0.010\mathrm{kg}$, $M=1.0\mathrm{kg}$, $r=0.050\mathrm{m}$, $L=0.10\mathrm{m}$, and torsion constant $\kappa =1.0\times 10^{-8}\mathrm{N}m/rad$. Use $G=6.67430\times 10^{-11}{\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{s}}^{-2}$, the central value in NIST's 2022 CODATA table; that table gives standard uncertainty $0.00015\times 10^{-11}{\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{s}}^{-2}$. This is an input for the example, not a measurement made with these assumed parameters. ^[ref]This sub-nanonewton force produces a measurable twist in the Cavendish torsion balance. The displayed digits make the toy calculation reproducible; they do not describe experimental precision.

Doppler spectroscopy detects an exoplanet through the host star's changing Doppler shift. Both bodies orbit their shared astronomical barycenter: the star undergoes stellar reflex motion even when the exoplanet itself is unseen. Its velocity along the line of sight shifts its spectral lines; at speeds small compared with the speed of light, $\Delta \lambda /\lambda \simeq v_r/c$, with positive $v_r$ for recession. Repeated measurements can fit a Kepler orbit, rather than merely detect one shifted spectrum. Starspots and other changes in stellar spectral lines must be distinguished from actual orbital motion.

For one companion in an isolated two-body problem, the stellar radial-velocity semi-amplitude isHere $G$ is the gravitational constant, $P$ the orbital period, $e$ the orbital eccentricity, and $M_{\star },M_p$ the star and exoplanet masses. The orbital inclination $i$ is measured relative to the sky plane: $i=0$ is face-on and $i=90^{\circ }$ is an edge-on orbit. The exact binary mass function measured from the fitted orbit isWith known $M_{\star }$ and $M_p\ll M_{\star }$, this is usually reported as $M_p\sin i$. Taking $\sin i=1$ gives the minimum companion mass; a smaller orbital inclination allows a larger mass. A detected exoplanet transit constrains $i$ close to $90^{\circ }$ for a compact planet orbiting well outside the star's surface, substantially resolving this ambiguity. Combining Doppler spectroscopy and exoplanet transit photometry then supplies a mass and radius for exoplanet interior studies.

For a computed circular example, let $q=M_p/M_{\star }=0.001$ and choose a relative orbital speed $V_{\mathrm{rel}}=100\mathrm{km}{\mathrm{s}}^{-1}$: this is the speed of the planet relative to the star, not its speed about the astronomical barycenter. Conservation of momentum givesThe barycentric velocities point in opposite directions, so their speeds add to $V_{\mathrm{rel}}$. Here $e=0$ and $K=V_{\star }\sin i$: $K=99.9001\mathrm{m}{\mathrm{s}}^{-1}$ at $i=90^{\circ }$, $49.9500\mathrm{m}{\mathrm{s}}^{-1}$ at $i=30^{\circ }$, and zero at $i=0$. The star still orbits in the face-on case, but its orbital velocity has no line-of-sight component. The plotted circular signal is $v_r(t)-\gamma =K\sin (2\pi t/P)$ for a chosen phase origin and constant system velocity $\gamma$.

For a test mass $M$ and isotropic background particles of mass $m$, the hyperbolic two-body problem uses the relative deflection scale $b_{90}=G(M+m)/w^2$. The test body's longitudinal velocity change in one encounter is $-2mw/[(M+m)(1+b^2/b_{90}^2)]$. Integrating the encounter rate over impact parameter gives a logarithm; replacing its velocity-dependent value by a common Coulomb logarithm in stellar dynamics gives the displayed systematic drag. The approximation $M+m\simeq M$ requires $M\gg m$, not merely $M>m$. Velocity diffusion and exact cutoff dependence are distinct effects omitted by this leading-log drag model.

An orbit around an isolated source constrains its standard gravitational parameter $\mu =GM$ rather than independently measuring the source's mass and the gravitational constant. In the test particle approximation, a circular orbit of radius $a$ and orbital period $T$ satisfiesThe left equation equates the required centripetal acceleration to the gravitational acceleration. More generally, a Kepler orbit depends on the same $\mu$. In the exact relative two-body problem, replace $GM$ by $G(M+m)$.

For any positive dimensionless number $q$, replacing $G$ by $qG$ and $M$ by $M/q$ leaves $GM$ and every predicted test particle trajectory unchanged. This is the gravitational mass degeneracy: more observations of the same orbit can refine $GM$ without separating its factors. A laboratory source whose mass is calibrated independently supplies the missing information.

Consequently an astronomical mass expressed in kilograms as $M=\mu /G$ inherits measurement uncertainty from $G$, even if $\mu$ is known much more precisely. To first order,When the relative standard uncertainty of $\mu$ is negligible, that of $M$ is approximately $u(G)/G$. Conversely, the ratio of two source masses inferred using the same $G$ is $M_1/M_2={\mu }_1/{\mu }_2$, so this common factor cancels. That distinction explains how excellent orbit determination can coexist with less precise absolute astronomical masses.

A repeating Doppler shift contains more than an amplitude. For an isolated two-body problem, let $\nu$ be the true anomaly, ${\omega }_{\star }$ the stellar argument of periapsis, $\tau$ the periapsis time and $\gamma$ the constant system line-of-sight velocity. The companion's argument of periapsis differs by $\pi$ from ${\omega }_{\star }$, so mixing the two conventions reverses the orbital signal. With positive velocity for recession, choose the stellar line-of-sight coordinate positive away from the observer. Kepler's equation supplies the clock:Here $M$ is the mean anomaly and $E$ the eccentric anomaly. The time coordinate is uniform in $M$, not in $\nu$: an eccentric Kepler orbit runs fastest near periapsis.

To derive the velocity law without guessing a sinusoid, write the stellar orbital coordinates as $X=a_{\star }(\cos E-e)$ and $Y=a_{\star }\sqrt{1-e^2}\sin E$. Project toDifferentiating $z$ and substituting the true anomaly identities givesThe constant $e\cos {\omega }_{\star }$ term makes the time average of the orbital velocity zero: $z$ returns to its original value after one orbital period. For $e=0$ this reduces to a sinusoid with an arbitrary phase origin; periapsis itself is then undefined. For $e>0$, the largest and smallest velocities are $\gamma +K(e\cos {\omega }_{\star }\pm 1)$, so $K$ is still half the peak-to-peak range, although the extrema need not be symmetric about $\gamma$.

For an original worked example, choose $M_{\star }=1M_{\odot }$, $P=10$ days, $K=50\mathrm{m}{\mathrm{s}}^{-1}$, $e=0.4$ and ${\omega }_{\star }=60^{\circ }$. Taking one day as $86400\mathrm{s}$ and $G=6.67430\times 10^{-11}{\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{s}}^{-2}$, the binary mass function is $f=1.98270\times 10^{20}\mathrm{kg}$. Its units check directly: $[P{K}^3/G]=\mathrm{kg}$. With adopted solar mass $M_{\odot }=1.98847\times 10^{30}\mathrm{kg}$ and Jupiter mass $M_J=1.89813\times 10^{27}\mathrm{kg}$, solving the exact binary mass function gives

All three systems have the same chosen $P,K,e,{\omega }_{\star },\tau$ and therefore the same Doppler spectroscopy time series, despite different true masses. Exactly what remains fixed is $M_p\sin i/(M_{\star }+M_p{)}^{2/3}$; $M_p\sin i$ alone is only the small-companion approximation. The $90^{\circ }$ value is the minimum mass. For this example the extrema are $+60$ and $-40\mathrm{m}{\mathrm{s}}^{-1}$ relative to $\gamma$, and the periapsis value is $+35\mathrm{m}{\mathrm{s}}^{-1}$. This is a chosen eccentric system, not a reconstruction of 51 Pegasi b.

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.

Newton's law of universal gravitation gives the attractive force between two point masses. With source mass $M$ at the origin and relative position $\mathbf{r}$ of mass $m$, write $r=|\mathbf{r}|>0$:Here $G$ is the gravitational constant. The minus sign points toward $M$; the magnitude follows an inverse-square law, so doubling the separation quarters the force. The opposite force on $M$ satisfies Newton's third law. ^[ref]

Both masses accelerate. Applying Newton's second law to each in the isolated two-body problem givesThe relative standard gravitational parameter includes both masses; replacing it by $GM$ requires the test particle approximation $m\ll M$.

Historically, Newton's Book III, Proposition VII argues that gravitational attraction applies to all bodies and scales with their masses; its second corollary states the inverse-square law for individual particles. ^[ref]

By Newton's third law, the force on the second particle is $-F_{12}$. For $M=m_1+m_2$ and centre of mass $R=(m_1{r}_1+m_2{r}_2)/M$, Newton's second law gives $M\ddot{R}=F_{12}-F_{12}=0$. ThusThe centre moves on a straight line with constant velocity; zero velocity is allowed. For the relative position $r=r_1-r_2$, subtraction of the two equations givesHence the two-body problem reduces to a single particle of reduced mass $\mu$, together with the elementary centre motion. Reconstruct the positions as $r_1=R+(m_2/M)r$, $r_2=R-(m_1/M)r$.

Yes: the two equal masses can follow the same fixed circle in diametrically opposite positions. Put the stationary centre of mass at the circle's center and choose tangential velocities with the same sense of rotation. If the circle has radius $a$, the separation is $2a$ and the mutual gravitational force is directed toward that center, with magnitude $G{m}^2/(4a^2)$. The circular orbit condition is $ma{\omega }^2=G{m}^2/(4a^2)$, so an equal-mass circular binary hasThese initial data maintain the opposite positions and give the same constant angular speed for both particles.

Use the separation vector $\mathbf{r}$ from the star to the comet, and let $m_{\mathrm{red}}=m{M}_{\star }/(m+M_{\star })$. The two-body problem separates into uniform centre of mass motion andTaking a cross product with $\mathbf{r}$ proves conservation of angular momentum; taking the scalar product with $\dot{\mathbf{r}}$ proves conservation of energy. The relative constants areThe total relative angular momentum and energy are $m_{\mathrm{red}}\mathbf{h}$ and $m_{\mathrm{red}}\epsilon$, respectively. The conserved eccentricity vector is $\mathbf{e}=\dot{\mathbf{r}}\times \mathbf{h}/\mu -\hat{\mathbf{r}}$. Its scalar product with $\hat{\mathbf{r}}$ gives $r=h^2/[\mu (1+e\cos f)]$, with $f$ the true anomaly. The periapsis and apoapsis distances are therefore $h^2/[\mu (1+e)]$ and $h^2/[\mu (1-e)]$. Their sum is twice the semi-major axis, soAt either apsis $v=h/r$; substituting an apsidal radius gives $\epsilon =-\mu /(2a)$. Thus the specific orbital energy implies the vis-viva equationThese are exact relative-coordinate formulas for an elliptic orbit, $0\le e<1$. In the barycentric frame the comet's speed is $M_{\star }/(M_{\star }+m)$ times the relative speed; the printed identification with the comet's stellar orbital speed uses $M_{\star }\gg m$.

Let $r_0$ be the separation of the two centres at release. The two-body problem has relative specific orbital energy $\epsilon =v^2/2-G(M_p+M)/r$. Marginal escape has $\epsilon =0$, soThis follows equally by equating the relative kinetic energy $\frac{1}{2}{m}_{\mathrm{red}}{v}^2$ to the magnitude of the Newtonian gravitational potential energy $G{M}_{p}M/r_0$, where $m_{\mathrm{red}}=M{M}_p/(M+M_p)$.

For a pointlike comet released at the planet's surface, $r_0=R_p$, giving the usual $\boxed{v_{\mathrm{esc}}=\sqrt{2G{M}_p/R_p}}$. If the finite comet touches that surface, its centre is at $r_0=R_p+R$; the same formula uses this larger separation. The assumption $M_p\gg M$ by itself does not imply $R\ll R_p$. Subsequent formulas written in terms of the planet-only escape velocity use the additional small-comet approximation.

This is local escape from an isolated planet. In a stellar Newtonian gravitational field, reaching the edge of the Hill sphere and escaping the stellar system are different energy thresholds.

Let $\mathbf{r}={\mathbf{r}}_2-{\mathbf{r}}_1$ and $r=|\mathbf{r}|$. Newton's second law for the two planetesimals givesSubtracting yields the forced relative two-body problemThe internal gravitational forces cancel from the centre of mass equation but add in the relative equation; external forces affect the internal orbit through their difference in acceleration.

For the relative position $\mathbf{r}={\mathbf{r}}_1-{\mathbf{r}}_2$, Newton's law of universal gravitation gives the two-body problem in its reduced one-body form,Taking the dot product with the relative velocity $\mathbf{v}=\dot{\mathbf{r}}$ givesIntegration therefore yields conservation of specific orbital energy:

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.

The reduced mass is the effective mass for the two-particle relative coordinate of two bodies with constant positive masses $m_1,m_2$:Both bodies can respond to their mutual force, so their relative acceleration is greater than if one were fixed. Consequently $0<\mu <\min (m_1,m_2)$. This explains why using either individual mass, or their sum, generally gives the wrong internal oscillation rate. The same reduced mass describes a two-body problem, a free two-mass spring, and the approximate molecular vibration of a diatomic molecule.^[ref]

For equal masses $m_1=m_2=m$, $\mu =m/2$. Two $1\mathrm{kg}$ bodies therefore have $\mu =0.5\mathrm{kg}$: in their stationary centre of mass frame a relative speed $2m/s$ means individual speeds $1m/s$ and total kinetic energy $1\mathrm{J}$. With $m_1=1\mathrm{kg}$ and $m_2=100\mathrm{kg}$, instead $\mu =100/101\mathrm{kg}\approx 0.990099\mathrm{kg}$. At the same relative speed, the individual speeds are $200/101$ and $2/101m/s$, and the total kinetic energy is $200/101\mathrm{J}$. These are calculations for stipulated masses and velocities.

For fixed $m_1$ and $q=m_2/m_1>0$, $\mu /m_1=q/(1+q)$. The heavy-partner limit $q\to \infty$ gives $\mu \to m_1$ from below; the relative error incurred by replacing $\mu$ with $m_1$ is $(m_1-\mu )/\mu =1/q$. At $q=100$ this is $1\%$. Treating the heavy body as fixed is thus a controlled approximation to internal motion, rather than an exact consequence of its being heavier.

The product $\mu =GM$ of the gravitational constant and a central mass sets the strength of the exterior Newtonian gravitational field. Its dimensions are length cubed divided by time squared. In the exact relative two-body problem, the parameter is $G(M_1+M_2)$; using $GM$ for one source assumes the test particle approximation.

The combination is useful because orbital and falling-body observations constrain $\mu$ directly. A gravitational circular orbit of radius $a$ has oscillation period $T=2\pi \sqrt{a^3/\mu }$, so measuring $a$ and $T$ gives $\mu =4{\pi }^2{a}^3/T^2$. That measurement alone does not separately determine $G$ and $M$. Similarly, radial free fall from rest has a natural time scale $\sqrt{r_0^3/\mu }$.

Other interactions require a different model. Inverse-square radiation pressure can reduce the effective attraction for a dust Kepler orbit; calling that effective coefficient a gravitational parameter must not hide the added force.

The motion of a star around the astronomical barycenter caused by an orbiting companion. In an isolated two-body problem with relative semi-major axis $a$, the stellar semi-major axis is $a_{\star }=M_{p}a/(M_{\star }+M_p)$. Doppler spectroscopy measures its line-of-sight velocity and astrometry measures its projected position on the sky.

Choose positions in one inertial frame and let $M=m_1+m_2$. The centre of mass and two-particle relative coordinate areThe lighter body is farther from the centre of mass. Reconstructing both positions from these equations is part of solving the two-body problem; $\mathbf{r}$ alone describes their separation, not either body's laboratory position.^[ref]

Substitute the reconstructed velocities into the kinetic energy. The result is the exact two-particle instance of kinetic-energy decomposition about the centre of mass:The mixed dot product has coefficient $-m_1{m}_2/M+m_2{m}_1/M=0$; the coefficient of $|\dot{\mathbf{r}}{|}^2$ is $(m_1{m}_2^2+m_2{m}_1^2)/M^2=\mu$. This algebra needs constant positive masses but no particular force law. In the stationary centre of mass frame, the fraction of internal kinetic energy carried by body 1 is $m_2/M$, and by body 2 is $m_1/M$: the lighter body carries more.

For a closed pair with no external forces, write $\mathbf{F}$ for the force on body 2 and $-\mathbf{F}$ for that on body 1, as required by Newton's third law. Adding and subtracting their Newton's second law equations yieldsThus the centre of mass moves uniformly while the two-particle relative coordinate behaves like one body of reduced mass. Equal-opposite internal forces suffice for these equations. For an internal central force, $\mathbf{F}=f(r)\hat{\mathbf{r}}$ with $r=|\mathbf{r}|>0$, the internal torque vanishes and conservation of angular momentum gives constant ${\mathbf{L}}_{\mathrm{rel}}=\mu \mathbf{r}\times \dot{\mathbf{r}}$. If the central force comes from a time-independent potential energy $U(r)$, then $f(r)=-dU/dr$ and the internal mechanical energy $E_{\mathrm{rel}}=\mu |\dot{\mathbf{r}}{|}^2/2+U(r)$ is conserved. The kinetic energy split alone does not establish either conservation law.