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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 $\gamma$.

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.

The most direct stellar mass determination uses orbital dynamics in a binary star. A resolved orbit and an independent distance give the relative semimajor axis $a$ from its angular size; measuring the period then gives the total mass by Kepler's third law:Individual barycentric orbital sizes, or the ratio of the two radial-velocity amplitudes, give $M_1/M_2=a_2/a_1=K_2/K_1$. A photocenter orbit alone needs the luminosity ratio to infer the barycentric orbit.

For a double-lined spectroscopic orbit, the line-of-sight amplitudes determine $a_i\sin i=K_{i}P\sqrt{1-e^2}/(2\pi )$. Combining them givesAn eclipsing binary light curve constrains the inclination and relative radii, turning these projected quantities into individual masses. Eclipse and orbit modeling must allow for eccentricity, surface brightness, limb darkening and possible tidal distortion.

If only one spectrum is measured, the binary mass function isWithout an inclination and an estimate of the visible star's mass it does not fix the unseen mass uniquely, but it provides useful constraints and lower limits.

Less direct estimates combine a spectroscopic surface gravity with a radius inferred from distance, bolometric flux and effective temperature: $R^2=L/(4\pi {\sigma }_{\mathrm{SB}}{T}_{\mathrm{eff}}^4)$ and $M=g{R}^2/G$. These depend on atmosphere modeling. Locating a star on an evolutionary track or using a calibrated main-sequence mass-luminosity relation also estimates mass, but assumes composition and evolutionary state; giants and mass-transfer products need not obey a single main-sequence relation. Dynamical binary masses provide the empirical calibration against which such model-dependent methods are tested.

Let $a_{\mathrm{ns}}$ be the neutron star's distance from the centre of mass. In a circular orbit,Combining this with Kepler third law,gives the binary mass functionSince $\sin i\le 1$ and $(M_{\mathrm{ns}}+M_{\mathrm{wd}}{)}^2>M_{\mathrm{wd}}^2$,

When a low-mass red giant undergoes stable Roche-lobe overflow, the neutron star receives matter with substantial specific angular momentum, usually through an accretion disk. The accretion torque spins it up and weakens its external magnetic field, producing a recycled pulsar with a millisecond period. Removal of the donor's envelope exposes its helium core as a low-mass white dwarf.

At detachment, the donor mass and core mass both become $M_{\mathrm{wd}}$. Its luminosity is fixed by the red-giant core-mass--luminosity relation, and the supplied radius law therefore makes its final radius $R_L$ a function only of $M_{\mathrm{wd}}$. Cubing the Roche-lobe relation givesEliminating $a$ with Kepler third law yieldsThis is the white-dwarf mass--orbital-period relation: dependence on the neutron-star mass cancels.

For randomly oriented binaries, the isotropic binary inclination distribution isThe observed period gives $M_{\mathrm{wd}}$ from the preceding relation. If $M_{\mathrm{ns}}=1.35M_{\odot }$ is adopted, the measured mass function then givesso the inclination is inferred without astrometry.

Two plausible causes of an apparent shortage of edge-on systems are:

Scatter or bias in the core-mass--period relation can reinforce the second effect.

The semimajor axis $a_1$ of the pulsar's orbit about the binary barycenter, projected along the line of sight and divided by the speed of light. The orbital inclination $i$ is zero for a face-on orbit. Thus $x$ is a time, often quoted as the numerical number of light-seconds in $a_1\sin i$, not the true binary separation. The relative-orbit semimajor axis is instead $a=a_1+a_2$; conservation of momentum gives $a_1=a{m}_2/(m_1+m_2)$.

Combining Kepler's third law with that barycenter relation gives the timing form of the binary mass function, also valid for eccentric Kepler orbits:Choose $x=2\mathrm{s}$ and $P_b=8\mathrm{h}$ for an original example: $a_1\sin i=599,585\mathrm{km}$ and $f\simeq 0.0773M_{\odot }$, where $M_{\odot }$ is a solar mass. For a specified pulsar mass $m_1=1.4M_{\odot }$, the smallest allowed companion mass occurs at $\sin i=1$ and is approximately $0.698M_{\odot }$. A different assumed $m_1$ changes this bound. One measured $f$ does not determine both masses and orbital inclination.

Orbital dynamics gives the most direct stellar masses: a resolved binary orbit and distance determine the relative semimajor axis, and Kepler's third law gives the total mass. The barycentric size or radial-velocity amplitude ratio gives individual masses. An eclipsing binary inclination removes the sine-cubed projection factor from spectroscopic masses; a single-lined binary mass function alone does not uniquely determine both masses. Surface-gravity/radius and stellar-track estimates are model-dependent and are calibrated against dynamical binary measurements.