= Keplerian radial-velocity curve
{c}
{title2=$v_r(t)=\gamma+K[\cos(\nu+\omega_\star)+e\cos\omega_\star]$}
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:
$$
M=\frac{2\pi(t-\tau)}P=E-e\sin E,\qquad
\cos\nu=\frac{\cos E-e}{1-e\cos E},\qquad
\sin\nu=\frac{\sqrt{1-e^2}\sin E}{1-e\cos E}.
$$
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 to
$$
z=(X\sin\omega_\star+Y\cos\omega_\star)\sin i,\qquad
\frac{dE}{dt}=\frac{2\pi/P}{1-e\cos E}.
$$
Differentiating $z$ and substituting the <true anomaly> identities gives
$$
v_r-\gamma=\dot z=K[\cos(\nu+\omega_\star)+e\cos\omega_\star],\qquad
K=\frac{2\pi a_\star\sin i}{P\sqrt{1-e^2}}.
$$
The 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 $\gamma+K(e\cos\omega_\star\pm1)$, 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 $M_\star=1\,M_\odot$, $P=10$ days, $K=50\,\mathrm{m\,s^{-1}}$, $e=0.4$ and $\omega_\star=60^\circ$. Taking one day as $86400\,\mathrm s$ and $G=6.67430\times10^{-11}\,\mathrm{m^3\,kg^{-1}\,s^{-2}}$, the <binary mass function> is $f=1.98270\times10^{20}\,\mathrm{kg}$. Its units check directly: $[PK^3/G]=\mathrm{kg}$. With adopted <solar mass> $M_\odot=1.98847\times10^{30}\,\mathrm{kg}$ and <Jupiter mass> $M_J=1.89813\times10^{27}\,\mathrm{kg}$, solving the exact <binary mass function> gives
|| <Orbital inclination> $i$
|| Companion <mass> $M_p/M_J$
| $90^\circ$
| $0.485930$
| $30^\circ$
| $0.972160$
| $10^\circ$
| $2.802479$
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 $-40\,\mathrm{m\,s^{-1}}$ relative to $\gamma$, and the periapsis value is $+35\,\mathrm{m\,s^{-1}}$. This is a chosen eccentric system, not a reconstruction of 51 Pegasi b.
\Image[/keplerian-radial-velocity-worked-example.png]
{title=A chosen eccentric radial-velocity curve and the exact mass–inclination ambiguity}
{description=The left panel compares eccentric and circular models with the same period and semi-amplitude. Every mass and inclination on the right gives the eccentric model on the left for the stated fixed stellar mass.}
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