Radial Velocity Semi-Amplitude

Use the RV semi-amplitude relation to estimate detectability or recover m sin(i) from observed Doppler wobble.

Worked example — Radial Velocity Semi-Amplitude

RV semi-amplitude K	0.2968 m/s
Minimum mass m sin(i)	1 Earth masses (0.00315 Jupiter masses)
Peak-to-peak RV swing	0.5936 m/s
sin(i)	1
Stellar orbit radius	40.814 km

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Radial velocity can only ever give you a minimum mass 🖖

The wobble tells you the star’s motion along our line of sight, and nothing about its motion across it. So what falls out of the measurement is not the planet’s mass but m sin i, where i is an orbital tilt nobody can read off the same data. Seen edge-on at i = 90° the value is the true mass. At 30° the real planet is twice as heavy as reported; at 10° it is 5.8 times heavier; at 5° it is 11.5 times. A signal announced as Neptune-sized can therefore be a brown dwarf lying nearly face-on, and radial velocity alone can never exclude that. It is exactly why a transit — which only happens when the orbit is close to edge-on, fixing i — turns a minimum mass into a real one.

How a star's wobble reveals a hidden planet 🖖

A planet does not simply orbit its star — both bodies circle their shared center of mass, so the star traces a small loop of its own. As it swings toward us its light is blueshifted; as it recedes, redshifted. K is the top line-of-sight speed of that wobble. Jupiter tugs the Sun at about 12.5 m/s, a fast cyclist's pace, while Earth manages only 0.09 m/s.

The wobble doesn't fade with distance 🖖

Almost every astronomical signal weakens with distance — brightness drops with the square of it. The RV semi-amplitude K does not: it is baked into the orbit itself, so a star 10 or 1000 parsecs away wobbles at exactly the same speed. Distance only limits how many photons you can gather for the spectrum. Instruments like ESPRESSO now reach about 10 cm/s — sensing a star drift slower than a strolling human across trillions of kilometres.

Problem solved in full

Star wobble at 0.2968 m/s from an Earth-mass planet on a 10-day orbit 5 steps
An Earth-mass planet on a 10-day orbit makes its star wobble at 0.2968 m/s. That is walking pace. Work out how far the star actually moves — and why hot Jupiters were found first.
1. Both bodies orbit their common centre of mass, so the star traces a small ellipse of its own. The semi-amplitude K is the speed of that motion projected onto our line of sight, which is all a spectrograph can measure.
  
  $K = (\frac{2 π G}{P})^{1/3} \frac{m _{p} sin i}{( M _{⋆} + m _{p} ) ^{2/3}} \cdot \frac{1}{1 - e ^{2}}$
2. Everything in the formula is known here except the planet: a solar-mass star, an edge-on circular orbit, and one Earth mass at ten days.
  
  $P = 10 d, m_{p} = 1 M_{\oplus}, M_{⋆} = 1 M_{⊙}, i = 9 0^{\circ}, e = 0$
3. The result is 0.2968 m/s. A spectrograph must therefore resolve a Doppler shift of one part in a billion — which is why the field waited on instrument stability rather than on telescope size.
  
  $K = 0.2968 m/s ⟹ peak-to-peak = 0.5936 m/s$
4. The star's orbit is smaller still, and finding it gives a second route to the same speed. Kepler's third law turns ten days around a solar mass into a planetary orbit of 1.359 × 10¹⁰ m, and the star's own circle is smaller than that by the mass ratio, one part in 333,000: 40.8 km, a circle you could drive across, traced by an object 1.4 million kilometres wide. One lap of it in ten days works out at 0.2968 m/s, which is the semi-amplitude reached without the compound formula.
  
  $a_{⋆} = a \frac{m _{p}}{M _{⋆} + m _{p}} = 40.814 km$
5. The scalings explain the discovery order. K falls only as the cube root of the period, so the same planet at one year still gives 0.0895 m/s — but K rises in direct proportion to mass, so a Jupiter at three days gives 141 m/s.
  
  $K \propto P^{- 1/3} : P = 365 d \Rightarrow 0.0895 m/s; 318 M_{\oplus} at 3 d \Rightarrow 141 m/s$
Answer

The tool prints K = 0.2968 m/s, a peak-to-peak swing of 0.5936 and a stellar orbit radius of 40.814 km. The ratio is the whole history of the field: a hot Jupiter produces about 475 times the signal of this Earth, and roughly 1600 times that of an Earth at one year. That is why the first exoplanets found were massive planets on short orbits — not because they are common, but because they are the only ones the instruments of 1995 could see. Set the period to 365 days and watch K fall to the 0.0895 m/s that a real Earth produces.

$K \propto \frac{m _{p}}{M _{⋆}^{2/3}} P^{- 1/3} : Earth moves the Sun 0.3 m/s$

References (1)

The measurement this tool models, and the m·sin i limit it carries: M. Mayor and D. Queloz, "A Jupiter-mass companion to a solar-type star." Nature 378, 355–359, 1995 — 51 Pegasi b, found by exactly this wobble.

By the 42Math team · Updated 2026-07-05

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

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Example problems

Earth-Sun analog - An Earth on a one-year orbit moves the Sun at 0.0894 m/s - under 10 cm/s, right at the limit ESPRESSO reaches. The hardest detection on this page.
Hot Jupiter - A Jupiter on a three-day orbit gives 140.7 m/s, about 1,570 times the Earth-Sun signal. Short periods and heavy planets are why the first exoplanets found were hot Jupiters.
Mini-Neptune - 17 Earth masses on a 12-day orbit around a 0.8-solar star: 5.5 m/s. Sixty times the Earth-Sun signal and twenty-five times smaller than the hot Jupiter - a realistic survey detection.
Infer from measured K - The inverse mode returns m sin i, never m. Tilt the same orbit to 30° and the real planet is twice the mass this reports; at 5° it is 11.5 times.