Why does the Doppler method for detecting exoplanets only give the minimum mass for the exoplanet and not the actual mass?

Is it to do with whether the exoplanet is a face on orbit or a side on orbit?


1 Answer 1


If the exoplanetary orbital plane is "face-on", then the velocity of the orbited host star, as viewed from Earth, will not change at all. If it is viewed "side-on", then the amplitude of the velocity variations will be equal to the orbital speed of the host star around the system centre of mass. At angles in between then the velocity amplitude gives a component of the host star speed, which is usually labelled $v_r \sin i$, where $i$ is the "inclination" of the orbital plane (zero for face-on, and 90 degrees for side-on).

Armed with this velocity amplitude one can estimate the mass of the exoplanet using Kepler's third law, which when rearranged to use the system properties that can be measured, yields $$ \frac{(m \sin i)^3}{(m+M)^2} = \frac{P}{2\pi G} (v_r \sin i)^3,$$ where $m$ is the exoplanet mass, $M$ is the stellar mass (must be estimated somehow) and $P$ is the orbital period (which can also be estimated from the doppler velocity variation).

The things on the RHS of this equation are measured and allow you to infer what is on the LHS. But as you can see, the LHS still contains the unknown inclination angle. Since $\sin i \leq 1$, this means any estimate of $m$ is a lower limit.


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