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This is some data presented in a lecture on exoplanets that depicts the distribution of the sizes of super-Earths in comparison to the mass of Jupiter.
I would like to know what the argument of the sine function i.e. 'i' implies and why it was multiplied by the mass of Jupiter.


1 Answer 1


If you discover an exoplanet via the Doppler (radial velocity) method, then the amplitude of the radial velocity variations depends on the inclination, $i,$ of the exoplanet's orbital axis with respect to your line of sight. Conventionally, $i=90^{\circ}$ corresponds to viewing an orbit "edge-on", which maximises the velocity variations, while a face-on orbit with $i=0$ would not be detectable.

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This means that you cannot directly estimate the exoplanet mass from your radial velocity data, only $M \sin i$, as plotted on your histogram. $M \sin i$ is a lower limit to the mass.

The $M_{\rm Jup}$ is there to indicate that the axis is labelled in units of Jupiter masses.

  • $\begingroup$ Why do you say that Msini is a lower limit to the mass? Why not the upper, because there could be other planets contributing to this Doppler effect? $\endgroup$ Commented Nov 14, 2021 at 6:43
  • $\begingroup$ Wouldn't a face-on viewing mean that I = 90°, as the plane of the orbit would be making a 90° angle with the line of sight? $\endgroup$ Commented Nov 14, 2021 at 6:45
  • $\begingroup$ $M\sin i$ is a lower limit to $M$ because $\sin i \leq 1$. A face-on orbit has $i=0$; that is how $i$ is defined - the angle between the orbital axis and the line of sight. $\endgroup$
    – ProfRob
    Commented Nov 14, 2021 at 7:18

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