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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.


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? $\endgroup$ Nov 14 '21 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? $\endgroup$ Nov 14 '21 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
    Nov 14 '21 at 7:18

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