10

enter image description here

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.

Ambica Govind
  • 779
  • 3
  • 13

1 Answers1

20

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.

enter image description here source

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.

Daddy Kropotkin
  • 4,523
  • 1
  • 10
  • 33
ProfRob
  • 151,483
  • 9
  • 359
  • 566
  • 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? – lazearoundallday Nov 14 '21 at 06:43
  • 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? – lazearoundallday Nov 14 '21 at 06:45
  • $M\sin i$ 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. – ProfRob Nov 14 '21 at 07:18