View Full Version : Determining the Plane of the Orbit of Extrasolar Planets

2009-Feb-01, 07:52 PM
Hello there:

My question is this: many of the extra-solar planets have been discovered by radial velocity measurements of their host star. I would have thought that that the plane that an exo-planet has in relationship to our viewing point would make it difficult to measure the true size/velocity of that planet. For example, wouldn't a large planet that is near 90 degrees to our sight-line have the equivalent "tug" on the host star of a much smaller planet that is viewed nearly edge on?

I've looked around and not seen a good answer for this one, so am very curious about the answer.



P.S. I love listening to Astronomy Cast on my commute into work and have been a listener since episode #1 debuted!

2009-Feb-01, 08:19 PM
I think you are right. But the angle to our sight -line may be determined by observing the star , and taken into account to compute the planet mass.
I found this link which explain how you can determine a star orientation.http://kookaburra.phyast.pitt.edu/hillier/web/rotation.htm
But as you imply , for a near 90 to our sight line the sensibility would be much smaller. You could detecxt only big planets.

grant hutchison
2009-Feb-01, 10:34 PM
You're absolutely right. That's why the mass of the planet is quoted as M.sin(i), where i is the inclination to the plane of the sky. That's what you find listed at [url=http://exoplanet.eu/index.php]The Extrasolar Planet Encyclopaedia[/i], for instance. So the mass value given is the one required to give the observed velocity change in the star if the planet orbited at right angles to the plane of the sky (parallel to the line of sight).
The only extrasolar planets detected by radial velocity changes which have defined masses are those that transit the parent star, so we can measure their orbital inclination to the plane of the sky.

Welcome to BAUT, by the way. :)

Grant Hutchison