1
$\begingroup$

I heard about the radial speed method to discover exoplanets a few years ago, however, these questions kept confusing me. If we know a planet star of mess $m$ is orbiting a star by interpreting the radial speed of the central mass, how can we tell wether this is a even more massive object whose orbiting plane has a larger inclination? Further more, even if we know the angle of our sight and the axis of orbit, can we determine the direction of orbit?

Thanks in advance.

$\endgroup$
1
$\begingroup$

Most of the time you can't determine the inclination of a planet detected by the radial velocity method, but in certain favourable cases the dynamical interactions between two planets in the same system will allow you to constrain the inclination. This has been used to determine the inclination of the resonant planets in the Gliese 876 system, e.g. Rivera et al. (2005).

Otherwise you would need to combine radial velocities with another detection method, for example transits or astrometry.

$\endgroup$
1
$\begingroup$

I assume you are talking about the radial velocity method.

Analysis of a radial velocity curve yields the minimum mass of a planet $M \sin i$, where $M$ is the true planetary mass and $i$ is its orbital inclination (90 degrees would mean the orbital plane is in our line of sight).

In general, with no other information that's all you can say. In principle $M$ could have any value $\geq M\sin i$.

Of course increasingly small values of $\sin i$ are increasingly unlikely and the average $\sin i$ of a randomly oriented population is $\pi/4$.

If you see a transit then you know that $i$ is close to 90 degrees and can in fact estimate $i$ from the transit duration.

If you measure the star's rotation period and can estimate it's projected equatorial velocity from spectral line broadening, then you can estimate it's inclination to the line of sight. One could then assume that the planet orbits in the equatorial plane of the star and use the same inclination for its orbit.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.