0
$\begingroup$

So, I am making an essay explaining planet-hunting techniques. And while I do get how astronomers measure the semi-major axis of a planetary orbit (using Kepler Laws), I don't get how they measure the planet's mass.

$\endgroup$
3
$\begingroup$

The planet and the star both rotate around their center of mass. That's how star "wobbling" occurs. When the star in our line-of-sight moves away from us, photons reaching us are red-shifted and when the star moves towards us, its photons are blue-shifted - this is the so-called Doppler effect. The star mass $M$ is extracted from its spectrum, luminosity and other parameters. Then over the years, astronomers measure the radial velocity of star, which is extrapolated from the Doppler effect.

In the graph, the highest radial velocity of the star is marked as $K$. From the same observations, the wobbling period $P$ is also detected. This wobbling period is the time duration between adjacent maximum blue-shifts (or between maximum red-shifts). In essence, astronomers compile such a star radial velocity graph:

enter image description here

Given this data, one can calculate exoplanet mass (given that I haven't made any formulas substitution error):

$$ m = \frac{M^{2/3}K}{\sin(i)} \left(\frac{P}{2\pi G}\right)^{1/3} $$

where $i$ is inclination - angle between reference plane and orbital plane.

EDIT

The equation above is suitable only if the exoplanet orbit is circular. This is rarely the case. According to this article, ones needs to take into account orbit eccentricity when it is elliptic. We need to make yet another assumption, that $ \frac{m}{M} \ll 1$, for example $ m_{earth}/M_{sun} \approx 3.004×10^{-6} $. ("$\ll$" means A LOT smaller than). Assuming that, planet mass can be reduced to:

$$ m = \frac{M^{2/3}K\sqrt{1-e^2}}{\sin(i)} \left(\frac{P}{2\pi G}\right)^{1/3} $$

where $e$ is orbital eccentricity

$\endgroup$
5
  • $\begingroup$ Thanks for the answer. What equations did you use to arrive at this result? (Assume that the orbital plane of the planet is edge-on; that is, $i = \pi / 2$). $\endgroup$ – Heath Sep 19 '19 at 14:37
  • 1
    $\begingroup$ @UnstoppableTachyon You are wellcome (up-vote, huh ?). The ones mentioned in Doppler spectroscopy page. These calculations are just as an example. Because they makes an assumption, that planet rotates around star in circular orbit. Albeit in practice, planet orbits never are circular, but elliptic instead $ 0 < eccentricity < 1$. However this example is very nice to understand overall process $\endgroup$ – Agnius Vasiliauskas Sep 19 '19 at 15:54
  • $\begingroup$ Ok. But how does this formula change for elliptical orbits? $\endgroup$ – Heath Sep 19 '19 at 20:25
  • 1
    $\begingroup$ Actually, I found a link that contains the full correct equation. home.strw.leidenuniv.nl/~keller/Teaching/ADA_2011/… $\endgroup$ – Heath Sep 19 '19 at 20:28
  • 1
    $\begingroup$ @UnstoppableTachyon I fixed formula for elliptic orbits. However I made some assumption, that $ m \ll M$, only then $m + M \approx M$, so that we could solve given equation in your reference for a planet mass. Otherwise solution will be very complex $\endgroup$ – Agnius Vasiliauskas Sep 20 '19 at 7:59
-2
$\begingroup$

I am not an astronomer, but I would measure the wobbling of the surounded star, the mass of the star known by its spectrum.

$\endgroup$
0

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.