A Jupiter-mass companion to a solar-type star

[darknessvirtu]

Hi, I'm a biology student...with barrrrely any physics knowledge. However I have to write a report on this article, A Jupiter-mass companion to a solar-type star. and in it, the author finds the upper limit to the mass of the planet, but i read that using the radial velocity method only the lower limit can be found. can anybody please explain this? also in the equation M2sini=mass of planet, what is M2? thanks so much!

 

[cepheid]

Welcome to PF darknessvirtu!

What you wrote, M₂sin(i) = mass of planet, doesn't make much sense. I think the point is that M₂ is the true mass of the planet (and presumably M₁ is the mass of the star it is orbiting). Unfortunately, you can't measure M₂. You can only measure M₂sin(i) (from Kepler's Third Law). If you don't know the inclination angle of the system, then this sine factor is unknown.

If the system is perfectly face-on (i = 0), then sin(i) = 0 and there is no radial velocity to measure, and you have no data.

If the system is perfectly edge-on (i = 90°), then sin(i) = 1, and the measurement you are getting is the true mass of the planet.

The truth is probably somewhere in between. The orbital plane of the system is probably inclined at some angle between 0° and 90°, meaning that 0 < sin(i) < 1. Therefore, your best estimate of the mass, M₂sin(i), is LESS THAN the true mass. So the only thing that you can say with certainty is that the mass of the planet must be AT LEAST equal to your estimate, but that it is probably larger. Therefore, I agree that what you compute is a LOWER limit on the mass of the planet.

Can you post a link to the article in question?

 

[darknessvirtu]

Welcome to PF darknessvirtu!

What you wrote, M₂sin(i) = mass of planet, doesn't make much sense. I think the point is that M₂ is the true mass of the planet (and presumably M₁ is the mass of the star it is orbiting). Unfortunately, you can't measure M₂. You can only measure M₂sin(i) (from Kepler's Third Law). If you don't know the inclination angle of the system, then this sine factor is unknown.

If the system is perfectly face-on (i = 0), then sin(i) = 0 and there is no radial velocity to measure, and you have no data.

If the system is perfectly edge-on (i = 90°), then sin(i) = 1, and the measurement you are getting is the true mass of the planet.

The truth is probably somewhere in between. The orbital plane of the system is probably inclined at some angle between 0° and 90°, meaning that 0 < sin(i) < 1. Therefore, your best estimate of the mass, M₂sin(i), is LESS THAN the true mass. So the only thing that you can say with certainty is that the mass of the planet must be AT LEAST equal to your estimate, but that it is probably larger. Therefore, I agree that what you compute is a LOWER limit on the mass of the planet.

Can you post a link to the article in question?

hey, thanks for the reply, here's a link to the article

http://www.nature.com/nature/journal/v378/n6555/abs/378355a0.html

in the article the author calculates the upper limit by speculating that the observed projected rotational velocity is equivalent to the equitorial velocity, is there anything wrong with doing that?

also, i was wondering if anybody knew what the alpha sign stood for under the section "Jupiter or stripped brown dwarf" i can't seem to find it anywhere :P, but thanks again, I'm slowly starting to understand the math involved in this paper..and it's pretty exciting!

 

[Nik_2213]

If you're lucky and the planet transits its sun, then that constrains Sin(i). As a lot of new extrasolar planetary candidates are being discovered by transit method, it makes estimating their mass easier...

( Uh, what's the proportion of planets that may transit ? ~10% ?? )

 

[cepheid]

darknessvirtu,

Sorry for not getting back to you sooner. I'm reading the Nature paper now. I don't have a subscription to Nature, so I had to figure out how to connect to my University's virtual private network so that I could read it for free. I'm not complaining though. It's pretty interesting reading the paper about the first exoplanet discovered around a sun-like star, and one of the first examples of a class of planets that came to be known as "Hot Jupiters." The field has taken off so much in the past 15 years.

Anyway, we were right that the radial velocity method was used to find a *lower* limit on the mass, which is the first equation: M₂sin(i) = 0.47 M_J (Jupiter masses). In the next section, the authors point out that even if you consider the range of all possible masses that can be inferred from this number, the object still has to be much less massive than a star, or even a typical brown dwarf. The argument goes like this: the fact that we're even seeing any part of the orbital motion in the radial (line of sight) direction suggests that the orbital plane is probably close to edge on. They expand on that with a simple statistical argument - if you consider the probability distribution of the inclination angles (which range between 0 and 90 degrees), you'll find that it's very unlikely that the planet is more massive than 4 M_J. Here's the basic idea:

If Msin(i) = 0.47, then M = 0.47 / sin(i)

If the system is perfectly edge-on, then sin(i) = 1, and 0.47 is the true mass. The less inclined the system gets (the smaller i becomes), the smaller sin(i) becomes, and the larger the true mass is. So, the statistical argument can be used to say that it's unlikely the orbital inclination is less than a certain amount, which leads directly to a statement that it is unlikely the mass greater than the corresponding amount. For example: 4 M_J corresponds to sin(i) = 0.1175, which corresponds to i = 6.75 degrees. So the authors are claiming that there's a less than 1% chance that i ≤ 6.75 deg. Now, in making this argument, I don't think the authors are assuming that all orbital inclinations are equally likely, which would be a uniform probability distribution. (I get more like 7% of the inclinations being < 6.75 degrees if I do that). It could be that the authors are assuming that the probability distribution of the orbital inclinations is shaped like a bell curve, centred on some mean (average) inclination. This is called a Gaussian probability distribution. Inclinations closer to the mean are therefore more likely than inclinations very far away from the mean.

If we could just get a handle on what sin(i) actually is, then we place a more stringent limit on the mass of the companion (rather than just talking about likelihood). Later on in that section, the authors use the properties of the parent star itself to do just that. They talk about the star's rotation (spin), saying that it is likely that the star's spin axis is perpendicular to the orbital plane of the planetary companion. In other words, the star's equatorial plane is the same as the planet's orbital plane. This is a reasonable assumption in the typical formation scenario in which everything that orbits the star was formed in a flattened disk of material that was originally accreting (gathering) onto that star. Anyway, what this means is that the spin speed V, is projected onto the sky by the same geometric factor as the orbital speed. In other words, the portion of the spin speed that we measure to be in the radial direction (as a Doppler shift) is equal to Vsin(i). Now it gets kind of complicated -- the authors use theoretical arguments, combined with observations of the star's chromosphere, to get an idea of what "V" is, independently of the measurement of Vsin(i) coming from the spectroscopy. The chromosphere is a specific layer of the star's atmosphere, and most of the "activity" in this region is driven by the star's magnetic field, which accelerates charged particles in that region. Now, a star's magnetic field comes from its rotation (this is the idea of the magnetic "dynamo" that they refer to). So, the faster the spin, the stronger the magnetic field, and the more chromospheric activity you expect to see. Based on that, and what we know about G type stars, they determine V (to a certain degree of observational uncertainty, of course). Since they have estimates of both V and Vsin(i), they can solve for sin(i). Doing just that is what gives them the value of 1.2 (or at most 2) M_J for the planet's mass.

I'll have to read more later, and get back to you.