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- Thread starter darknessvirtu
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cepheid

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What you wrote, M

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, M

Can you post a link to the article in question?

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What you wrote, M_{2}sin(i) = mass of planet, doesn't make much sense. I think the point is that M_{2}is the true mass of the planet (and presumably M_{1}is the mass of the star it is orbiting). Unfortunately, you can't measure M_{2}. You can only measure M_{2}sin(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, M_{2}sin(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!

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

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cepheid

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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: M

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

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

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

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