# More exoplanet equation problems :/

1. Mar 18, 2005

### Dracovich

Hey guys, i made a thread a couple of weeks ago about how to derive equations for exoplanet data, and so far i´ve managed to derive an equation for the radial velocity as:

$$V_{rad}=K[cos(\theta + \omega) + ecos(\theta + \omega)$$

Which i'm guessing would be the function to be plotting against the radial velocity data i get, although i have not been able to derive how $$\theta$$ is a function of time yet. But that's not what i'm trying to derive at right now. Right now i'm having troubles finding a equation to give me the mass of the planet from the given data i get from the plot. I'm guessing i get $$\omega, e, P,a$$ and the mass of the sun is already known, where e is eccentricity, P the period and a the semi-major axis.

This is pretty much where i'm stuck, i thought today i'd found a way. Since $$K=\frac{2\Piasin(i)}{P(1-e^2)^{1/2}}$$ and a can be written as $$a=\frac{L^2}{(1-e^2)GMm^2}$$, then i saw a formula (must admit that i have not trie deriving this formula yet, i just wanted to see if i could use it to get writ of L since that's a crossproduct and i wanted something with known constants insted of L), $$L=\frac{2\Pia^2(1-e^2)^{1/2}mM}{P(m+M)}$$, so i was thrilled and figured i could now write a as $$a=\frac{4 \pi^2 a^4M)}{GP^2(M+m)}$$ and could substitute that into my K equation to get $$K=\frac{8 \pi^3 a^4Msin(i)}{P^3G(M+m)^2(1-e^2)^{1/2}}$$ From which i could simply isolate m and get a nice equation for my mass, but this doesn't seem to give right results (if i just take data from exoplanets.org and put it in it does not agree with what they get), although checking the units i do seem to end up with Kg's.

2. Mar 18, 2005

### SpaceTiger

Staff Emeritus
You seem to have made two mistakes that made the units work out. According to your substitution, $$(M+m)$$ should not be squared. Further, the thing you substituted:

$$a=\frac{4\pi^2 a^4M}{GP^2(M+m)}$$

is just a convoluted version of Kepler's Law with an extra M in the numerator. You should solve for a:

$$a=(\frac{P^2G(M+m)}{4\pi^2})^{1/3}$$

and then plug it into K so that you have it in terms of observables. You can get the period from (obviously) the period of the oscillations and eccentricity from the shape of them. As said before, the mass comes from the spectral type of the star. You can't determine the inclination from radial velocity measurements alone, so you'll have to settle for finding $$msin(i)$$.



Last edited: Mar 18, 2005
3. Mar 18, 2005

### Dracovich

D'oh! Don't know why i didn't think of using that substitution, much more straightforward, and i just derived the elliptical orbit yesterday lol, oh well. But i still seem to be having trouble isolating m, i'll go through my algebra step by step so you can see where i do something wrong if you have the time :)

$$K=\frac{2 \pi asin(i)}{P(1-e^2)^{1/2}}$$

$$a=((\frac{P^2G(M+m)}{4\pi^2})^{1/3}$$

$$K=\frac{2 \pi sin(i)}{P(1-e^2)^{1/2}}(\frac{P^2G(M+m)}{4\pi^2})^{1/3}$$

The only way i see to get m out of this is to put everything to the power of three, which becomes reasonably pretty:

$$K^3=\frac{8 \pi^3 sin^3(i)}{P^3(1-e)^{3/2}}\frac{P^2G(M+m)}{4\pi^2}$$
$$K^3=\frac{2 \pi sin^3(i) G(M+m)}{P(1-e^2)^{3/2}}$$

Then moving everything over except sinus and the masses:

$$\frac{PK^3(1-e^2)^{3/2}}{2 \pi g}=sin^3(i)(M+m)$$

So the only solution i can see would be

$$\frac{PK^3(1-e^2)^{3/2}}{2 \pi g}-sin^3M(i)=sin^3(i)m$$

But that can't be right, since #1 i have sin in the third power, and the equation basicly has one huge number in the M part, and the other part becomes a lot smaller (haven't checked the units in the second part yet since it doesn't look right at all).

4. Mar 18, 2005

### SpaceTiger

Staff Emeritus
I think that the "a" in your K equation should actually be:

$$a_*=\frac{m}{M}a$$

where a is from Kepler's law (the star-planet distance), and a_* is the radius of the tiny orbit undergone by the star (as a result of the planet's influence). Once you've made this substitution, you can solve for msin(i) (setting $$m+M \sim M$$).

Last edited: Mar 18, 2005
5. Mar 18, 2005

### Dracovich

Ahh yes i remember reading about that, that $$a_1 m_1=a_2 m_2$$. I wondered about taht since our data is from the star, the semi-major axis would be the semi major axis for the stars orbit around the center of mass, not the planets orbit right?

Anyway :) I got it now, thanks a bunch, i really appreciate it.