Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I have this problem: They give me, from a satellite that is in orbit in earth, a value for the period, and the closest height to earth surface, the ask me what the eccentricty of the orbit is. I have no idea how to do this. I've tried using Binet's equation, and the equation that comes for these movements ($p/r=1+e\cos(\theta-\theta_0)$) or the conservation of angular momentum to try to get some relations between quantities, but I can't get anything.

For example, the angular momentum is conserved, so: $$l=mr^2\dot \phi$$ I can get from here: $$\int_0^T\frac{l}{mr^2}dt=2\pi$$ Being T the period, but I don't know $r(t)$ Or the other way: $$T\frac{l}{m}=\int_0^{2\pi}r(\phi)d\phi$$ Now I now the function $r(\phi)=\frac{p}{1+e\cos\phi}$, where $p=\frac{l^2}{\mu k}$, being $l$ the angular momentum and $k$ the constant of the gravity potential: $U=-k/r$. I don't know how to integrate that if it's possible or how to use the known value of the closest point. Some help?

share|improve this question
1  
Hi MyUserIsThis - remember, this is a site for conceptual questions about physics, so can you edit your question to ask about the specific physics concept that is giving you trouble? It's very close, since you've already listed some things you tried, but what problems exactly did you run into when you tried to get relations between the quantities? In other words, why couldn't you get anything? Once you put some of that detail in I'll be happy to reopen this. –  David Z Jan 27 '13 at 22:14
    
@DavidZaslavsky Done –  MyUserIsThis Jan 27 '13 at 22:21
add comment

1 Answer

up vote 2 down vote accepted

You're trying all of the right things, but the problem is actually much simpler---you just need to use better relations for this problem. In particular, think about kepler's law of orbital periods; and some of the general ellipse equations, like:

$r(\theta) = \frac{ a \, (1 \, - \,e^2) }{1 \, - \, e \cos \theta}$ (which is the equation you have, just expressed a little differently), and

$e = \frac{ r_{max} \, - \, r_{min} }{r_{max} \, + \, r_{min}} = \frac{r_{max} \, - \, r_{min} }{2a}$

Here, $a$ is the 'semi-major axis', $e$ the eccentricity, $\theta$ the angle of the orbit, and $r_{max}$ and $r_{min}$ are the maximum and minimum separations respectively (also referred to as 'apocenter' and 'pericenter').
Does that help?

share|improve this answer
    
Thanks for the help, I'm now doing $r(\theta)=R_{max}$, substuting $\theta=0$, the point of the maximum of that function, and getting $e$ out of the equation, I have the rest of the values, but I'm getting a result for $e=3\cdot 10^6$. That's imposible. (I found a mistake in my calculations, let me revise them) –  MyUserIsThis Jan 27 '13 at 22:50
    
Great, still a step in the right direction. How did you find the value of $a$? –  zhermes Jan 27 '13 at 22:52
    
I'm stupid, I confused your $a$ with the $a$ of the wiki page you sent me, so I find your $a$, which will be the max value of the function, right? –  MyUserIsThis Jan 27 '13 at 22:53
1  
Not quite, $a$ is the semi-major axis, which is half of the widest axis of the ellipse. For eccentricity of zero (i.e. a circle), the semi-major axis is the same as the radius. The semi-major axis is always somewhere inbetween the minimum and maximum separations (the pericenter and apocenter distances). –  zhermes Jan 27 '13 at 22:59
1  
It is true that $r_{max} = r(\theta = \pi)$, but that is not the same as the semi-major axis. $r_{max} + r_{min} = 2a$ --- which you can see in the second equation In my answer. It'll be much more clear after some sleep ;) –  zhermes Jan 27 '13 at 23:00
show 4 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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