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celestial-mechanics:analysis-of-orbit-equation [2016/05/25 21:52]
trinh
celestial-mechanics:analysis-of-orbit-equation [2016/05/26 00:49] (current)
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 ====== Analysis of the orbit equation ====== ====== Analysis of the orbit equation ======
  
 +Previously we found the orbit equation, which describes the distance of a second mass from the first mass located at the origin of a shifted coordinate system. The equation was,
 +\begin{equation}
 + r = \frac{h^2}{\mu} \frac{1}{1 + e\cos\theta},​
 +\end{equation}
 +where $\mu$ is the gravitational parameter, and $h$ is the magnitude of 
 +\begin{equation}
 + \mb{h} = \mb{r} \times \dot{\mb{r}}
 +\end{equation}
 +i.e. the relative angular momentum of $m_2$ per unit mass. The parameter $e$ is the eccentricity and $\theta$ is the true anomaly (the angle between the vector $\mb{e}$ and the position vector $\mb{r}$. ​
 +
 +  * Note there are six constants of integration from the original Newton'​s laws (3 x 2 dimensions). In the orbit equation, we involve $\mb{h}$ (three) plus $\mb{e}$. However, $\mb{h}$ is perpendicular to $\mb{e}$ which removes one. To recover the sixth constant, we have to bring time back into the picture. ​
 +
 +  * The direction along $\mb{e}$ is the **apse** line. Note that from orbit equation, $m_2$ is closest to $m_1$ when $\theta = 0$. This marks the **periapsis** point. ​
 +
 +  * Separating velocity into components parallel and perpendicular to the position vector, we can show that the radial velocity is 
 +  \begin{equation}
 +  v_r = \frac{\mu}{h} e \sin\theta.
 +  \end{equation}
 +
 +  * We can write down a local energy at a point in the orbit, given by sum of kinetic and potential energies. It must be constant, and is:
 +  \begin{equation}
 +  \varepsilon = \frac{v^2}{2} - \frac{\mu}{r}
 +  \end{equation}
 +
 +==== Circular orbits ($e = 0$) ====
 +
 +Setting $e = 0$ gives 
 +\begin{equation}
 + r = \frac{h^2}{\mu}
 +\end{equation}
 +and hence circular orbits. Velocity along the orbit is entirely angular, and since $h = rv_\bot$, then solving for $h$ and combining with above gives
 +\begin{equation}
 + v_\bot = \sqrt{\mu/​r}.
 +\end{equation}
 +
 +Thus one period of an orbit is given by 
 +\begin{equation}
 + T = \frac{2\pi r}{\sqrt{\mu/​r}} = \frac{2\pi}{\sqrt{\mu}} r^{3/2}.
 +\end{equation}
 +
 +==== Elliptical orbits ($0 < e < 1$) ====
 +
 +**Periapsis** (P) is minimum point of orbit from $m_1$ at $\theta = 0$. **Apoapsis** (A) is the maximum point of orbit at $\theta = \pi$. The mass $m_1$ lies at a focal point, $F$. After work, you can show that (with $x$ directed down the apse line):
 +\begin{equation}
 + \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1,
 +\end{equation}
 +where
 +\begin{equation}
 + a = \frac{h^2}{\mu} \frac{1}{1 - e^2} \qquad b = a\sqrt{1 - e^2}.
 +\end{equation}
 +
 +Can also show that the period is 
 +\begin{equation}
 + T = \frac{2\pi}{\mu} a^{3/2},
 +\end{equation}
 +that is, the period is independent of the eccentricity of the orbit. This yields Kepler'​s Third Law.
 +
 +  * **Kepler'​s Third Law**: the period of a planet orbit is proportional to $a^{3/2}$ where $a$ is the semimajor axis. 
 +
 +==== Parabolic trajectories ($e = 1$) ====  ​
 +
 +When $e = 1$, we can show that (with $x$ down the apse line):
 +\begin{equation}
 + x = \frac{p}{2} - \frac{y^2}{2p},​
 +\end{equation}
 +where $p = h^2/\mu$ is the **parameter of the orbit**.
 +
 +==== Hyperbolic trajectories ($e > 1$) ====  ​
 +
 +For $e > 1$, we can show that 
 +\begin{equation}
 + \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1,
 +\end{equation}
 +where 
 +\begin{equation}
 + a = \frac{h^2}{\mu} \frac{1}{e^2 - 1} \qquad b = a \sqrt{e^2 - 1}.
 +\end{equation}
 +
 +==== Summary ====  ​
 +
 +Summary from p.104 of "​Orbital mechanics for engineering students"​ (Curtis)
 +
 +{{ :​celestial-mechanics:​curtis_summary.jpg?​direct&​500 |From Curtis}}