This shows you the differences between two versions of the page.

Both sides previous revision Previous revision | |||

celestial-mechanics:analysis-of-orbit-equation [2016/05/25 20:52] trinh |
celestial-mechanics:analysis-of-orbit-equation [2016/05/25 23:49] (current) |
||
---|---|---|---|

Line 1: | Line 1: | ||

====== 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}} | ||