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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) Both sides previous revision Previous revision 2016/05/25 21:52 trinh 2016/05/25 21:51 trinh created 2016/05/25 21:52 trinh 2016/05/25 21:51 trinh created 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, + + r = \frac{h^2}{\mu} \frac{1}{1 + e\cos\theta},​ + + where $\mu$ is the gravitational parameter, and $h$ is the magnitude of + + \mb{h} = \mb{r} \times \dot{\mb{r}} + + 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 + + v_r = \frac{\mu}{h} e \sin\theta. + + + * 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: + + \varepsilon = \frac{v^2}{2} - \frac{\mu}{r} + + + ==== Circular orbits ($e = 0$) ==== + + Setting $e = 0$ gives + + r = \frac{h^2}{\mu} + + 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 + + v_\bot = \sqrt{\mu/​r}. + + + Thus one period of an orbit is given by + + T = \frac{2\pi r}{\sqrt{\mu/​r}} = \frac{2\pi}{\sqrt{\mu}} r^{3/2}. + + + ==== 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): + + \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, + + where + + a = \frac{h^2}{\mu} \frac{1}{1 - e^2} \qquad b = a\sqrt{1 - e^2}. + + + Can also show that the period is + + T = \frac{2\pi}{\mu} a^{3/2}, + + 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): + + x = \frac{p}{2} - \frac{y^2}{2p},​ + + where $p = h^2/\mu$ is the **parameter of the orbit**. + + ==== Hyperbolic trajectories ($e > 1$) ====  ​ + + For $e > 1$, we can show that + + \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1, + + where + + a = \frac{h^2}{\mu} \frac{1}{e^2 - 1} \qquad b = a \sqrt{e^2 - 1}. + + + ==== Summary ====  ​ + + Summary from p.104 of "​Orbital mechanics for engineering students"​ (Curtis) + + {{ :​celestial-mechanics:​curtis_summary.jpg?​direct&​500 |From Curtis}}