# Differences

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

 celestial-mechanics:lagrange-coefficients-and-orbit-equation [2016/05/26 01:25]trinh created celestial-mechanics:lagrange-coefficients-and-orbit-equation [2016/05/26 01:44] (current) 2016/05/26 01:25 trinh created 2016/05/26 01:25 trinh created Line 1: Line 1: ====== Solution of the orbit equation from initial conditions ====== ====== Solution of the orbit equation from initial conditions ====== + + How is the orbit equation solved given initial position and velocity? ​ + + First, we create a coordinate system through the perifocal frame. Let $(\bar{x},​\bar{y},​\bar{z})$ be a new coordinate system where $\bar{x}$ is directed through the periapsis. Let $\bar{y}$ be directed at 90 degrees true anomaly (so in the plane of the orbit), and $\bar{z}$ taken perpendicular to both. + + We then have + \begin{gather} + \mb{r} = \bar{x} \hat{p} + \bar{y} \hat{q} \\ + \mb{v} = \dot{\bar{x}} \hat{p} + \dot{\bar{y}}\hat{q} ​ + \end{gather} + for unit vectors in the plane. ​ + + It can be shown that, given initial conditions, $\mb{r}_0$ and $\mb{v}_0$, ​ + \begin{gather} + \mb{r} = f \mb{r}_0 + g \mb{v}_0 \\ + \mb{v} = \dot{f} \mb{r}_0 + \dot{g} \mb{v}_0 ​ + \end{gather} + where + + f = \frac{\bar{x}\dot{\bar{y_0}} - \bar{y}\dot{\bar{x_0}}}{h} \qquad ​ + g = \frac{-\bar{x}\bar{y_0} + \bar{y}\bar{x_0}}{h} + + and $h$ is the constant magnitude of the angular momentum, now related by + + h = \bar{x_0}\dot{\bar{y_0}} - \bar{y_0}\dot{\bar{x_0}}. + + + Basically, after a lot of work, you can show that $f$ and $g$ and their derivatives can instead be written in terms of + + f = f(\Delta \theta; r, \mb{r}_0, \mb{v}_0) \qquad ​ + g = g(\Delta \theta; r, \mb{r}_0, \mb{v}_0), + + where $r$ is known from the solution of the orbit equation, and $\Delta \theta$ is the desired change in the true anomaly from its initial value.