How to derive the polar form of the equations of motion?

Question

As part of an optimal control problem (see linked problem), I need the polar form of the equations of motion (EOM) defining the orbit of a spacecraft. In e.g. Bryson and Ho (1969) and Vallado (2007) the following EOM are presented:

$$\dot{r}=v_r$$ $$\dot{\theta}=\frac{v_{\theta}}{r}$$ $$\dot{v}_r=\frac{v_{\theta}^2}{r}-\frac{\mu}{r^2}+\frac{T\sin{\beta}}{m_0-|\dot{m}|t}$$ $$\dot{v}_{\theta}=-\frac{v_{r}v_{\theta}}{r}+\frac{T\cos{\beta}}{m_0-|\dot{m}|t}$$

where $r$ is the radial distance of the spacecraft from the attracting center, $v_r$ is the radial-velocity component, $v_{\theta}$ is the tangential/transverse-velocity component, $\mu$ is the gravitational constant of the attracting center, $T$ is thrust (constant), $\beta$ is the in-plane control angle (measured from the local horizontal to the thrust vector), $m$ is the mass of the spacecraft and:

$$\dot{m}=-\frac{T}{I_{sp}g_0}$$

is the (constant) fuel consumption rate or mass flow rate in $kg/s$, with $I_{sp}$ being the (constant) specific impulse and $g_0$ the standard gravity (Earth).

Unfortunately the full derivations for the latter two equations, radial ($\dot{v}_{r}$) and tangential acceleration ($\dot{v}_{\theta}$), are not given. I don't necessarily need these derivations for my calculations, but I just like to know how they were done.

The equation for the radial acceleration ($\dot{v}_{r}$) can be derived by starting with:

$$\ddot{r}-r\dot{\theta}^2=-\frac{\mu}{r^2}+\frac{T\sin{\beta}}{m_0-|\dot{m}|t}$$

However for the derivation of the equation for tangential acceleration ($\dot{v}_{\theta}$), I'm not completely sure where to start. Probably with:

$$r\ddot{\theta}+2\dot{r}\dot{\theta}=\frac{T\cos{\beta}}{m_0-|\dot{m}|t}$$

Is this correct or am I missing a term? Does anyone have an idea how derive the equation for $\dot{v}_{\theta}$ as stated at the top of this post?

Thanks!

Extra edit: Maybe the product rule offers a solution here?

$$[r\dot{\theta}]^{'}=\dot{r}\dot{\theta}+r\ddot{\theta}$$ $$r\ddot{\theta}=[r\dot{\theta}]^{'}-\dot{r}\dot{\theta}$$ $$r\ddot{\theta}=[v_{\theta}]^{'}-\dot{r}\dot{\theta}$$ $$r\ddot{\theta}=\dot{v}_{\theta}-\dot{r}\dot{\theta}$$

Substituting this in the equation above would yield the same result presented in the literature. But if this is true, can someone explain me why $r\ddot{\theta}\neq\dot{v}_{\theta}$ (see e.g. Wikipedia)?

Answer 1

There's probably a simpler way to do it, but here is how you can obtain these equations from the basics. Let's start with equations in Cartesian coordinates: \begin{align} \dot{x} & = v_x;\\ \dot{y} & = v_y;\\ \dot{v_x} & = \frac{F_x}{m};\\ \dot{v_y} & = \frac{F_y}{m}. \end{align} The relations between Cartesian and the polar coordinates are \begin{align} x & = r\cos\theta;\\ y & = r\sin\theta, \end{align} or \begin{align} r & = \sqrt{x^2 + y^2};\\ \theta & = \arctan \frac yx. \end{align} (Actually, this equation for $\theta$ is valid only if $x>0$, but let's assume that it is so. The final result is the same anyway.)

Besides, in the polar coordinates, instead of $v_x$ and $v_y$, we would like to use $v_r$ and $v_\theta$, the projections of the velocity on the vertical and the horizontal axes. These axes are rotated by angle $\theta$ relative to $x$- and $y$-axis, so \begin{align} v_r & = v_x\cos\theta + v_y\sin\theta;\\ v_\theta & = v_y\cos\theta - v_x\sin\theta. \end{align}

Similarly, instead of $F_x$ and $F_y$, we would rather use the vertical and the horizontal components of the force: \begin{align} F_r & = F_x\cos\theta + F_y\sin\theta;\\ F_\theta & = F_y\cos\theta - F_x\sin\theta. \end{align}

Taking the derivative of the expressions for $r$ and $\theta$ by time, we get \begin{align} \dot{r} & = \frac{2\dot{x}x + 2\dot{y}y}{2\sqrt{x^2 + y^2}} = \frac{v_x r\cos\theta + v_y r\sin\theta}{r} = v_x\cos\theta + v_y\sin\theta = v_r;\\ \dot{\theta} & = \frac{\dot{y}x - \dot{x}y}{x^2}\cdot\frac{1}{1+(y/x)^2} = \frac{v_y\cos\theta - v_x\sin\theta}{r} = \frac{v_\theta}{r}. \end{align}

Taking the derivative of the expressions for $v_r$ and $v_\theta$ by time, we get \begin{align} \dot{v_r} & = \dot{v_x}\cos\theta - v_x\dot{\theta}\sin\theta + \dot{v_y}\sin\theta + v_y\dot{\theta}\cos\theta = \frac{F_x}{m}\cos\theta + \frac{F_y}{m}\sin\theta + \dot{\theta}(v_y\cos\theta - v_x\sin\theta) = \frac{F_r}{m} + \frac{v_\theta^2}{r};\\ \dot{v_\theta} & = \dot{v_y}\cos\theta - v_y\dot{\theta}\sin\theta - \dot{v_x}\sin\theta - v_x\dot{\theta}\cos\theta = \frac{F_y}{m}\cos\theta - \frac{F_x}{m}\sin\theta - \dot{\theta}(v_y\sin\theta + v_x\cos\theta) = \frac{F_\theta}{m} -\frac{v_\theta v_r}{r}. \end{align}

If we leave out the thrust for now, then $F_r = -\frac{\mu m}{r^2}$, $F_\theta = 0$, and substituting these values, we get \begin{align} \dot{v_r} &= -\frac{\mu}{r^2} + \frac{v_\theta^2}{r};\\ \dot{v_\theta} & = -\frac{v_\theta v_r}{r}. \end{align}

You say that $\beta$ is the angle between the velocity vector and the thrust vector; however, to obtain the formulas you have given we need to assume that $\beta$ is the angle between the horizontal positive direction and the thrust vector. Then $F_r = -\frac{\mu m}{r^2} + T\sin\beta$, $F_\theta = T\cos\beta$, and (substituting $m = m_0 - |\dot{m}|t$) we get the formulas in the original post.