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:


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:


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


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?


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)?


1 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.

  • $\begingroup$ This is great, thanks for your efforts! Indeed, I actually meant that $\beta$ is the angle between the horizontal component of the velocity vector and the thrust vector. $\endgroup$
    – woeterb
    Oct 12, 2017 at 8:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.