I created a rocket trajectory simulator and optimizer using the following equations of motion for a non-rotating frame of reference

$$\dot{v} = \frac{T \cos \varepsilon - D}{m} - g \sin \gamma$$

$$v \dot{\gamma} = \frac{T \sin \varepsilon + L}{m} - \left(g - \frac{v^2}{r} \right) \cos \gamma$$

(screeshot to be deleted as soon as MathJax is complete)

enter image description here

where $v$ is velocity, $\gamma$ is flight path angle, $h$ is altitude, $r$ is distance to planet center, $\theta$ is angle along a great circle path of the Earth, $m$ is mass, $T$ is thrust, $D$ is drag, $L$ is lift and $\varepsilon$ is angle of attack.

I would like to do this for a rotating planet in 3-D, in an inertial frame of reference. I think it is simpler than having to include all the non-inertial acceleration terms that come with a rotating frame of reference. But to compute the aerodynamic forces you would need to have the relative velocity between the rocket and atmosphere, atmosphere rotating fixed to Earth.

I can do that, but I am wondering if it's a bad idea. I am thinking there would be numerical problems especially at the start of the trajectory, where you are taking the difference of two vectors that are relatively close in magnitude - the velocity of the rocket and that of the atmosphere, both in the inertial frame. Subtracting two big things close in magnitude generally gives you an inaccurate result. Maybe the accuracy is good enough that it doesn't matter, but was curious if anyone had experience trying this.

  • $\begingroup$ Can you update your question and rewrite your equations in MathJax? Screen shots of equations are discouraged here. I'll do the first two to get you started, Welcome to Space SE! $\endgroup$
    – uhoh
    Jan 21, 2021 at 10:15

1 Answer 1


For the aerodynamic calculations only, you should convert the rocket's position, velocity, etc. from inertial to rotating-surface-relative coordinates before doing the work. This way the atmosphere's velocity is always low -- zero, in fact, if you assume the atmosphere rotates with the Earth instead of modeling winds. The lift and drag calculated in the surface-relative frame will still be valid in the inertial frame.

Immediately after launch, you'll be working with very small values for both rocket and wind speed, so any errors you have will be tiny. If you work with both rocket and wind speed in the inertial frame, you'll be working with two quantities near 400 m/s near liftoff -- not all that large, as Organic Marble notes -- and will lose a little precision, but in practice even that is probably okay -- lift and drag are proportional to the square of airspeed, so they'll be very small near liftoff compared to their peak values.


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