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I am wondering if I can exclude coriolis effects and centrifugal forces if I calculate my position relative to an inertial reference frame with as origin the launch site at the moment of lift off. Directly after lift off the reference frame no longer follows the earth rotation.

If I do this, can I measure the position of the rocket by simply adding omega_earth * R_earth to the initial speed of the rocket? And would the rocket equation remain:

$$ {M \Delta V \over \Delta t}=T-D-Mg $$

With:

  • $M$ = mass of the rocket
  • $V$ = velocity of the rocket
  • $T$ = thrust of the rocket
  • $D$ = drag caused by the atmosphere
  • $g$ = gravity constant
  • $t$ = time
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Welcome to the site ThaNoob!

I believe the answer to the question you are asking is yes, just pay attention to your choice of coordinate system. You can eliminate the fictitious coriolis and centrifugal force terms through your choice of an inertial reference frame no matter the coordinate system. However, there may still be a "coriolis" and "centrifugal" look-alike term in the equations of motion if you choose spherical coordinates. They aren't pseudo-forces, just byproducts of the mathematics of coordinate systems.

I interpret your inertial reference frame to be the one that is stationary in space relative to the earth's center. The actual choice of coordinate system is arbitrary, but perhaps a spherical coordinate system may be preferred for its simplicity. But that is beside the point. You can use $xyz$ cartesian coordinates just as well (and is probably easier to implement numerically). If you choose a spherical coordinate system, then parts of the equations of motion will look similar to the fictitious forces, but they are not. If you choose $xyz$ cartesian coordinates, then it will look like straight-forward Newton's laws.

To answer your second question:

If I do this, can I measure the position of the rocket by simply adding $\omega_{earth}\cdot R_{earth}$ to the initial speed of the rocket?

Yes you can. But I want to point out something about the equation you are using. All the quantities save mass and time are vector quantities. So the equation is

$$M\frac{d\vec{V}}{dt} = \vec{T} + \vec{D} + M\vec{g}$$

so that the proper components of the thrust, drag and gravity add correctly. So in reality you actually have 3 equations, one each for $xyz$ (or $r\theta\phi$ if you please). It is a simple differential equation that can be integrated based on your ascent profile. In the integration, you simply add the initial vector quantity $\vec{\omega}_{earth}\cdot R_{earth}$ as the constant of integration. Note that it too is a vector quantity and the components will have to be correctly carried through.

To sum it all up, by choosing an inertial reference frame you do eliminate the added fictitious forces that are the coriolis and centrifugal forces in a rotating reference frame. However depending on your choice of coordinate system (i.e. spherical), there will still be "coriolis" and "centrifugal" look-alike terms in the equations of motion, but they are not the fictitious forces you mention. No such imposters exist in $xyz$ coordinates.

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  • $\begingroup$ Excellent answer! Do the look-alike terms in spherical coordinates come from the definitions of the differential operators? $\endgroup$ – uhoh Aug 19 '19 at 23:53
  • $\begingroup$ Thank you very much for the elaborate reply! Although it answers my question I quickly realized that this will not make my life any easier. As in this coordinate system calculating the pitch angle tan-1(vz/vx) becomes very hard since this should be done with respect to the plane perpendicular to the line that goes throught the instantaneous position of the rocket and the earth center. I guess I will have to deal with the coriolis effects. For those having the same issue this source might help: [link]math.hmc.edu/~dyong/math164/2008/zitter/finalreport.pdf $\endgroup$ – ThaNoob Aug 20 '19 at 7:15
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    $\begingroup$ @uhoh Not directly. The operators themselves don't introduce the components, but instead the Jacobian terms (1/r*d/dr(rv_theta) as an example) and so on. These carry product rule terms that end up looking like the centrifugal and coriolis terms. $\endgroup$ – Quietghost Aug 20 '19 at 16:48
  • $\begingroup$ @ThaNoob That is why I would recommend using a spherical coordinate system. Then the pitch angle is just arctan(vr/(vtheta^2 + v_phi^2)), and no rotation transformation in xyz is needed $\endgroup$ – Quietghost Aug 20 '19 at 16:50
  • $\begingroup$ Thanks for the recommendation but I couldn't find a site where they explain the rocket motion equation for spherical coordinates. For carthesian coordinates it is well explained here: rhef.net/docs/Sim_modeling/… $\endgroup$ – ThaNoob Aug 20 '19 at 20:27

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