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How to express the angular momentum in non inertial Earth frame like Earth Fixed frame so that angular momentum is conserved ?

Is H = R x mV still valid when R and V are taken in the rotating frame ?

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closed as off-topic by Phiteros, peterh says reinstate Monica, Heopps, Rob, Organic Marble Aug 24 '18 at 20:21

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is about other space sciences (physics, weather, astronomy, etc), and does not directly pertain to space exploration as outlined in the help center." – Phiteros, peterh says reinstate Monica, Heopps
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Welcome on the Space SE! Your question is not really about the space exploration, I suggest to try physics.stackexchange.com . $\endgroup$ – peterh says reinstate Monica Aug 24 '18 at 15:50
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    $\begingroup$ This question is better suited to Physics.SE, and off topic for this site. $\endgroup$ – Rob Aug 24 '18 at 16:12
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    $\begingroup$ It's a good question for sure. Before you ask there, I'd recommend you look there for existing answers. $\endgroup$ – uhoh Aug 24 '18 at 17:22
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Yes, $\vec H = \vec r \times (m \vec v)$ is still valid in a rotating frame. That's the definition, after all. Whether this definition has any meaning is a different question. Angular momentum is not necessarily conserved in a rotating frame. Consider, for example, a spacecraft in a geosynchronous orbit that has a non-zero inclination with respect to the equator and a non-zero eccentricity. The angular velocity vector as defined by $\vec r \times (m \vec v)$ points all over the place.

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  • $\begingroup$ I'm glad you were able to squeeze this in just before closure, having a good answer here is a net benefit to the site. I'm not sure this should have been closed. $\endgroup$ – uhoh Aug 25 '18 at 2:00

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