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 ?
$\begingroup$
$\endgroup$
3
-
$\begingroup$ Welcome on the Space SE! Your question is not really about the space exploration, I suggest to try physics.stackexchange.com . $\endgroup$– peterhCommented Aug 24, 2018 at 15:50
-
1$\begingroup$ This question is better suited to Physics.SE, and off topic for this site. $\endgroup$– RobCommented Aug 24, 2018 at 16:12
-
1$\begingroup$ It's a good question for sure. Before you ask there, I'd recommend you look there for existing answers. $\endgroup$– uhohCommented Aug 24, 2018 at 17:22
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
1
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.
answered Aug 24, 2018 at 19:54
-
$\begingroup$ 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. $\endgroup$– uhohCommented Aug 25, 2018 at 2:00