I'm modelling an orbit of a satellite that has gravitational, drag and lift forces acting on it. Am I correct in stating that the orbit speed $V = \sqrt{GM/r}$ is just the tangential velocity of the satellite and the lift/drag forces are what produce the radial component of the velocity?
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$\begingroup$ A stable orbital velocity is when the centripetal force balances the gravitational force. $\endgroup$– Cinaed SimsonCommented Nov 23, 2019 at 4:39
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$\begingroup$ The velocity vector is tangential by definition. $\endgroup$– Qmechanic ♦Commented Nov 23, 2019 at 7:03
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2 Answers
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That's the tangential velocity for a circular orbit of radius $r$ subject only to gravitational force. If you have an elliptical orbit, then the gravitational force is changing both the radial and tangential components of the velocity. If the path is hyperbolic or parabolic, the same is true.
Also, if you have drag, it's going to affect the tangential component as well as the radial.
So, the answer to your question is no for a general orbit.
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The gravitational force implies radial acceleration so it only changes the radial velocity.
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$\begingroup$ Lots of things that are true and easy to explain about circular orbits are not true about non-circular orbits and the simple explanations generally fall apart in those cases. Consider the velocity at perigee and apogee for a elliptical orbit. They are different in magnitude, in both cases they are purely in the angular direction, and it is gravity that causes them to be different. $\endgroup$ Commented Nov 23, 2019 at 15:57
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$\begingroup$ @dmckee Indeed. So if the tangent velocity has a component in the radial direction, that component changes. $\endgroup$– my2ctsCommented Nov 23, 2019 at 17:20
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$\begingroup$ Uh ... in a circular orbit the acceleration is radial, but the radial velocity remains fixed at zero. The rule you are trying to state only works in Cartesian coordinates. $\endgroup$ Commented Nov 24, 2019 at 18:22
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$\begingroup$ @dmckee Are you trying to say that the rule I am attempting to establish, aka Newtonian gravity, is coordinate dependent? $\endgroup$– my2ctsCommented Nov 24, 2019 at 19:44
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$\begingroup$ The problem is the statements about kinematics independent of what causes the force. You are expressing them in coordinate language that is only correct for Cartesian coordinates but using the language of non-Cartesian description. And that is simply wrong. $\endgroup$ Commented Nov 25, 2019 at 2:32