MHB Derivative of the flight path angle

AI Thread Summary
The discussion centers on the relationship between the flight path angle and the equations governing orbital mechanics, specifically the expression for angular velocity, where the angular momentum \( h \) is defined as \( h = rv_{\perp} \). It is clarified that \( v_{\perp} \) represents the tangential speed, while \( v_r \) denotes the radial speed. The equation \( \dot{\nu} = \frac{h}{r^2} \) is derived from these definitions, linking angular momentum to the radius and angular velocity. A visualization provided in the linked resource aids in understanding this relationship. The conversation emphasizes the distinction between tangential and radial speeds in the context of orbital dynamics.
Dustinsfl
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Why is this true?
$$
h = rv_{\perp} = r(r\dot{\nu})\Rightarrow\dot{\nu} = \frac{h}{r^2}
$$
Look at the last page http://www.mathhelpboards.com/f49/orbital-mechanics-notes-3682/#post16317 to see a visualization.
 
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dwsmith said:
Why is this true?
$$
h = rv_{\perp} = r(r\dot{\nu})\Rightarrow\dot{\nu} = \frac{h}{r^2}
$$
Look at the last page http://www.mathhelpboards.com/f49/orbital-mechanics-notes-3682/#post16317 to see a visualization.

If $\displaystyle v_{\perp}$ is the radial speed is $\displaystyle v_{\perp}= r\ \dot{\nu}$ where $\nu$ is the angle in radians...

Kind regards

$\chi$ $\sigma$
 
chisigma said:
If $\displaystyle v_{\perp}$ is the radial speed is $\displaystyle v_{\perp}= r\ \dot{\nu}$ where $\nu$ is the angle in radians...

Kind regards

$\chi$ $\sigma$

Wouldn't $v_{\perp}$ be the tangential speed? $v_r$ I would think is the radial speed.
 
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