MHB Derivative of the flight path angle

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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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