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[SOLVED] Derivative of the flight path angle

dwsmith

Well-known member
Feb 1, 2012
1,673
Why is this true?
$$
h = rv_{\perp} = r(r\dot{\nu})\Rightarrow\dot{\nu} = \frac{h}{r^2}
$$
Look at the last page here to see a visualization.
 

chisigma

Well-known member
Feb 13, 2012
1,704
Why is this true?
$$
h = rv_{\perp} = r(r\dot{\nu})\Rightarrow\dot{\nu} = \frac{h}{r^2}
$$
Look at the last page here 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

$\chi$ $\sigma$
 

dwsmith

Well-known member
Feb 1, 2012
1,673
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.