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Angular momentum of a planet about its apogee is maximum at __________

Now, I do know that

  • Angular momentum of a planet around the focus of the elliptical orbit (the sun) is conserved due to gravity being a central force. (Side note: this gives Kepler’s second law.)
  • The formula of speed of a planet at any point on its orbit which is a distance $r$ away from the star ($a$ is the length of the semimajor axis): $$v(r)=\displaystyle\sqrt{\frac{GM(2a-r)}{ar}}$$
  • Angular momentum about a point is defined as $\vec L=m\vec v\times \vec r$.

So I can’t see any easy way out to determine the answer. I was thinking that it may be at perigee because of the large distance but got confused because of the decreasing speed as the planet moves from apogee to perigee. So is the only way out for me now to do some coordinate-bashing of the ellipse? Or is there any easier way? Intuitive (but convincing) reasoning is more than welcome.

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    $\begingroup$ I added, as an answer, the explanation that I just found out from a source. If any of you knows of a more rigorous solution, please add it :)) $\endgroup$ Jul 26, 2022 at 11:40

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I just found out an explanation: For local extrema of $\vec L$, we must have $\dfrac{d\vec L}{dt}=0$ but we know $\dfrac{d\vec L}{dt}=\vec \tau$ so that $$\tau_{\text{about apogee}}=0$$ which happens at only the apogee and perigee. It is easy to notice that apogee gives the minimum (because $\vec r=\vec 0$) so that $\vec L_{\text{apogee}}$ is maximised for the perigee.

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