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We know velocity of a planet in an elliptical orbit is given by: $$v^2 = GM * (\frac{2}{r} - \frac{1}{a})$$ in an elliptical orbit. [Here r is distance between particle and sun] source

We also know, areal velocity in an elliptical orbit is given by $$\frac{dA}{dt} = \frac{1}{2}vr$$

By putting value of velocity in this equation we find that areal velocity is dependant on r, and thus ever changing(since distance between sun and a planet is also changing)

But keplers second law states that areal velocity of a particle is always constant.

How do I resolve this contradiction? What am I doing wrong? I assumed r would cancel out leaving only contants behind. If I am putting the value of velocity wrong what is the correct one?

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  • $\begingroup$ Here's a nice animated diagram showing the relationship between the radius & velocity vectors in an elliptical orbit, plus a few other goodies. gregegan.net/SCIENCE/LRL/LRL.html $\endgroup$
    – PM 2Ring
    Commented Apr 23, 2023 at 14:33
  • $\begingroup$ Definately looks cool, it's apparant here now that r and v are really not perpendicular. $\endgroup$
    – abhitruechamp
    Commented Apr 23, 2023 at 14:59
  • $\begingroup$ You may like this answer of mine on the connection between the magnitudes of the tangential velocity, radial velocity, and angular momentum. Also check out the eccentricity vector. $\endgroup$
    – PM 2Ring
    Commented Apr 23, 2023 at 15:09

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The areal velocity is $$ \frac {dA}{dt}= \frac 12 |{\bf r}\times {\bf v}| $$ and the vector product ${\bf r}\times {\bf v}$ has magnitude $rv$ only if ${\bf r}$ is at right-angles to ${\bf v}$.

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  • $\begingroup$ As far as I am aware though velocity vector is always perpendicular to distance between the sun and planet. Its certainly true for circular orbits, and it seems it applies to elliptical orbits too. $\endgroup$
    – abhitruechamp
    Commented Apr 23, 2023 at 13:14
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    $\begingroup$ No. It is not true for ellipses. The velocity vector is necessarily tangent to the ellipse, and the tanget is everywhere perpendicular to the radius only for a circle. $\endgroup$
    – mike stone
    Commented Apr 23, 2023 at 14:14
  • $\begingroup$ Thanks, that's definitely the error then, as even though areal velocity does depend on the radius when we calculate it for a circle(directly proportional to root r) it doesn't matter as circile had constant radius. While I am not able to find (in general terms)the perpendicular component of the line joining foci with point of contact of tangential vector in a ellipse, my main concern was with the apparant contradiction, I consider the query solved since the alternative proof(dA/dt = L/2m) exists. $\endgroup$
    – abhitruechamp
    Commented Apr 23, 2023 at 14:53

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