we know that Kepler's third law says that the Period of the planet (time elapsed for a planet to perform a complete rotation around its sun) to the power of 3 is proportional to the orbit's semi-major axis to the power of 2. so we can say $P^2=a^3$ or $T^2=a^3$.

we know $T=\frac{x}{V}$ where $x$ is the orbit's perimeter and $V$ the speed of the planet.

so the Kepler's third law can be rewritten as $x=aV\sqrt{a}$

$a$ and $V$ are constant for each planet and we can find the orbit's perimeter for each planet with the above equation. right?

another question I have is that can we use the last equation to get an approximately good perimeter for each given ellipse? (for example we draw an ellipse with the semi-major and semi-minor axes of 5 and 3 meters respectively; then we put an object like a ping pong ball on the ellipse to move with the speed of 0.1 $\frac{m}{s}$ on it. can we use that equation to find the perimeter of that ellipse?)

  • $\begingroup$ Planets don't move at constant speed: Kepler's second law implies they are faster when they are nearer to the Sun. $\endgroup$ – Intelligenti pauca Dec 4 '19 at 18:21
  • $\begingroup$ please change that equation so that it have the acceleration too. is it necessary to put angular speed too? @Aretino $\endgroup$ – aminabzz Dec 4 '19 at 18:36
  • $\begingroup$ Sorry but I don't understand your comment above. $\endgroup$ – Intelligenti pauca Dec 4 '19 at 18:42
  • $\begingroup$ I want the correct equation. you said $x=aV\sqrt{a}$ isn't correct. $\endgroup$ – aminabzz Dec 5 '19 at 17:20
  • $\begingroup$ The correct equation involves elliptic integrals: I don' think that's the kind of stuff you want. $\endgroup$ – Intelligenti pauca Dec 5 '19 at 18:17

Keplers second law tells you that $$ r^2\dot\varphi=ab\omega $$ where $r,φ$ are the polar coordinates around the gravity center, $a,b$ are the half-axes of the ellipse (relative to the center of the ellipse) and $ω=\frac{2\pi}{T}$ is the average angular speed. If one knew a nice formula for the perimeter, one could make a proportion law as proposed, but it would be artificial.

The velocity along the orbit can now be compactly computed from $$ \dot x+i\dot y=\frac{d}{dt}(re^{iφ})=(\dot r+ir\dotφ)e^{iφ}, $$ Using the formula of the first law, with $b^2+c^2=a^2$, $c=ea$, $$ r=\frac{b^2}{a+c\cosφ} \implies \dot r=\frac{b^2c\sinφ\dotφ}{(a+c\cosφ)^2} =\frac{acω}{b}\sinφ $$ gives the square of the speed as $$ \dot r^2+r^2\dotφ^2=\frac{(acω)^2}{b^2}\sin^2φ+\frac{(aω)^2}{b^2}(a+c\cosφ)^2 =\frac{a^2ω^2}{b^2}\left(c^2+a^2+2ac\cosφ\right) $$ so that $$ v=\frac{aω}{b}\sqrt{c^2+a^2+2ac\cosφ} $$

The third Kepler law follows from the gravity equation $\ddot z=-\frac{GMz}{|z|^3}$, which has the radial component $$ \ddot r+\frac{GM}{r^2}-\frac{(abω)^2}{r^3}=0 $$ which integrates after multiplication with $2\dot r$ to the first integral $$ \dot r^2-\frac{2GM}{r}+\frac{(abω)^2}{r^2} =-\frac{2GM}{r_{\rm peri}}+\frac{(abω)^2}{r_{\rm peri}^2} =-\frac{2GM}{r_{\rm apo}}+\frac{(abω)^2}{r_{\rm apo}^2} $$ so that $$ 2GM\left(\frac1{a(1-e)}-\frac1{a(1+e)}\right) =b^2ω^2\left(\frac1{(1-e)^2}-\frac1{(1+e)^2}\right) \\~\\ GM=a^3ω^2 $$


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