1
$\begingroup$

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?)

$\endgroup$
7
  • $\begingroup$ 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
0
$\begingroup$

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 $$ 2GM\left(\frac1{a(1-e)}-\frac1{a(1+e)}\right) =b^2ω^2\left(\frac1{(1-e)^2}-\frac1{(1+e)^2}\right) \\~\\ GM=a^3ω^2 $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.