can we say $x=aV\sqrt{a}$ in Kepler's third law? What else can we say? 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?)
 A: 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
$$
2GM\left(\frac1{a(1-e)}-\frac1{a(1+e)}\right)
=b^2ω^2\left(\frac1{(1-e)^2}-\frac1{(1+e)^2}\right)
\\~\\
GM=a^3ω^2
$$
