Stuff about math!
There are 2 planets,one is big (mass M1 kg),and one is small (mass M2 kg),They are R meters apart.The big planet can not move.What is,to maintain a stable circluar orbit, the speed of the small planet needed in the tangent of the circle?
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Math category, so let's do some math and physics.
Centrifugal force
$$$F_r=M_2\omega^2R$$$
Gravitational force
$$$F_g=G\frac{M_1*M_2}{R^2}$$$
$$\omega$$  angular speed
$$G \approx 6.7 * 10^{−11}$$  gravitational constant
Equilibrum condition $$ F_r = F_g $$ =>
$$$\omega^2 = G\frac{M_1}{R^3}$$$
$$$\omega = \sqrt {\frac{G*M_1}{R^3}}$$$
Linear speed
$$$v=\omega * R$$$
so
$$$v = \sqrt {\frac{G*M_1}{R}}$$$
But its realworld speed. It must be scaled to pixel.
BTW: you asked to get a response? Or just to show yours?
That seems like the start of a question, and it ends with a question mark, so he did (implicitly) ask to get a response.
My answer was v=sqrt(m/r) pixels per sec
I set GC to 1 for easyness
It took me 2 hour of derivitiving and intergraling bc I was thinking abt speed and calculus
And snapenik yes,I was finding what to fill in the"math" var in Snap! 6.8.1 Build Your Own Blocks
looks like that it didn't work well it's great
One extra question:What if I want an elliptical orbit and tell you the eccentricity E?
For eliptic orbit $$$v_i = \sqrt {(G*M_1)(\frac{2}{r_i}\frac{1}{R}})$$$ where
$$v_i$$  velocity at given point
$$r_i$$  distance to the $$M_1$$ at any given point ($$M_1$$ must be placed at focal point)
R  considered a semimajor axis (we are deforming minor axis)
semiminor axis given the eccentricity "e" => $$ b= R\sqrt{1e^2}$$
Sources : Eccentricity and the SemiMajor/SemiMinor Axes — Orbits Tutorial
What about eccentricity bigger than 1?
Also where did you get these imgs?I can't copy them.
??? Do you think of KaTeX expressions for Discourse?
$$$v_i = \sqrt {(G*M_1)(\frac{2}{r_i}\frac{1}{R}})$$$
$$$v_i = \sqrt {(G*M_1)(\frac{2}{r_i}\frac{1}{R}})$$$
Wiki:Eliptic orbit
The velocity equation for a hyperbolic trajectory has either $$+ \frac{1}{a}$$, or it is the same with the convention that in that case a is negative.
Looks like I forgot that
ok
oohkay
so i started playing ksp at may
and now i have more understanding of this yay