Circular Orbit

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?
:black_circle: ---------->

Math category, so let's do some math and physics.

Centrifugal force

Gravitational force

$$\omega$$ - 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

$$$v=\omega * R$$$


$$$v = \sqrt {\frac{G*M_1}{R}}$$$

But its real-world 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 :frowning:
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 semi-major axis (we are deforming minor axis)

semi-minor axis given the eccentricity "e" => $$ b= R\sqrt{1-e^2}$$

Sources : Eccentricity and the Semi-Major/Semi-Minor 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