Orbit Simulator

This project aims to simulate how orbital mechanics work

Examples

Earth around the Sun	Mercury, Venus, Earth and Mars orbiting the Sun	Geostationary orbit around earth	Two Massive objects orbiting a fixed point

3D animation of 3 planets in orbit

Execution

usage: orbit_simulator [-h] [-3d] [-c] preset_name

positional arguments:
  preset_name

options:
  -h, --help    show this help message and exit
  -3d, --3d
  -c, --coords

Execution Command Example

Run preset SOLAR_SYSTEM with 3d rendering

python3 orbit_simulator solar_system -3d

Measurement Units

Planet mass: $kg$
Distance: $m$
Speed & Velocity: $m/s$

Assumptions and Formulas

Gravitational Constant

This is an aproximate value of $G$ based on Committee on Data for Science and Technology (CODATA) with the article "CODATA recommended values of the fundamental physical constants: 2018". By Eite Tiesinga, Peter J. Mohr, David B. Newell, and Barry N. Taylor:
$G= 6.6743 \times 10^{−11} m^3 kg^{−1} s^{−2}$

Distance between celestial bodies

In a 2d matrix, celestial bodies have $x$ and $y$ coordinates. The formula obtains the distance between the axes of the 2 celestial bodies by interpreting it as a right triangle, resulting in the value of the hypotenuse, in meters.

$\sqrt{|x_1-x_2|^{2}+|y_1-y_2|^{2}}$

And for a 3d matrix, you simply need to add the $z$ axis to the formula

$\sqrt{|x_1-x_2|^{2}+|y_1-y_2|^{2}+|z_1-z_2|^{2}}$

Gravitational Acceleration

To calculate the gravitational acceleration I decided to use Newton's universal gravitational law, since it's simpler and works well for the scenarios addressed by the simulator (Stars, planets, ships and natural or artificial satellites)

$G$ = Gravitational Constant
$M$ = Mass of the celestial bodies
$r$ = Distance between the celestial bodies

$F=G{M_1M_2\over r^2}$
Using that equation, and considering that $F=g*M$, we can also directly calculate $g$ by the following formula:

$g={G{M_1M_2\over r^2}\over M_2}={G{M_1\cancel{M_2}\over r^2}\over \cancel{M_2}}=G{{M_1} \over r^{2}}={G\over 1}*{M_1\over r^2}$

Therefore, $g = {{GM} \over r^{2}} $

Speed distribution between axes

Imagine you have a celestial body on the following XY coordinates: $[10, 5]$.

That means the distance from the $[0, 0]$ is $d=\sqrt{10^2+5^2}\approx11.18$.

That also means that $[10, 5]\over 11.18$ retuns a rotation matrix (with values between -1 and 1) directing precisely the speed distribution ratio we need to use $[0.89, 0.44]$.

Now let's take into consideration that the speed of the celestial body is $2m/s$. After 1 second, the new distance should be $9.18$, right?

The approximate distance it should move on each axis after 1 second is then $[0.89, 0.44] * 2 = [1.78, 0.88]$

These are the distances we should substract from the coordinates, resulting into new coordinates of $[8.21, 4.10]$. Using these coordinates, the new distance is $d=\sqrt{8.22^2+4.10^2}\approx9.18$

The simulator uses the same trigonometry principles, where the target celestial body position $P_2$ is transformed into the center of the matrix, creating the new matrix $M$. This is achieved by the following formula: $ M=-(P_1-P_2) $

Also, the simulator uses a 3D matrix to simulate the $Z$ axis. The formulas mentioned above will not change much. All it needs to do is add the $Z$ coordinate to the equations.

$x=x_1-x_2$
$y=y_1-y_2$
$z=z_1-z_2$

$M_1=[x,y,z]$

$d_0=\sqrt{x^2+y^2+z^2}$

$R_0=M_1 \div d_0$

$\Delta D_0=R_0*\Delta v$

$M_2=M_1+\Delta D_0$