The code of this universe
This is a Simulation Theory that I have been working on for a while now. I decided to share my fantasies to amuse you. I’m a builder, no college degree, so don’t expect differential equations …
It all started with Tycho Brahe, who spent his life in DDoS attacks on the solar system … he carefully placed the obtained data in tables.
Johannes Kepler studied these tables, generalized and searched for the music that the Lord wrote with the help of these heavenly spheres. So he discovered the dependence of periods on radii. The squares of the periods of the planets are related to each other, like cubes of the arithmetic mean radii of their elliptical orbits. The arithmetic mean radius of the ellipse is equal to half of the major axis, and this is how the name “semi-major axis” appeared. Kepler’s third law is contained in one paragraph, in his book “The Harmony of the World” pages 189-190.
In 2013, Vadim Lovchikov gave the third law a more convenient form for application – “Generalized formula of energy”. Since we are usually only interested in one revolution of a celestial body, he expressed the period T in terms of the angular velocity and in terms of the orbital velocity, he called this constant energy E.
E = W² • R³ or E = V² • R
W is the angular velocity of the satellite.
V is the orbital speed of the satellite.
R is the semi-major axis of the satellite.
E – the energy of a planet that has a satellite.
Vadim Lovchikov also showed how this formula reveals LT – the system on which Robert Ludwigovich Bartini worked all his life.
The essence of the LT-system is to express all existing units of measurement in meters and seconds.
Here’s where I started. To start laying out the first line of code, I needed to understand what is the meaning, the meaning of life … there must be some kind of plan along which the code is built.
I assumed that our world is registered on some decentralized payment network, such as a blockchain, and it is profitable to own such a universe.
Let’s call this network “Aesthetics” and the coin “admiration”. How does this network work?
I’ll give you a very crude analogy … You have a piece of wire in your hand and you twist it clockwise. The wire has a length and a speed of rotation, we can say that you have energy that can be calculated using the Lovchikov formula … but you do not feel this energy. You approach another person with the same spinning wire and begin to feel some kind of electrical impulses. Together, you are like an electrophoretic machine with one petal on each disc. If you put ten such people in front of you, then you will have a lot of energy. The amount of admiration depends on the number of connections … This is a crude analogy, but it allows you to build some kind of primitive design.
Let’s say we have one developer, he codes browsers with the least perception through the command line, and he will be connected with all participants, will have the most energy. Further, let’s say there are two planetary planners. They have very primitive minds, like the first version of Linux. Their energetic power will depend on the number of stars, planets, moons that they will build, and how popular their creations will be. The developer hands them the planetary tool.
I will briefly tell you how it works. It must be kept in a circle forward. Planetary Builder:
Decides what to create a star or planet.
Adjusts the size of the celestial body.
Sets the tilt of the axis.
Sets the height of the stationary orbit (the line connecting the surface with the arc of the sixth radius – in the picture Mars). The height of a stationary orbit determines the weight of objects at the equator. The template does not allow establishing a stationary orbit below the surface of the planet, and objects at the equator cannot have negative weights.
When setting the altitude of a stationary orbit, the time is automatically entered, but I have not figured out this algorithm yet. In this case, 6000 seconds are entered for Mars. If we divide the circumference of Mars by 6000 seconds, then we get the first space velocity of Mars. In the case of the Earth, at 6.618 radii of the height of the stationary orbit, T = 5000 seconds is entered. So the celestial body receives its main physical and dynamic characteristics.
Presses ENTER or Amen!
Next, let’s say there are 8 miners. Their minds are bulky like Windows 2000. For them, this template works completely automatically, like a pre-installed application controlled by a neural network. I will explain gradually. A fresh planet has only a surface.
The content of the planet is undefined … the content is in superposition.
The universe is designed for a very large number of miners, they have great graphics and analytical powers. They are supposed to be inventive and active, begin to compete with each other, unite into clans, conquer galaxies and establish their own rules and all that … It was expected that they will frolic and make a big mess, turning planets and constellations into dust. But something went wrong … They are the main consumers of chemical elements. And so the miner goes to extract metals. The neural network evaluates what the miner needs and, using a template, displays those elements that are suitable in terms of properties for the implementation of the miner’s plan. In the interface, he just digs and finds everything he needs, he just gets lucky and that’s it. He is lucky until he gets to the planet that was dug up all before him, there his luck will not work, everything is dug up and sifted, there is what is and there is not what you want, there is nothing to define. Dig deeper or look for a fresh planet. The energy power of a miner depends on the amount of matter that he has identified, organized, shaped … but they don’t really chase after energy, they like games, secrets, barriers, dominate … They quickly learned how to extract energy by simply creating many connections … invented concerts, public speaking, corporations, cities … so as not to dig.
Now let’s analyze the template itself and fish out its secrets.
Let’s consider it in the L-system. Here we are dealing only with lengths – arc length, radius length. Here we observe that the product of the square of any arc and the cube of its radius is always the same number. It looks like some kind of field with decreasing intensity as you move away from the center. Knowing how the template works, we will quite rightly look for M-mass here. For convenience, I took an arc that is a full circle:
M = (2πR) ² • R³ = 4π² R⁵
We know the formula for the volume of a sphere:
V = 4 / 3πR³
Residue density formula
ρ = 3πR²
Do not be confused by the density in square meters, perhaps this is true. When we break a stone, we see more surfaces, break more and more, the surface area becomes larger and larger, and our confidence that we have learned something about the contents of this stone also grows.
M-mass – dimension m⁵
R-radius – dimension m
V- volume – dimension m³
ρ- density – dimension m2
1m⁵ = 106.1032953945969 kg
M (earth) = 4.14e35 m⁵ = 4.4e37 kg
See, my Earth is a little heavier …
Now let’s look at the same template in the LT-System. Now the arcs will become orbital velocities, and the radii will remain radii.
If we draw an analogy with the field strength, then the characteristic of the field will be the orbital velocity anywhere in the field, which is easily calculated by the formula
E = V² • R
Orbital velocity anywhere is calculated as follows:
V = √ E / R
Let me show you with an example: the height will be measured by the radii of the Earth.
Let’s say you are flying your spacecraft around the Earth in a highly elongated orbit. At perigee 2 radii, at apogee 10 radii. The weight of the device is 500 kg. You want to get a circular orbit at an altitude of 10 Earth radii with one pulse.
Let’s calculate E of the Earth, use the parameters of the Moon:
E (earth) = V² (moons) • R (moons) = 1023² • 384399000 = 4e14 m³ / s²
Let’s calculate the orbital velocity at perigee and apogee.
V (perigee) = √ E / 2R = = √ (4e14 / (2 • 6371000) = 5602m / s
V (apogee) = √ E / 10R = = √ (4e14 / (10 * 6371000) = 2506m / s
I know that you are not used to such formulas, you can skip them.
To find out with what acceleration our spacecraft is moving in an elliptical orbit, we need to find out the average speed, and then the orbital period.
V (average) = √ E / 6R = 3235m / s
T = 2π6R / V (average) = 74244sec
We are only interested in half of the period
0.5T = 37122sec
a = (V (perigee) -V (apogee)) / 0.5T = 0.0834006788427348 m / s²
We are now ready to calculate the strength of the momentum to eliminate acceleration and obtain uniform motion in a circular orbit for a 500 kg vehicle. Sir Isaac Newton wrote the formula for us:
F = a • m = 0.0834006788427348 • 500 = 41.7 kg • m / s²
For fun, I will translate the magnitude of the impulse into the LT-system.
F = 0.0834006788427348 • (500 / 106.1032953945969) = 0.393 m⁶ / s²
Now we need to wait until our own spacecraft reaches the maximum distance from the center of the Earth at apogee and throw out our neighbor’s boots towards the Earth, so we will extinguish the orbital acceleration and get a circular orbit at an altitude of 10 Earth radii.
Now let’s consider the interaction of the fields of the Sun and the Earth.
We will find out where the border of the transition to the Solar federal speed limit is.
R is the semi-major axis of the planet’s orbit.
r is the radius of the local speed limit. The formula is:
r = R • ³√ (E (earth) / 3E (sun)) = 1.5e11 • ³√ (4e14 / (3 • 1.33e20)) = 1 501 252 087m ≈ 1.5 million km
Here I will note that anomalies will be observed at the border of the fields, at the Lagrange points L1 and L2. This is due to the fact that the orbital speed in the solar federal field is about 150m / s higher at the L1 point and 150m / s lower at the L2 point than at the Earth. This difference coincides with the speed of rotation around the Earth, which is 150 m / s.
Getting to these points, objects hang in them and dull. They do not know what code to execute, whether to continue rotating around the Sun, or to roll up and start rotating around the Earth.
Now a hypothetical question: Can a spacecraft, in a highly elongated elliptical orbit, go beyond the Earth’s field (Hill’s sphere) and remain a satellite of the Earth.
I think yes. A satellite can leave the Hill sphere with impunity if its average orbital speed is higher than 150 m / s, and such a satellite will not hover at points L1 and L2.
I also wrote a small formula for calculating the acceleration of gravity on the surface of planets:
V₁ – Keplerian orbital velocity on the planet’s surface (we take from Wikipedia “the first space velocity”)
V₂ – the speed of rotation of the planet’s surface (we take from Wikipedia “the speed of rotation at the equator”)
g – acceleration of gravity
R-radius of the planet
We will calculate the g of the Earth at the equator.
g = (V₁²-V₂²) / R
g = (7919²-465²) / 6378 136 = 9.7982m / s²
I will also give a formula for calculating the height of stationary orbits using the example of Mars:
R= ³√(( E • T²)/(2π)²)
R is the radius of the stationary orbit.
E is the energy of the planet
E (mars) = 4.3e13 m³ / s²
T is the time of one complete revolution of the surface around its axis
T (mars) = 88500 sec
R (mars) = ³√ ((4.3e13 • 88500²) / (2π) ²) = 20 432 982 m
This will make up 6 radii of Mars.
Now a little about the mass. The formula for mass, which I wrote, gives mass only inertia and no more energetic properties. A celestial body will have an energy field if it is created in a standard way and its contents are not defined.
If the planet is broken, parts of matter that have not been determined will retain the energetic properties of the mother planet, stones will revolve around them.
Platinum and gold asteroids were most likely mined on planets and their contents determined, indicating that they should not be expected to exhibit any gravitational properties.
You see, everything is much better than just living on the server of a more highly developed civilization, which is impossible to comprehend, as a cabbage can not comprehend a goat.