Stability of the Solar System

History of the question

The Solar System exists already several billions of years. All this time the planets are rotating around the Sun, and revolving around their own axes. The Solar System does not exist in the ideally empty space. Planets are constantly bombarded by meteorites. It is possible that in the billions years history they were even hit by asteroids. Every such collision must change the planet's orbits. The Solar System must be transformed from the regular system into chaotic system. Some planets must be rejected from the Solar System, and some of them must be swallowed by Sun. Nevertheless the order exists in the Solar System. The radiuses of planets rotation approximately satisfy the Titius-Bode rule. With the help of this rule the asteroids belt was discovered.

The Titius-Bode rule:

a = 0.1(3*2n+4) astronomical units.

Where, a is an average distance of a planet to the Sun, measured in astronomical units.
nMercury = " - infinity";
nVenus = 0;
nEarth = 1;
nMars = 2;
nasteroids, or former Phaeton = 3;
nJupiter = 4...

If in the Solar System there is an order, then there is some force which supports this order. If some influence had changed the orbit of some planet, then this force must return the planet into it's place.

Gravity velocity and Laplace

Marquis de Laplace (1749 -1827) had clearly realized that the Solar System is very sensitive to perturbations. At his times it was already known that the velocity of light was finite. And this means that the visible and the real positions of the Sun in the sky do not coincide.

Let's investigate the simplest cases. Two objects, A and B, having the same masses, rotate clockwise around the common center of masses, C. Let in the moment t=0 they lie on the line.


Because the velocity of light is finite, the observer A does not see B in it's real position. He sees it there, where B was some time before. Let's denote that point B'. The same conclusion is valid for the observer B.




Both observers see each other a little ahead from the position of the center of masses. If gravity forces are spreading with the same velocity as light do, then every of these objects will be subjected by the accelerating force. That means that the system must be accelerated and disperse.

That is why; Laplace had come to conclusion about the almost infinite velocity of spreading of gravity forces. Thus, we can see, that already two hundred years ago, in order to save the Solar System, the great Laplace had made the erroneous unnatural conclusion because he could not see yet the opposite compensating force.

At the beginning of the twentieth century there was proved that there are no velocities bigger than the velocity of light. As a result the conclusion of Laplace must be changed. It was necessary to find the new way to save the Solar System from the dispersion. And again the unnatural conclusion was made. The leading scientists have concluded that the gravity and visible positions of the Sun do not coincide.

Cunning tricks in cosmology

According to the contemporary cosmology the Universe expands -- the distances between the galaxies are constantly growing. The question arises: At what scales there is no expansion? But Big Bang cosmologists are cunning persons. In order to hide their false conclusions they assert that there is no expansion inside the galaxies. But why? According to your model galaxies are rushing away from each other, and accordingly to your new observations the expansion has the acceleration, such that the Hubble constant becomes a real constant and does not depend from the age of the Universe. But galaxies are not expanding...

No, dear bigbangers! In this case your Big Bang is a Big Grenade.

If we speak about expansion then we must say that space expands everywhere: between the galaxies and inside the galaxies. In this case the stars inside the galaxies do not move on the circular orbits, but move on the spiral orbits on the expanding space. Thus the stable galaxies and stable star systems will have the approximately constant shape and approximately constant dimensions. Unstable galaxies and star systems will be torn on parts by space expansion or gathered into the more compact object. Planets are trending to their attracting star. Their trajectories are also spirals on the expanding space. The Hubble expansion brakes the velocity of a planet. The force found by Laplace accelerates the planets. The Hubble force is compensated by the Laplace force. If Laplace would know about the existence of the Hubble force, he would not make his unnatural conclusion about the infinite velocity of gravity forces. If Einstein would not listen to the lovers of the expanding model of Universe, he would not make his biggest error and he would not reject the static model of the Universe. His first correct model contained the gravity and antigravity. Contemporary cosmologists again invent the antigravity but their attempts are funny. They must reject Big Bang model as nonsense.

Now we will prove that in order to explain the stability of the Solar System, the two opposite mutually compensating forces must exist: the Hubble force and the Laplace force. Thus we will receive one more precise independent method of Hubble constant deduction.

The cause of Solar System stability

Let's suppose that the space expands everywhere with the constant rate:


The space is constantly being absorbed by the massive objects. This looks as gravity force directed to the massive objects. The direction of gravity force to attractive object coincides with the direction of its visible position in the sky. (Here we have rejected the unnatural conclusions of Laplace and of the leading scientists of twentieth century)

The expansion of space gives the planets the negative acceleration. The shifted ahead position of gravitating object gives the positive acceleration. If the sum of these accelerations is equal to zero then the planet moves on stable orbit.

Let planet with mass m rotates on the circumference of radius r with velocity v. Then the Sun with mass M rotates on the circumference of radius R with velocity V around the common center of masses C.


The forces, accelerating the Earth and Sun are equal, accordingly:

FSun = Fv/c; FEarth = FV/c.

The power, spent on acceleration is equal to the product of force and velocity:

PSun = FvV/c; PEarth = FVv/c.

After substitution we'll have:

v =sqr(rF/m); V=sqr(RF/M); F=GMm/(r+R)2;

P = GMm/(r+R)2 * sqr(rF/m) * sqr(RF/M) / c.

P = G2M3/2m3/2/(r+R)4 * sqr(rR)/c.

Let's investigate two cases: 1 - masses are equal; 2 - mass of the star are much more the mass of a planet.

1. m=M; r=R,


2. If M is much more than m, then r is much more than R. R = r(m/M); r~R+r. Consequently:

P = G2M3/2m3/2/(r+R)4 * sqr(rR)/c = G2M3/2m3/2/r4 * sqr(r2m/M)/c = G2Mm2/r3/c.

Every planet of Solar System receives the power, which we will name the Laplace power:

PLaplace = G2Mm2/(r3c).

Let's now investigate the Hubble force.

The space expands: v = Hr. The potential energy of every interacting pair will grow:

dE = GMm/r - GMm/(r+dr).

PHubble = dE/dt = GMmdr/r2/dt = GMmdr/r2/dt = GMmv/r2 = GMmHr/r2 = GMmH/r.

Now we will equal the powers of Laplace and Hubble, and then we will receive the formula for Hubble constant.

PHubble = PLaplace,

GMmH/r = G2Mm2/(r3c).

H = Gm/(r2c).

In this work the exact value of Hubble constant was received by several different methods. By taking this value we'll try to find the radiuses of stable orbits for planets of the Solar System:

r = sqr(Gm/(Hc))

In the bottom there is a table of observable distances of planets to the Sun, and several distances of satellites to their planets. Distances and masses were taken from the NASA web site: One must be careful, because the information about the masses of planets in most old books are different because of the recent change of the recommended value of the gravity constant. Useful information one can also find here:

Object Mass of Object (*1024kg) Average distance to the Sun (*109 m). In brackets perihelion/aphelion.
For satellites: average distance to their planets. In brackets - eccentricity.
Ratio of computed radius to the observable radius.
Mercury 0.3302 57.91 (46.00 / 69.82) 3.038 ~ 3
Venus 4.8685 108.21 (107.48 / 108.94) 6.2421~ 2p
Earth 5.9736 149.60 (147.09 / 152.10) 5.0014 ~ 5
Mars 0.64185 227.92 (206.62 / 249.23) 1.0760 ~ 1
Phaeton ... ????????? ...
Jupiter 1 898.6 778.57 (740.52 / 816.62) 17.132
Saturn 568.46 1433.53 (1352.55 / 1514.50) 5.0914 ~ 5
Uranus 86.832 2872.46 (2741.30 / 3003.62 ) 0.99308 ~ 1
Neptune 102.43 4495.06 (4444.45 / 4545.67) 0.68925
Pluto 0.0125 5869.66 (4434.99 / 7304.33) 0.00583
Moon 0.07349 0.3844 (0.3633 / 0.4055) 215.9
Io 0.08933 0.4216 (0.004) 217.0
Europa 0.04797 0.6709 (0.009) 99.94
Ganymede 0.1482 1.070 (0.002) 110.1
Callisto 0.1076 1.883 (0.007) 53.33
Titan 0.13455 1.22183 (0.33) 91.901
Triton 0.02147 0.35476 126.4

Only Mars and Uranus had confirmed the naked formula for the stable radius. If the formula was erroneous then the dispersion of results would be unsystematic. But it clearly visible that received results dropped very near to the numbers 1, 3, 5... This leads us to the conclusion that the Solar System is quantisized. Consequently, the deduced naked formula for stable radiuses must be transformed into the form:

r = sqr(Gm/(Hc))/n.

Where n is quantum (or resonance) number of a planet:
nMercury = 3;
nVenus = 2p or unknown; May be Venus changes its orbit now;
nEarth = 5;
nMars = 1;
nJupiter = unknown; Jupiter is unstable, growing planet-star;
nSaturn = 5;
nUran = 1;
nNeptun = unknown; May be Neptune is leaving the Sun System;
nPluto = unknown; Pluto can hardly named a planet, compare it's mass with the masses of planet's satellites.

Quantum numbers of planet's satellites are not known exactly, but it is visible that they also have some order. Besides, it is known that the planet's satellites do not move on stable orbits.

If somebody think that here we have a simple coincidence, let he solve the following problem:

How many times one must through a handful of nine stones in order they dropped as close to the points 1, 2, 3, 4, 5, 6... as we have received: 3.038; 5.0014; 1.0760; 5.0914; 0.99308; any; any; any; any?

If he compute this probability then he inevitably conclude, that some parameter, which is very close to the value of Hubble constant, is really working in the Solar System. And now tell me, why I must name this value as "some parameter" if I used in the deduction just the Hubble constant; if I used just the properties of the Hubble constant in the deduction? Why I must believe the supporters of the Big Bang model, who says that the Hubble constant does not work inside the galaxies? Now we can see that it works even inside the Solar System; and sustain it from dispersion. And we will see later that it also works inside the atom, because the Hubble constant is one of the main intrinsic characteristics of the space-time.

The power income and expenditure of planets, satellites and the Sun we will compute at the page Dark Energy. Today, 03.21.2002, there is only Russian language page.

This page was last updated: 21 March 2002, by Ivan Gorelik

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