• The solution to Newton's GEXACT VALUE & FORMULA
  • The theory controlling planetary spinTHE MATHEMATICAL LAW
  • The pressure at the centre of a massEARTH'S CORE PRESSURE (calculation procedure)
  • Proof of the non-exitence of Dark MatterDOES NOT EXIST
  • The atom as Newton and Coulomb describe itNO NEED FOR A UNIFICATION THEORY
The solution to Newton's G1 The theory controlling planetary spin2 Pressure at the centre of the Earth3 Proof of the non-exitence of Dark Matter4 The atom as Newton describes it5
Useful Stuff Algebra Trig Functions Calculus Max-Min Differentiation Rules Differentiation Trig Differentiation Logs Integration Methods Standard Integrals Stiffness & Capacity Mohr's Circle Earth's Atmosphere Earth's Properties Stars & Planets Laws of Motion Solar System Orbits Planetary Spin Rydberg Atom Planck Atom Newton Atom The Atom Dark Matter? Core Pressure The Big Bang Brakes and Tyres Vehicle Impacts Speeding vs Safety Surface Finish Pressure Classes Hardness Conversion Thermodynamics Steam (properties) Heat Capacity Work Energy Power Constants

Planetary Spin {© 14/03/17}
(a theory for the spin of stars and planets)

This paper, which was released by Keith Dixon-Roche (one of CalQlata's Contributors) on the 7th of April 2017 may well become one of the most important astronomical theories since Isaac Newton's Principia.
Whilst Newton's famous work told us how planets and moons orbited suns and planets, this paper tells us how and why they spin.

It not only accurately predicts all aspects of the spin of every planet and the sun in our solar system, it does so using Newton's own theories and is the only study we (at CalQlata) know on this subject that manifestly works.

Note: All the input data in these calculations has been provided by CalQlata's Solar System Orbits
All the calculations are the sole copyright priority of Keith Dixon-Roche © 2017
Keith Dixon-Roche is also responsible for all the other web pages on this site related to planetary motions
A 'pdf' version of this paper can be found at: Planetary Spin - The Paper


The purpose of this paper is to answer the following questions:

1) What causes a star or a planet to rotate on its axis?

2) What defines the magnitude and direction of this rotation?

(Refer to Mathematical Symbols & Units for an explanation of the terminology, mathematical symbols and units in this web page)


Spin is only induced in a force-centre or an orbiting satellite by each other.
These calculations, which comply with Newton’s laws of motion, successfully predict ...
● The angular velocity of the sun and every planet in our solar system
● The directional switch (prograde to retrograde) of Venus, Uranus & Pluto
● The rotational direction for Neptune despite its largest moon orbiting in the opposite direction
● The correct orientation for magnetic North in the earth and Venus
● Verification of results via an alternative calculation method
Pluto's local orbit

Angular velocity induced by rotational kinetic energy in each layer of a rotating body is defined by the polar moment of inertia of the body’s inner core.

Angular velocity induced by potential energy in each layer of a rotating body is defined by the polar moment of inertia of the layer concerned.

These formulas can now be used to estimate the structural properties of any planet or star from its angular velocity or vice versa.

The author questions the currently accepted value for the mass of Mars

Further Work

Determine the accuracy of the structure of the bodies in our solar system, especially Mars.

The basic solar system of planetary spin

Fig 1. The Basic System

The Basic System

The basic system (Fig 1) comprises;
a force-centre (e.g. a star); and
an orbiting satellite (e.g. a planet); and
a secondary orbiting satellite (e.g. a moon)


The following procedure was used to establish the controlling formulas for planetary spin using our solar system for verification:

1) The rotational kinetic energy of all its satellites was applied to the sun to define a radial factor (Δ) that best represents its polar moment of inertia based upon its observed angular velocity.

2) Mercury, being the simplest planet, with no dedicated satellites and minor [axis] tilt, was used to identify its angular velocity 'ωₒ + ωᵣ' by applying the sun's induced angular velocity to its own orbital angular velocity. An almost exact result (ε = 6.02192E-10) was achieved when using the average potential energy generated between Mercury and the sun along with the sun's polar moment of inertia.

3) The same formulas were applied to Venus, being the next simplest planet (with no dedicated satellites), which also produced an exact result and identified its inverted axis.

4) The same formulas were applied to the earth, which now has only one unknown 'ωᵢ'.

Using the recognised properties of the planet's inner and outer cores, the rotational kinetic energy of its moon was applied to its three principal layers (inner core, outer core and mantle). An exact result was found for the earth on the basis of the following assumptions:
a) The inner core's average density is ≈13000kg/m³ (the generally accepted values is 12600kg/m³ to 13100kg/m³)
b) The inner core's radius is ≈1234980m (the generally accepted value is 1220000m)
c) The angular velocity induced in all of the layers of a planet by its orbiting satellite(s) is identical in all of its layers to that of its inner core (see Further Work above)

5) The same formulas were applied to Mars, which also produced an exact result assuming:
a) The planet has a solid iron inner core of radius ≈1500m

6) Given the unknown structure of the gas and ice planets, the same procedure was applied to these as was applied to the sun. The resulting 'Δ' values were then used to predict the surface density for each planet, all of which proved to be representative

7) Rotational Inducements

As a result of the above, it is possible to define the induced rotational velocities as follows:

There is one rotational inducement in a solitary primary force-centre (star):
The sum of the rotational kinetic energies (KE) in all its satellites (planets)

There are two rotational inducements in an orbiting planet with no dedicated satellites (moons):
The angular velocity of its orbit; plus,
The angular velocity induced by the average potential energy (PE) between itself and its force-centre

There are three rotational inducements in an orbiting planet with dedicated satellites (moons):
The angular velocity of its orbit; plus,
The angular velocity induced by the average potential energy (PE) between itself and its force-centre; plus,
The sum of the rotational kinetic energies (KE) in all its satellites (moons)

Calculation Results

Input values provided by CalQlata
Input values provided by NASA fact sheets
Calculated values

Table 1 shows the relevant properties of the bodies in our solar system:

The Sun3.78412E+349.2182E+451.9885E+306957100000.154739797
Table 1
⁽¹⁾ 'Δ' accommodates the use of the relationship: J = 2/5.m.(ΔR)² for non-homogeneous stars and planets.
If actual values for the polar moment of inertia (J) of a planet are available, this variable is unnecessary
⁽²⁾ Refer to: Venus; Earth; Mars for an alternative/comparative calculation method for polar moment of inertia
⁽³⁾The rotational kinetic energies from 189 moons were used for these calculations.

Table 2 shows the calculated spin properties of the planets in our solar system:

Bodyθωᵢωₒωᵣω [calculated]ω [actual]
The Sun2.86533E-06
Mercury0.0001745N/A8.26678E-07 4.13449E-071.24013E-061.24E-06
Venus3.0962141N/A3.23639E-07-6.22876E-07 -2.99237E-07-2.9924E-07
Earth0.4084077.19215E-051.99099E-07 8.00633E-077.29212E-057.29212E-05
Mars0.4398237.05181E-051.05859E-07 2.58412E-077.08824E-057.08824E-05
Jupiter0.05410521.67468E-041.67849E-08 8.36776E-060.0001758530.000175853
Saturn0.4660031.60756E-046.75905E-09 3.0256E-060.0001637880.000163788
Uranus1.706932-1.01113E-042.36992E-09 -1.26629E-07-0.000101238-0.000101238
Neptune0.4939281.07625E-041.20823E-09 7.12151E-070.0001083380.000108338
Pluto2.138028-1.1382E-058.03026E-10 -4.41609E-09-1.13856E-05-1.13856E-05
Table 2


Claim 1: An orbiting satellite causes its force-centre to spin, not vice-versa

Claim 2: The rotational velocity component that has been induced by the kinetic energy in its satellites, is identical in each layer of a force-centre to that of its core.

Claim 3: The rotational velocity component in each layer of a force-centre that has been induced by the potential energy between itself and its satellites, is according to the polar moment inertia of the layer concerned.

Claim 4: A force-centre is caused to rotate by the difference between; the sum of the kinetic energies generated by all of its orbiting satellites at their perigee, and the sum of the kinetic energies generated by the orbiting satellites at their apogee.
The formula governing this rotation is:
ωᵢ = √[2.KE / J₂] . Sign(Cos(θ))

Claim 5: A satellite orbiting its force-centre synchronously will present the same face to its force-centre causing it to rotate about its axis at the same angular velocity as its orbit according to the relationship:
ωₒ = 2π / tₒ

Claim 6: The average potential energy between a force-centre and a satellite orbiting non-synchronously will induce rotation in the satellite according to the following relationship:
ωᵣ = √[2.PE / J₁] . Cos(θ)

Claim 7: Where an orbiting body, such as a planet, is not orbited by satellites, it will rotate according to the relationship:
ω = ωₒ + ωᵣ
Where an orbiting body, such as a planet, is itself orbited by satellites, it will rotate according to the relationship:
ω = ωᵢ + ωₒ + ωᵣ

Mathematical Symbols & Units

A mass orbiting a force-centre will generate a positive kinetic energy (KE) and a negative potential energy (PE) between the force-centre and the orbiting body. The sum of the two is Newton's combined energy (E). Refer to Laws of Motion.

The potential energy (PE) between two or more bodies is also gravitational energy.

'ΣKEᴾ' is the sum of the kinetic energies of (all) a force-centre's orbiting bodies at their perigee {J}

'ΣKEᴬ' is the sum of the kinetic energies of (all) a force-centre's orbiting bodies at their apogee {J}

'KE' is the difference between 'ΣKEᴾ' and 'ΣKEᴬ' (ΣKEᴾ - ΣKEᴬ) {J}

'PE' is the average potential energy between a force centre and its orbiting body ½(PEᴾ + PEᴬ) {J}

'θ' is the angle of inclination of a satellite relative to the axis of its force-centre {radians}

'J₁' is the polar moment of inertia of the primary force-centre (star) {kg.m²}

'J₂' is the polar moment of inertia of the secondary force-centre (planet) {kg.m²}

'tₒ' is the period (of time) a satellite takes to complete one orbit around its force-centre {s}

'ωᵢ' is the angular velocity induced in a secondary force-centre by its satellites (moons) {radians/second}

'ωₒ' is the angular velocity of a planet's orbit around its force-centre (star) {radians/second}

'ωᵣ' is the angular velocity induced in a planet by its force-centre (star) {radians/second}

'Δ' radial modifier (factor) for the polar moment of inertia of a rotating body

'₁' refers to the primary force-centre (star)

'₂' refers to the secondary force-centre (planet)

For the purposes of this document, the terms 'rotational' and 'angular' are interchangeable; all such velocities shall be interpreted has having magnitude and direction.

Refer to Laws of Motion for a detailed explanation of Newton's laws of planetary motion and Solar System Orbits for planetary orbit details

Refer to CalQlata's Definitions for an explanation of the terminology used in this paper

Chicken & Egg?

It is generally believed that a sun rotates under its own steam pulling its planets around with it.

If this were the case, it would need a suitable energy source to do so, moreover, the same claim must also apply to rotating planets. Whilst this claim may (or may not) be made for our sun and even the earth itself, it cannot be made for planets such as Pluto, which is a solid lump of rock and ice with no internal energy source. Moreover, Pluto's local orbit could not be explained by such an internal energy source.

It is therefore claimed (by the author) that an orbiting body induces spin in its force centre and not vice-versa.

Moreover, it must also be the case that the initiation of a solar system is due to a force induced in matter surrounding an already formed force-centre, that was not initially rotating, causing the matter to orbit the force-centre. Any extant rotational kinetic energy in the orbiting matter will [subsequently] cause the force-centre to rotate.

Rotational kinetic energy is generated in orbiting matter by its orbital angular velocity and its orbiting satellites (moons). It will therefore be some time after the birth of a solar system that a primary force-centre will begin to rotate.

Polar Moment of Inertia (Δ)

The basic formula for the polar moment of inertia (J) of a sphere is:
J = ⅖.m.R²
Where 'm' is the mass of the sphere and 'R' is its radius (Fig 2)

In order to apply this formula accurately, the sphere must comprise the same homogeneous material throughout its structure
Planets, however, are anything but homogeneous and gravitational energy generates very high densities at the surface of their cores (Fig 3)

Various planetary spheres

Fig 2. Planetary Structures

The above formula must therefore be modified to account for this variation in material density. Because we know very little about the structure of the various planets in our solar system, we need to establish a representative value for 'J' for each planet from known data.

This can be done by using a variable 'Δ' in the above formula to provide us with an equivalent representative radius of an homogeneous sphere with the same mass and the correct polar moment of inertia (J). The formula now looks like this:
J = ⅖.m.(Δ.R)²

'Δ' is calculated thus:

A) The angular velocity of a primary force-centre (star) can be calculated as follows:
ω = √[2.KE / J₁]
in which J₁ = ⅖.m.(Δ.R)²

B) The angular velocity of a primary force-centre's satellite (planet) with no satellites (moons) of its own can be calculated as follows:
ω = ωₒ + ωᵣ
ωᵣ = √[2.PE / J₁] . Cos(θ)
ωₒ = 2π / tₒ
in which all of the variables are known

C) The angular velocity of a primary force-centre's satellite (planet) with satellites (moons) of its own can be calculated as follows:
ω = ωₒ + ωᵣ + ωᵢ
in which ωₒ + ωᵣ are calculated as shown in B) above and where ...
ωᵢ = √[2.KE / J₂] . Sign(Cos(θ))

We can now solve for the only unknown in the formula (Δ) from: J₂ = ⅖.m.(Δ.R)²
The calculated variable; 'Δ' for each planet with satellites is provided in the following table.

Δ0.19381970.0004499 0.0033845544 0.004957197 0.0022880510.0025934221.77642672
Table 3


It is assumed that due to its size and volcanic activity that Venus comprises the same basic structure as the earth. Therefore, applying the same calculation method as that adopted for the earth, Venus should look like this:

Inner CoreOuter CoreMantle
Radius (m)149949641707066051800
Density (kg/m³)12348.6723511140.47952345.415752
Mass (kg)1.7454E+233.2314E+241.4614E+24
%age mass3.58%66.32%30.09%
ωₒ (radians/second)3.23639E-073.23639E-073.23639E-07
ωᵣ (radians/second)-2.14926E-07-9.24685E-07-6.22876E-07
ω (radians/second)1.08713E-07-6.01046E-07-2.99237E-07
Table 4

The densities of Venus' inner and outer cores have been estimated as being less than those of the earth according to the ratio of their radii. As seen in the above table, Venus must have a larger core than the earth in order to generate its known angular velocity.

The above angular velocities (ω) show a variance between mantle and core of -4.0795E-07 radians per second, in which the mantle is rotating faster than the core and this variance is a little more than half that of the earth. Contrary to popular belief therefore, Venus must have a magnetic field albeit less than half the earth's strength when taking into account Venus' lesser mass.

Using the right-hand-rule for electro-magnetism, the magnetic North on Venus should point downwards in its axis (which is indeed the case).


These calculations constitute an alternative calculation method for 'ωᵢ' that corroborates the principle method using the radial modifier 'Δ'

The earth comprises 3 principle and fairly distinct layers all of which are permitted to rotate relatively independently due to the molten nature of the outer core:

Inner CoreOuter CoreMantle
Radius (m)123498034349806371000.685
Density (kg/m³)1300011728.081324339.003285
Mass (kg)1.0257E+231.8985E+243.9634E+24
%age mass1.72%31.83%66.45%
J (kg.m²)6.2574E+349.3407E+362.6412E+36
ωᵢ (radians/second)7.192E-057.192E-05 (8.0078E-05)7.192E-05 (6.8806E-05)
ωₒ (radians/second)1.99099E-071.99099E-071.99099E-07
ωᵣ (radians/second)1.28797E-075.54127E-078.00633E-07
ω (radians/second)7.22404E-057.26657E-057.29212E-05
Table 5
The red 'ωᵢ' values are not used as the angular velocity due to KE in all layers is defined by the inner core (Claim 2)

Given the now known and proven values of 'ωᵣ' & 'ωₒ', the unknown velocity requirement for 'ωᵢ' in the mantle (which includes the planet's crust) must be 7.29212E-05 radians per second, which happens to coincide with the calculated value of its core.

It is assumed therefore, that the angular velocity induced in all layers of the earth by the rotational kinetic energy in its satellite (the moon) is defined, by the angular velocity induced in its core (Claim 2).

The above angular velocities (ω) show a variance between mantle and core of 6.71836E-07 radians per second, in which the mantle is rotating faster than the core. Using the right-hand-rule for electro-magnetism, the magnetic North on Earth should point upwards in its axis (which is indeed the case).


Mars is not an iron planet, but is assumed to contain a small amount of iron in its structure given its apparent density. Furthermore, early in its life it must have had an active (hot) interior that contributed towards its volcanic activity but as it is now inactive it probably has no molten outer core.

Mars is, however, a genuine anomaly in that it appears to be rotating much faster than its polar moment of inertia should permit. Towit:
The sun and Mercury together prove the theories for ωᵢ (force-centre) and ωₒ + ωᵣ (satellites)
Venus and the earth prove the application of Cos(θ) to satellites
The earth proves ωᵢ (ωᵢ = ω - ωₒ - ωᵣ) for satellites
All of which provide exact results for the planets concerned.
The gas planets and Pluto confirm the above.

Yet either ωᵢ for Mars must have a smaller polar moment of inertia than can be intimated by its observed angular velocity based upon the available rotational kinetic energy (KE) from Deimos and Phobos or it is under the influence of a greater rotational kinetic energy than can be applied by its two moons.
A smaller polar moment of inertia could be found by reducing its mass, which would not affect the size and shape of its orbit but would result in a considerably lower average density than is currently attributed to it (see Discussion below). Or; a greater rotational kinetic energy would have to be supplied by another moon or orbiting satellite, which has not yet been found, i.e. a substantial comet with a very elongated orbital path.

Whilst either of the above may or may not be the case, it is suggested (by the author), that perhaps there is something a little odd with regard to the properties of Mars.

Using the same calculation method as that adopted for the earth, Mars should look like this:

Radius (m)1518.933389500
Density (kg/m³)7870⁽¹⁾3934.080868
Mass (kg)1.1553E+146.4171E+23
%age mass0.000000018003%<100%
J (kg.m²)1.0661E+201.0694E+35
ωᵢ (radians/second)7.07973E-057.05178E-05 (1.666E-07)
ωₒ (radians/second)1.05859E-071.05858E-07
ωᵣ (radians/second)3.46722E-122.58412E-07
ω (radians/second)7.09031E-057.08824E-05
Table 6
1) If the same calculation technique is used to determine Mars’ core density as that used for Venus, it would be less than 7870 kg/m³ therefore, if Mars has a small iron core, it is assumed that its density must be 7870 kg/m³, however, calculations using numerous alternative values have shown that altering it has negligible effect on the outcome

If Mars does indeed have an iron core, it must be very small given the planet’s average density (3934.081 kg/m³); according to the calculations probably no larger than 3000m in diameter.

See Discussion

Stars, Gas & Ice Planets

The gas planets

Gas pressure variation with planetary radius

Fig 3. Gravity vs Radius

As can be seen in Fig 3 the pressure/density of gas increases exponentially under gravity and inversely proportional with radius indicating that the vast majority of the mass in a gas planet must be at its centre.

Jupiter, for example, comprises hydrogen, helium, ammonia, methane and water vapour. Whilst their relative concentrations are unknown, the average of these densities (>4 kg/m³) at the surface temperature of Jupiter (<127K) multiplied by the planet's volume (1.43128E+24 m³) accounts for only 0.001% of its total known mass. Even taking into account the increase in density with radial depth, more than 99% of Jupiter's mass must be in its core, the dimensions of which are unknown. It is therefore expected that 'Δ' for the gas planets will be very small when compared with the iron and rocky planets and smaller even than the ice planets

Relative Densities

As we can only guestimate the structures of our sun or the ice or gas planets, we can only guestimate their polar moments of inertia. To do this, we may use the known values for ωₒ and ωᵣ in conjunction with the known [surface] angular velocity of each to establish a representative radial modifier 'Δ'

We can then use 'Δ' to estimate the expected surface density for each planet based upon its average density.

For each ‘Δ’ to be representative, it must reflect the structure of the planet concerned. A reasonable estimate can be made from the average densities of each planet.

By way of illustration, it is possible to estimate for most planets from their relative densities
‘ρˢ = Δ.ρᵅ’
Where: ρᵅ is the average density and ρˢ is the surface density

Using this argument for the planets in our solar system with moons, the surface densities of each are estimated as follows:

ΔDensity (kg/m³)
Table 7

Given their respective surface temperatures and despite the unknown nature or composition of each planet’s inner material(s), with the exception of Mars and Pluto, each is representative of its expected surface materials.


Pluto's local orbit due to Charon

Fig 4. Pluto's Local Orbit

Pluto's principal moon; Charon, is so large (>12% of the Pluto's mass) it is pulling Pluto into a local orbit (Fig 4) and is the reason why its effective radial modifier (Δ) is greater than 1

Pluto is the only planet in these calculations with a 'Δ' value greater than 1 and the only planet being pulled by its moon into a significant localised orbit, thereby vindicating a value of; 'Δ>1' and the use of this variable in these calculations.


As stated in Scientific American; the mass of any planet is estimated by its gravitational influence on other bodies. But this raises a couple of problems;

1) As can be seen in Table 8, using Newton’s laws of motion it is possible to put Jupiter, for example, in Earth’s orbit without altering the orbital shape. The observable orbit is not therefore an indicator of a planet’s mass.

PropertyEarth in Earth OrbitJupiter in Earth OrbitUnits
Orbital Shape:
@ Perigee:
@ Apogee:
PE-5.20412966E+33 -1.65619819608E+36N.m
KE2.55858655E+33 8.14262268768E+35N.m
E-2.64554311E+33 -8.41935927310E+35N.m
Table 8
The above calculations have been carried out using Newton’s laws of planetary motion

2) You cannot see a force, nor can you measure it remotely without using known datums against which such forces can be verified.

3) I.e. if you are going to calculate the mass of a planet from the force it exerts on a satellite (or vice versa), you first need to know the properties of the satellite.

4) Accurate mass values could be found from the gravitational acceleration impressed on a man-made orbiting satellite at a known [accurate] distance from the planet’s centre. In fact, the most accurate method of calculating a planet’s mass (without actually standing on its surface) is by measuring its angular velocity and knowing its polar moment of inertia.

5) How can we, therefore, know the mass of any of the planets in our solar system if we don’t know their structure?

Without knowing how to calculate the polar moment of inertia of the sun, it is difficult to be sure of its mass. And without being sure of the sun’s mass it is difficult to then apply mathematical arguments to the estimation of any other masses in the solar system.

As we don’t know the mass of Mars, nor do we know the masses of Phobos and Deimos, and to date there has been no way to calculate the causes of spin in planets, stars, moons, comets, etc. the author suggests that the mass of Mars needs further investigation. After all, one of smallest planets in the solar system with the largest volcano must tell us something. Perhaps much of its internals were blasted out of the planet very early in its life leaving significant internal voids and the planet’s water, having seeped through the body material, may have filled these voids. Subsequent impacts from meteorites, asteroids and or comets may have penetrated the crust releasing this water, which created its surface canals and subsequently seeped back into the planet. This would mean that, perhaps, Mars is not as dense as we might think (seepage); and may contain a larger than normal percentage of water, both of which would significantly affect its mass.

Magnetic Reversal

Because we now know that the sun is responsible for the angular velocity of each layer in the planet, and the planet's orbiting bodies are responsible for the angular velocity of the planet's iron core, it should be possible, from the above theory, to determine the cause of magnetic reversal in the earth.

Further Reading

You will find further reading on this subject in reference publications(55, 60, 61, 62, 63 & 64)

      Go to our store
CalQlata™ Copyright ©2011-2017 CalQlata info@calqlata.com Site Map Terms of website use Our Store