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# The Laws of Motion (planetary orbits)

Whilst CalQlata has adopted 'U' to symbolise energy, with due respect to Henri Poincaré we have adopted his symbol 'E' for this page

Fig 1. Isaac Newton's Contribution to Science
image created by Eléonore Dixon-Roche and displayed with kind permission from the NSTA

The four principle agents for the theories of planetary motion were Copernicus, Kepler, Galileo and Newton. Between them, they defined the behaviour of orbiting satellites, moons and planets that remain valid 400 years later.

A calculator is now available for Isaac Newton's laws of motion

## Nicolaus Copernicus (1473 to 1543)

Copernicus stated that; contrary to religious doctrine, the sun does not orbit the earth, but all the planets in the solar system orbit the sun. He was so concerned for his safety⁽¹⁾ regarding this claim, however, that he arranged for the publication of his findings to be deferred until after his death (1543).

Fig 2. Principal Theory of Motion

## Johannes Kepler (1571 to 1630)

Kepler used Tycho Brahe’s (1546 to 1601) observational data to show that the planets not only orbited the sun, just as Copernicus had previously claimed, but that their orbital paths were ellipses. Kepler also stated that the time taken to traverse between any two points (see Fig 2) on this elliptical curve is proportional to the swept area:
i.e; t₁/A₁ = t₂/A₂
Whilst he did not provide a mathematical proof for his swept area theory, it was later confirmed by Isaac Newton (below).

## Galilei Galileo (1564 to 1642)

Fig 3. Falling Body = Parabolic Curve

Galileo is best known for his physical evidence of celestial bodies (moons) orbiting other planets, revealed in his book; Dialogue Concerning the Two Chief World Systems (frequently referred to as the 'Dialogue'), therein declaring Copernicus correct and finally quashing over a thousand years of religious dogma that stated all celestial bodies orbit the earth. In return for his findings, he was put under permanent house arrest, but only after being threatened with death if he didn’t recant this claim⁽¹⁾.

However, it was during his confinement that Galileo completed his most important work, his laws of motion, one of which states that a body fired from the surface of the earth would follow a parabolic curve back to its surface.
This claim may be demonstrated by comparing the results from 'best-match' examples of Galileo's parabolic equations with a trajectory calculation⁽²⁾ (Fig 3), where:
v = initial velocity
A = initial horizontal velocity {i.e.; A=v.Cos(α)}
B = offset horizontal distance from t=0
C = initial vertical velocity {i.e.; C=v.Sin(α)}
D = offset vertical distance from t=0

x(t) = A.t + B
y(t) = C.t + D - ½.g.t²

If B and D are zero {i.e. v occurs at t=0}:

x(t) = v.Cos(α).t
y(t) = v.Sin(α).t - ½.g.t²

Whilst the parabolic path is not strictly correct, it is stunningly close as can be seen in the example shown in Fig 3 where two alternative parabolic curves, one of which passes through the same latus rectum and the other being the best parallel match, are superimposed on the path followed by the projectile

## Isaac Newton (1642 to 1727)

Along with [his explanation for] gravity, Newton used [his creation of] calculus⁽³⁾ to mathematically prove the theories previously generated by Copernicus, Kepler and Galileo. In 1687 he published his results under the heading Philosophiæ Naturalis Principia Mathematica (first of three issues), probably the most important scientific work ever produced.

In his Principia, Newton discusses the alternative curves that describe the elliptical paths followed by an orbiting body. However, the parabolic and hyperbolic curves can only be responsible for paths followed by a body (e.g. comet) travelling towards a force centre from well outside its influence, sufficiently close to fall under its influence, pass around the force centre and then travel back out of its influence (see Elliptical Curves). A complete orbit (i.e. that of a planet or moon or a returning comet {e.g. Halley’s}) must be an ellipse.

Fig 4. Gravitational Constant

As a result of this work, Newton defined the fundamental relationship (G) between attracting bodies (Fig 4) in which the gravitational force (F) is directly proportional to the inverse of the square of the distance (R) between the attracting bodies (see The Inverse Square Law below).

Whilst a value for 'G' was never established by Newton, despite it being of special importance to his theories, it has been estimated many times since the publication of the Principia, varying between 6.67E-11 and 6.76E-11 N.m²/kg² (see Gravitational Constant below)

Fig 5. Earth's Orbital Path

The minimum and maximum distances between the earth and sun (Fig 5: @ A & B respectively) are assumed to be as defined in the EarthSky fact sheet⁽⁴⁾. Therefore, using Newton's theories and value of 6.671282E-11N.m²/kg² for 'G', the principal properties of the earth's orbit are as follows:
a = 1.495945981E+11m (R + R)/2)
b = 1.495737135E+11 {√[a².(1-e²)]}
e = 0.01670914665 {a.e² + R.e + R - a = 0}
p = 1.495528319E+11m {a.(1-e²)}
ƒ = 1.47095E+11m {a.(1-e)}
x’ = 2.499598078E+09m {a-ƒ}
F = 3.658178805E+22N to 3.421649078E+22
v = 30286.008788376m/s to 29290.53557m/s
R = 1.47095E+11m to 1.520941962E+11m⁽⁵⁾

Newton’s creation of Calculus allowed him to generate formulas for non-linear versions of Galileo’s relationships for distance (s), time (t), velocity (v) and acceleration (a) (see Calqlata’s velocity and acceleration calculator) as follows:

s = ut + ½at²

δs/δt = v = u + at

δ²s/δt² = δv/δt = a

Fig 6. Proof of an Elliptical Orbit

### Proof (elliptical orbits)

By applying calculus, Newton was able to generate the non-linear formulas necessary to complete his theories concerning the elliptical (conic) path of orbiting bodies, which was proven as follows:

Assume an ellipse and the planet is passing the x-axis @ 'A' (y=0) (Fig 6)

x component = R {a}
y component = v/ω {b}

x(t) = R.Sin(ω.t)
y(t) = (v/ω).Cos(ω.t)

y(t)² / (v/ω)² = 1 - Sin²(ω.t)
Sin²(ω.t) = 1 – y(t)² / (v/ω)²

Fig 7. Euclidean Geometry

x(t)² / R² = Sin²(ω.t) = 1 – y(t)² / (v/ω)²

x(t)² / R² = 1 – y(t)² / (v/ω)²
x(t)² / R² + y(t)² / (v/ω)² = 1 (an ellipse!)

### Equal Area (Euclidean Geometry)

Whilst Kepler had already predicted the equal-swept-area-with equal-orbital-time theory, it had still not been mathematically proven by the time Newton was writing his Principia. Newton did this by using Euclidian geometry.
The areas of each triangle in Fig 7; A₁, A₂ & A₃ are all equal if the base widths; x₁, x₂ & x₃ are equal, which can be proven as follows:

Let y=6 and x₁, x₂ & x₃ all equal 3 (a)
The area of a triangle: A = x.y/2
A₁ = 6 x 3 ÷ 2 = 9
A₂ = 6 x (3+3) ÷ 2 - A₁ = 9
A₃ = 6 x (3+3+3) ÷ 2 - A₁ - A₂ = 9
Therefore, all the areas are equal (i.e. 9)
The same applies to triangles with equal bases between parallel lines (b)

### Proof (conservation of energy & equal time-swept area) {© 05/03/16}

Newton’s proposition diagram for his proof of Kepler’s ‘equal-areas-equal-time’ theory is shown in Fig 8, where the following instructions describe its construction (CalQlata’s words):

Fig 8. Newton's Proof

1) Divide time [of orbit] into equal parts [represented by equal swept areas {triangles}]

2) Assume the line A-B describes the linear path of the body if unconstrained by gravitational attraction

3) The same body would then continue to B-c

4) Assume that the body is attracted by a central-force (S) and diverted from its right line (B-c) in a direction parallel to V-B as far a C

5) Continue to generate similar triangles (S-A-B) following the points D, E, F, etc.

Note: The dimensions L, θ, X₀ & Y₀ in Fig 8 were not part of Newton’s original drawing. They have been added by CalQlata in order to assist with the correlation between Figs 8, 9 & 10

Newton was therefore stating that all swept areas (triangles SAB, SBC, SCD, SDE, SEF, etc.) must be equal.

The difficulty in generating the above diagram is knowing how far along the line C-c that C occurs in order to ensure that each subsequent area remains equal.

Fig 9. Unknown Variables

This can be achieved using the process described in Fig 9, where all the blue variables are entered (X₀, Y₀, L, θ {Fig 8}), all the green variables (X₁, Y₁, R₁, r₂, A₂, ε₂, X₀, α₁) can be easily calculated using the blue variables and the red variables (h, R₂, α₂, X₂, Y₂) may be determined using the formulas provided.

Newton claimed that if you reduce length L 'in infinitum' and join up the dots (X,Y co-ordinates) you produce a curved line thereby demonstrating that a centripetal force is continually acting on the body in the direction of the force centre and the triangular areas will always be proportional to the time passed by the body traversing each triangle; QED

This argument is less easily seen from his words and his fairly simple diagram (Fig 8) than if you actually complete his diagram and repeat it for ever smaller values of L

Fig 10. Newton's Diagram (Fig 8) in Completed Form

Calculations and associated plots were carried out by CalQlata using the following input data:
X₀,Y₀ = 260,0
L = 50
θ = 100°

As can be seen in Fig 10, the diagram does indeed produce a curve, exactly as Newton claimed

... and following this through a sequence of diminishing values for L from 50 to 0.1, the following X,(Y) co-ordinates are achieved immediately prior to reaching 360°:

 L X (Y) 50 119.8970215 (-6.310487628) 25 179.800507 (-11.66860541) 10 229.5181732 (-9.486414926) 1 257.0676077 (-0.349331185) 0.1 259.7089445 (-0.174048539)

Note: the 'Y' co-ordinate is in parenthesis because it is simply a resultant. The trend is demonstrated by the 'X' co-ordinate.

... from which it isn’t difficult to anticipate where X (& Y) will end up if L is diminished in infinitum [i.e.: X(,Y) = 260(,0)], making the final shape a circle and thereby proving that:
a) the path of the body is continuous (conservation of energy and angular momentum)
b) the orbital time passed by the body is proportional to the swept area (triangle)
c) his calculus can be used to determine the properties of the path {'in infinitum'}

This result does not mean that the orbital path is circular, simply that it is continuous.
The orbital path is calculated using the procedure provided in Elliptical Orbit above.

Corollary 1

Newton's first Corollary (to the above proof) states that the velocity of the body (v), represented by L, at positions A, B, C, D, E, F, etc. (Fig 8) is inversely proportional to the perpendicular distance of its tangent from the force centre (Fig 10; p)
Newton also stated that; v multiplied by p is a constant, i.e. his constant of motion (h), which is the angular momentum without the mass component.
Using the above 'in infinitum' argument it can be seen in the following table where these calculations have been carried out for successively reduced values of L between the start and end of the orbit (h₀ @ 0°, h < 360°), 'h' does indeed become a constant:

 L h₀ h 50 2201.214 7485.049 25 580.0837 1196.551 10 95.12179 123.8112 1 0.963516 0.992077 0.1 0.0970 0.0974

Newton's constant of motion 'h' is not to be confused with the perpendicular distance 'h' shown in Fig 9; they are neither the same nor in any way connected.

See Alternative Velocity Calculation (method) below

### Centripetal Force

Centrifugal acceleration (according to Christiaan Huygens {1629 to 1695}): α = R.ω²

Fig 11. Newton's Diagram for Proposition VI

where ω = 2.π/t
a = √[(R.ω²)² + (R.α)²]
with constant angular momentum; α = 0
a = R.ω²
a = R.(2.π/t)²

Centrifugal force:
F = m.a
F = m.R.(2.π/t)² = 4.π².m (R/t²)

Through his inverse rules, Newton shows that the centripetal force (F) between the orbiting body and the force-centre (Fig 11);
F = SP² . QT² / QR

PR = vᴾ . δt
where; vᴾ is the velocity of the body at P and δt is the time taken for the body to travel from P to Q

QR = (Fᴾ / 2m).δt²
where; Fᴾ is the centripetal force on the body at P
F = QR . (2m / δt²)
where; Fᴾ is the centripetal force on the body at P

δt = PR/vᴾ = PR / (h/SY) = PR . SY / h
where h is Newton's constant of motion (see Corollary 1 above)

Therefore, the centripetal force (F) can be calculated as follows:
F = QR . (2.m / (PR . SY / h)²)
= QR . (2.m / (PR² . SY² / h²))
= QR . (2.m.h² / (PR² . SY²))
= QR.2.m.h² / PR².SY²

Newton preferred the calculation in geometric form by setting 2.m.h² as a constant (k):
F = k.QR / (PR.SY)²

See Elliptical Curves for a simple method to determine the angle of a tangent (PY).

### Distance Between Bodies (R)

The separation (distance) between an orbiting body and its force centre, can be found by using general elliptical equation: R = a.(1-e²) / (1+e.Cos(θ))
where 'R' & 'θ' are as shown in (Fig 5) and 'e' is eccentricity

The force centre is not at the centre of an ellipse but at its focus (Fig 11; S)

### The Inverse Square Law

Proposition XI: “If a body revolves in an ellipse; it is required to find the law of the centripetal force tending to the focus of the ellipse

Using similar geometric arguments as above (Figs 7 to 11) Newton worked out that the force between an orbiting body and its force centre is proportional to the inverse of their separation (the distance between them): F ∝ 1/R²
i.e. F = K / R²
where:
the constant of proportionality: K = G.m₁.m₂
i.e. F = G.m₁.m₂ / R²
where G is a constant and m₁ and m₂ are the masses of the force centre and the orbiting body

This same relationship (F ∝ 1/R²) also applies to parabolas and hyperbolas as well as the ellipse

The above constant of proportionality (K) can also been written as; K = m.h²/p
Where 'h' and 'p' are defined in Corollary 1 above and m is the mass of the orbiting body
i.e. F = (m.h²/p).(1/R²)
In the first formula, you can resolve the problem knowing the mass of the bodies
In the second formula, you can resolve it knowing the velocity and mass of the orbiting body and the parameter of its curve (p)

Both of the above F calculations produce the same result;
e.g. the following centripetal force occurs in the earth's orbit, 0.000175° from the major semi-axis:

G = 6.67359232004332E-11 (gravitational constant)
m₁ = 1.9885E+30 (sun mass)
m₂ = 5.964519768E+24 (earth mass)
R = 1.5209420E+11 (distance between mass centres)
F = G.m₁.m₂ / R² = 3.421649078E+22 (centripetal)

h = 4.454920463E+15 (constant of motion - see Corollary 1 above)
m = 5.964519768E+24 (earth mass)
p = 1.495528319E+11 (ellipse parameter)
R = 1.5209420E+11 (distance between mass centres)
F = m.h² / p.R² = 3.421649078E+22 (centripetal)

### Orbital Period

Proposition XV: “The same things being supposed, I say, that the periodic times in ellipses are as the 3/2th power of their greater axes

This means that if the major semi-axis of an ellipse is 'a' (Fig 6) and the time taken for a body to orbit the elliptical path is 'T' then the relationship between the two is:
T ∝ (2.a)¹˙⁵ or T² ∝ (2.a
Therefore; T = K . a¹˙⁵
Where K is the constant of proportionality, which is dependent on the properties of the orbiting body.
This is actually Kepler's third law

## Magnificent Principia

Some of what has been generated above has been inferred from or based upon Colin Pask's "Magnificent Principia"⁽⁵³⁾, an invaluable book that we wholeheartedly recommend.

Everything under this heading down to the foot of this page is CalQlata's own contribution to Newton's magnificent work

### Constant of Proportionality

To determine 'K' (the constant of proportionality for T = K . a¹˙⁵ {Orbital Period above}) ...

K = T² / a³ {s²/m³}
now we know ...
... that the earth travels around the sun in 31558149s
... the earth's semi-major orbital x-axis is 1.495945981E+11m
therefore T² / a³ = 2.974914364E-19 {s²/m³}

G = 6.67359232E-11 {N.m²/kg² = kg.m.m² / s².kg² = m³ / s².kg}
m₁ = 1.9885E+30 kg (the mass of our sun)
1 / m₁.G = 7.535546116E-21 {s².kg / m³/kg = s²/m³}
2.975944645E-19 ÷ 7.538155846E-21 = 39.47841760436
√39.47841760436 = 6.2831853071796 = 2.π

Therefore:
K = (2π)² / G.m₁ = (2π)² ÷ 6.67359232E-11 ÷ 1.9885E+30
= 2.974914364E-19 s²/m³
i.e.;
K = (2π)² / G.mᶠᶜ
where mᶠᶜ is the mass of the force-centre

The above calculation, based upon NASA's data for the sun and the earth's orbit⁽⁴⁾, gives an error margin of 0

### Alternative Velocity Calculation

Fig 12. Alternative Velocity Calculation

A much simpler orbital velocity calculation method is based upon Kepler's 'swept-area = time' rule (Fig 12). Using the earth's orbit as an example:

Earth's total orbital area (A) is 7.029445371E+22m² and it takes 31558149s (T) to complete
The swept area (A) is equal to ½.R x 1m
The velocity of the orbiting body at any given distance between the centres of mass (P & S) is calculated as follows:
v = 2.A / T.R {m² / s.m = m/s}

By way of verification:

The earth's maximum velocity occurs when R = 1.47095E+11 m (@ A)
v = (2 x 7.029445371E+22) ÷ (31558118.4 x 1.47095E+11) = 30286.008788376 m/s
see Isaac Newton (1642 to 1727) above 30286.008788376 m/s (calculated using; h=v.R)

The earth's minimum velocity occurs when R = 1.52054683E+11 m (@ B)
v = (2 x 7.029445371E+22) ÷ (31558149 x 1.52103776E+11) = 29290.53557168 m/s
see Isaac Newton (1642 to 1727) above 29290.53557168 m/s (calculated using; h=v.R)

The above confirms Kepler's 'swept-area = time' rule and shows that v ∝ 1/R

or v = k/R
where k = 2.A / T

Fig 13. Centrifugal Force

### Centripetal Force

In any orbiting system, the centripetal force, i.e. Newton’s gravitational force (Fig 4), must be equal to the orbiting body’s centrifugal force, which may be calculated thus (Fig 13):
F = m₂.v₁² / R
Fc = m₂.v₂² / R
where
v₂ = √[G.m₁ / R]

@ the perihelion (perigee) of an ellipse; Fc = F . ƒ/p = F / (1+e)
@ the aphelion (apogee) of an ellipse; Fc = F . p/ƒ = F . (1+e)

Orbital velocity anywhere in an orbit may be calculated thus:
v₁ = 2π.a.b / R.T
where T is the satellite's orbital period

Centrifugal velocity anywhere in an orbit may be calculated thus:
α = √[⁴/₃.π]
ζ = √[ (ƒ.Sin(θ/2)α + p.Cos(θ/2)α) / (ƒ.cos(θ/2)α + p.Sin(θ/2)α) ]
v₂ = ζ.v₁

Refer to Formulas (below) for definitions of; G, m₁, a, b, R, T, ƒ & p

### The Formulas

The following table comprises the complete set of Newton's formulas, along with the calculation sequence (steps), that you need to determine the properties of any planetary orbit:
{Step 0 = input data}

 Step⁽¹⁾ Variable Formula Units Description Force Centre 0 G Input N.m²/kg² gravitational constant 0 m₁ Input kg mass Orbiting Body 0 m₂ Input kg mass 0 R₂ Input m radius Orbit Shape (Fig 5) 0 T Input s orbit period 1 a ³√[G.m₁ / (2.π/T)²] m major semi-axis 3 b √[a².(1-e²)] m minor semi-axis 2 e⁽²⁾ [-R̂ + √(R̂² - 4.a.{R̂-a})] / 2.a eccentricity 3 p a.(1-e²) m half-parameter 3 ƒ a.(1-e) m focus distance from Peri-helion/gee 4 x' a-ƒ m distance from focus to ellipse centre 4 L π . √[ 2.(a²+b²) - (a-b)² / 2.2 ] m ellipse circumference 1 K (2.π)² / G.m₁ s²/m³ force-centre constant 4 A π.a.b m² orbit total area Body Properties at Perihelion or Perigee 0 R̂ Input m distance between force centre and body 1 F̌ G.m₁.m₂ / R̂² N centripetal force on orbiting body 5 F̌c⁽³⁾ m₂.v̌²/R̂ . ƒ/p © N centrifugal force on orbiting body 1 g -G.m₁ / R̂² m/s² gravitational acceleration on body 5 v̌ h / R̂ m/s body velocity 4 h √[F.p.R̂² / m₂] m²/s Newton's motion constant 2 PE m₂.g.R̂ N.m potential energy 6 KE⁽⁴⁾ ½.m₂.v̌² © N.m kinetic energy (see note 3 below) 7 E PE+KE N.m total energy Body Properties at Aphelion or Apogee 11 Ř⁽⁵⁾ x' + a m distance between force centre and body 11 Ř⁽⁵⁾ (E - ½.m₂.v̂²) / m₂.g m (see note 3 below) 12 F̂ G.m₁.m₂ / Ř² N centripetal force on orbiting body 12 F̂c⁽³⁾ m₂.v̂²/Ř . p/ƒ © N centrifugal force on orbiting body 12 g -G.m₁ / Ř² m/s² gravitational acceleration on body 12 v̂ h / Ř m/s body velocity 8 h h m²/s Newton's motion constant{conservation of angular momentum} 13 PE⁽⁶⁾ m₂.g.Ř N.m potential energy 14 KE⁽⁶⁾ E-PE N.m kinetic energy 8 E E N.m total energy{conservation of energy} Notes: 1) Calculation sequence; variables with the same step (sequence) number can be calculated in any order within that step 2) If e is negative, the values for R̂ and/or T will be incorrect 3) The centripetal forces at the perigee (F̌ & F̌c) should be equal and the centripetal forces at the apogee (F̂ & F̂c) should also be equal. 4) In Newton's laws of motion, KE does not include polar moment of inertia, which is verified thus: Ř/Ř = 1.0 (exactly) 5) Any disparity between these figures is due to incorrect input data, not the calculation method. 6) conservation of energy: δKE + δPE = 0 (i.e. a constant) and KEᴾ-KEᴬ = PEᴾ-PEᴬ 7) KEᴾ+PEᴬ and KEᴾ-KEᴬ are used in spin theory {suffixes ᴾ & ᴬ are used to identify perigee or apogee}

From the above table it can be seen that the shape of the orbit is dependent upon the force-centre only, not the orbiting body. I.e. before the properties of the orbiting body can be calculated, at least one variable from the orbit must be known.
In other words, you cannot predict the aphelion or apogee of an orbiting planet simply from its energy at its perihelion or perigee.

We have used the above formulas to calculate the properties of a number of orbits (below). Where CalQlata's input data is exactly the same as that provided by NASA, CalQlata has used NASA's data.

### The Earth's Orbit

The following table contains the results for the earth's orbit around the sun using the above formulas:

 Variable Calculation Results Units NASA Fact Sheet Force Centre G 6.67359232E-11 N.m²/kg² - m₁ 1.9885E+30⁽⁵⁾ kg 1.9885E+30 Orbiting Body m₂ 5.964519768E+24 kg 5.9723E+24 R₂ 6371000.685 m Orbit Shape T 31558118.4 s 31558118.4 a 1.495945981E+11 m 149.6E+09 b 1.495737135E+11 m - e 0.01670914665 0.0167 p 1.495528319E+11 m - ƒ 1.47095E+11 m 147.09E+09 x' 2.499598078E+09 m - L 9.398649712E+11 m - K 2.974914364E-19 s²/m³ - A 7.029445371E+22 m² - Body Properties at Perihelion R̂ 1.47095E+11 m 147.09E+09 F̌ 3.658178805E+22 N - F̌c 3.658178805E+22 N - g -0.006133232761 m/s² - v̌ 30286.00879 m/s 30290 h 4.454920463E+15 m²/s - PE -5.38099811E+33 N.m - KE 2.73545500E+33 N.m - E -2.64554311E+33 N.m - Body Properties at Aphelion Ř 1.520941962E+11 m 152.1E+09 F̂ 3.421649076E+22 N - F̂c 3.420693766E+22 N - g -0.00573667153 m/s² - v̂ 29290.53556234 m/s 29290 h 4.454920463E+15 m²/s - PE -5.204129659E+33 N.m - KE 2.558586546E+33 N.m - E -2.645543113E+33 N.m -

### The Moon's Orbit

The following table contains the results for the moon's orbit around the earth using the above formulas:

 Variable Calculation Results Units NASA Fact Sheet Force Centre G 6.67359232E-11 N.m²/kg² - m₁ 5.96451977E+24 kg - Orbiting Body m₂ 7.34637741E+22 kg 7.346E+22 R₂ 1737494.51 m Orbit Shape T 2.36059142E+06 s 2360594.88 a 3.83006099E+08 m 3.844E+08 b 3.82284594E+08 m - e 0.0613517612 0.0549 p 3.81564449E+08 m - ƒ 3.59508E+08 m 3.633E+08 x' 2.34980987E+07 m - L 2.4042322E+09 m - K 9.91801091E-14 s²/m³ - A 4.59983612E+17 m² - Body Properties at Perigee R̂ 3.59508E+08⁽⁶⁾ m 3.633E+08 F̌ 2.26251399E+20 N - F̌c 2.26251399E+20 N - g -0.00307976825 m/s² - v̌ 1084.03417 m/s 1076 h 3.89718955E+11 m²/s - PE -8.13391879E+28 N.m - KE 4.31647452E+28 N.m - E -3.81744427E+28 N.m - Body Properties at Apogee Ř 4.06504197E+08 m 4.055E+08 F̂ 1.7693612337E+20 N - F̂c 1.7627012897E+20 N - g -0.00240882409 m/s² - v̂ 958.708317 m/s 964 h 3.89718955E+11 m²/s - PE -7.19355149E+28 N.m - KE 3.37610722E+28 N.m - E -3.81744427E+28 N.m -

### Jupiter's Orbit

The following table contains the results for Jupiter's orbit around the sun using the above formulas:

 Variable Calculation Results Units NASA Fact Sheet Force Centre G 6.67359232E-11 N.m²/kg² - m₁ 1.9885E+30⁽⁵⁾ kg 1.9885E+30 Orbiting Body m₂ 1.89819E+27 kg 1.89819E+27 R₂ 69911000 m 71492000 Orbit Shape T 3.74335690E+08 s a 7.78065217E+11 m 7.7857E+11 b 7.77158824E+11 m - e 0.0482545881 0.0489 p 7.76253488E+11 m - ƒ 7.4052E+11 m 7.4052E+11 x' 3.75452166E+10 m - L 4.88588087E+12 m - K 2.97491436E-19 s²/m³ - A 1.89965903E+24 m² - Body Properties at Perihelion R̂ 7.4052E+11 m 7.4052E+11 F̌ 4.593580846E+23 N - F̌c 4.593580846E+23 N - g -2.41997948E-04 m/s² - v̌ 13705.9020 m/s 13720 h 1.01494946E+16 m²/s - PE -3.40163849E+35 N.m - KE 1.78289158E+35 N.m - E -1.61874691E+35 N.m - Body Properties at Aphelion Ř 8.15610433E+11 m 8.1662E+11 F̂ 3.786686959E+23 N - F̂c 3.777869638E+23 N - g -12444.0470 m/s² - v̂ 12443.981 m/s 12440 h 1.01494946E+16 m²/s - PE -3.08846139E+35 N.m - KE 1.46971448E+35 N.m - E -1.61874691E+35 N.m -

### Halley's Comet Orbit

The following table contains the results for Halley's comet orbiting the sun using the above formulas:

 Variable Calculation Results Units NASA Fact Sheet Force Centre G 6.67359232E-11 N.m²/kg² - m₁ 1.9885E+30⁽⁵⁾ kg 1.9885E+30 Orbiting Body m₂ 2.2E+14 kg 2.2E+14 R₂ 5646.216173⁽⁷⁾ m - Orbit Shape T 2.3998896E+09 s 2399889600 a 2.68515495E+12 m 2.68378579E+012 {17.94 AU} b 6.8051429E+11 m - e 0.967352219 0.967 p 1.72466657E+11 m - ƒ 8.76643518E+10 m 8.76643518E+10 {0.587 AU} x' 2.5974906E+12 m - L 1.15513608E+13 m - K 2.97491436E-19 s²/m³ - A 5.74058927E+24 m² - Body Properties at Perihelion R̂ 8.76643518E+10 m 8.76643518E+10 {0.587 AU} F̌ 3.798935298E+12 N - F̌c 3.798935298E+12 N - g -0.0172678877 m/s² - v̌ 54572.2903 m/s - h 4.78404446E+15 m²/s - PE -3.33031201E+23 N.m - KE 3.27594836E+23 N.m - E -5.43636491E+21 N.m - Body Properties at Aphelion Ř 5.28264556E+12⁽⁸⁾ m - F̂ 1.11791096E+12 N - F̂c 1.019550091E+12 N - g -4.75534821E-06 m/s² - v̂ 905.615265 m/s - h 4.78404446E+15 m²/s - PE -5.52658020E+21 N.m - KE 9.02152909E+19 N.m - E -5.43636491E+21 N.m -

If your calculations result in a positive energy (E) and/or a negative velocity () at your apogee, the body will not complete an elliptical orbit, it will fly off into outer-space. In which case, the orbital shape will be parabolic.

Important: It is a fundamental conclusion of Isaac Newton's laws of motion that if positions 1 and 2 in any orbit are 180° apart ...
... the difference between the velocities of a satellite at these two orbital positions divided by their sum is always equal to the eccentricity of the orbit; i.e. e = (v₁-v₂)/(v₁+v₂)
... the difference between the potential energies of a satellite at position 1 is always equal to minus the difference between its kinetic energies at position 2; i.e. (PE₁ - PE₂) = -(KE₁ - KE₂)
... the potential energy anywhere in a circular orbit is always minus twice the kinetic energy anywhere in the same orbit; i.e. PE = -2.KE
... the total energy (E) of a satellite remains constant anywhere in its orbit

## Swapping Orbits

As can be seen in the Table below, 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.

 Property Earth in Earth Orbit Jupiter in Earth Orbit Units G 6.67359232E-11 6.67359232E-11 N.m²/kg² m₁ 1.9885E+30 1.9885E+30 kg m₂ 5.966585416E+24 1.89819E+27 Kg Rₐ 6371000.685 69911000 m Orbital Shape: T 31558118.4 31558118.4 s a 1.495945981E+11 1.49594598E+11 m b 1.495737135E+11 1.49573714E+11 m e 0.01670914665 0.016709146651 p 1.495528319E+11 1.49552832E+11 m ƒ 1.47095E+11 1.47095E+11 m x' 2.499598078E+09 2.49959808E+09 m L 9.398649712E+11 9.39864971E+11 m K 2.974914364E-19 2.97491436E-19 s²/m³ A 7.029445371E+22 7.02944537E+22 m² @ Perigee: R̂ 1.47095E+11 1.47095E+11 m F̌ 3.658178805E+22 1.16420411E+25 N g -0.006133232761 -0.006133232761 m/s² v̌ 30286.008788376 30286.00879 m/s h 4.454920463E+15 4.45492046E+15 m²/s PE -5.38099811E+33 -1.71248603492E+36 N.m KE 2.735455E+33 8.70550107606E+35 N.m E -2.64554311E+33 -8.41935927310E+35 N.m @ Apogee: Ř 1.520941962E+11 1.52094196E+11 m F̂ 3.421649078E+22 1.08892925E+25 N g -0.005736671536 -0.00573667154 m/s² v̂ 29290.53557 29290.53557 m/s h 4.454920463E+15 4.45492046E+15 m²/s PE -5.20412966E+33 -1.65619819608E+36 N.m KE 2.55858655E+33 8.14262268768E+35 N.m E -2.64554311E+33 -8.41935927310E+35 N.m The above calculations have been carried out using Newton's laws of planetary motion

### Calculating Planetary Mass

So, if you can't determine a satellite’s mass simply from its orbit and its velocity within it (see Swapping Orbits above), how do you calculate the mass of everything in the solar system with only a ball of known mass, a watch and Tycho Brahe's orbital data?

Galileo kicked us off with his discovery of gravitational acceleration, albeit his attempts to find an accurate value were hampered by his inability to measure time accurately.
Kepler gave us the next step with his definition of elliptical orbits and his law of equal orbital time and swept-Area.
Newton put it all together ...

... and it works like this:
By dropping our ball (or rolling it down a slope) and measuring the time it takes to cover a known height, we can find the gravitational acceleration on the surface of our planet, an accurate value for which is now available:
The earth’s mean radius: R₂ = 1.496E+11 m
The earth’s mean gravitational acceleration at sea level: g = 9.80663139027614 m/s²
Newton's gravitational constant: G = 6.67359232004332E-11 m³/kg/s² (see G below)

Using Newton's formula; g = G.m₂/R², we can now determine the earth's mass (m₂)

As we know the earth's orbital shape and its velocity (v) at any point within it, we can accurately determine its centrifugal force (Fc) at its perigee for example, from; Fc = ƒ/p . m₂.v²/R
and because we know that the gravitational force between the earth and sun is exactly the same as the earth's centrifugal force (i.e. they balance each other exactly), we can establish the mass of our sun (m₁) using; Fc = G.m₁.m₂ / R² ⁽⁹⁾
Knowing the mass of the sun and the earth along with the orbital shape of the other bodies in our solar system, we can calculate their masses by comparing force-triangles, one from a planet's theoretical orbit and the other as it passes close to one of which its mass is known, using its orbital deviation and the formulas: g = G.m₁/R² & m₂ = F.R² / G.m₁

So, from measuring the time it takes to drop a ball anywhere on our planet, along with the observed orbits for all the planets and moons in the solar system, Newton has given us all we need to calculate every one of their masses, including that of the sun.

In conjunction with their spin properties, you can also estimate a planet's internal composition.

### G

Newton's gravitational constant (G) has finally been defined.

Note: All the input data in these calculations has been provided by CalQlata's Constants
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 atomic theory
A 'pdf' version of this paper can be found at: G - The Paper
Reference is also made to Planck's Atom

You will see the units for this constant (G) written as: N.m²/kg², which were units of convenience originally assigned to reflect Newton's formula: F = G.m₁.m₂ / R²
because, until now, the formula for 'G' was unknown, hence its units were unknown.

From the relationship between Newton's Atom and Planck's Atom:
FN/FP ≡ Vᵨ.Vₑ = 3E-91 (exact)
where:
FN = G.mᵨ.mₑ/aₒ²
FP = c⁴/G
G = √[3E-91.aₒ².c⁴ / mᵨ.mₑ] = 6.67359232004334E-11 {√[m².m⁴ / s⁴.kg²] = m³ / s².kg}
giving us a value and units for G, moreover;
Vᵨ = mᵨ/ρᵤ
Vₑ = mₑ/ρᵤ
G.mᵨ.mₑ/aₒ² ÷ c⁴/G = mᵨ/ρᵤ . mₑ/ρᵤ = mᵨ.mₑ/ρᵤ²
G.mᵨ.mₑ/aₒ² ÷ c⁴/G = mᵨ.mₑ/ρᵤ²
G² / aₒ².c⁴ = 1/ρᵤ²
G² = aₒ².c⁴/ρᵤ²
G = aₒ.c² / ρᵤ {m⁶/kg/s²}
Because we know the units for G are m³/kg/s², there must be a per unit volume component in this last formula, and given that ρ.V = mass; Newton's gravitational constant must be gravitational acceleration multiplied by an area per unit mass
i.e. G = aₒ.c² / mᵤ {m³/kg/s²}
where:
FN = Newtonian force between a proton and an electron
FP = Planck's force
Vᵨ = volume of a proton
Vₑ = volume of an electron
mᵨ = mass of a proton
mₑ = mass of an electron
mᵤ = unit mass = 1m³ of substance of ultimate density (ρᵤ)
c = speed of light
G = Newton's gravitational constant

G = aₒ.c² / mᵤ
G = 5.2917721067E-11 x 299792459² / 7.12660796350450E+16 = 6.67359232004334E-11

If G = gravitational acceleration times spherical area per unit mass,
the following calculations show that, contrary to popular belief, gravitational force does not simply vary with the square of the distance from the centre of its source
Factors 1.5 & 4 below are exact values and will be explained in due course
G = 1·5.c² / A.ρᵤ = 6.67359232004334E-11 {m²/s² / (m².kg/m³) = m³ / s².kg}
A = 1·5.c² / G.ρᵤ = 2.83458918818674E+10 {m²/s² / (m³ / s².kg . kg/m³) = m²}
R = √[A / 4.π] = 47494.1512680647 {m}
V = ⁴/₃.π.R³ = 4.48754692288540E+14 {m³}
Rs = 2.G.m / c² = 47494.1512680647 {m³ / s².kg . kg / (m²/s²) = m} #
R = Rs
m = R.c² / 2.G = 3.19809876372352E+31 {m . m²/s² / (m³ / s².kg) = kg}
ρ = m/V = 7.12660796350450E+16 {kg/m³}
ρ = ρᵤ
G = Rs.c² / 2.m = 6.67359232004334E-11 {m . m²/s² / kg = m³ / s².kg}
Rs = 2.G.m / c² = 47494.1512680647 {m³ / s².kg . kg / (m²/s²) = m} #
g = G.m / R² = 9.46174592804013E+11 {m³ / s².kg . kg / m² = m/s²}
Fᴺ = G.m² / R² = 3.02595979551312E+43 {m³ / s².kg . kg² / m² = kg.m/s² = N}
Fᴾ = c⁴ / G = 1.21038391820525E+44 {m⁴/s⁴ / (m³ / s².kg) = kg.m/s² = N}
Fᴾ/Fᴺ = 4
R = ⁴√[9.Fᴾ / 16.π².G.ρ².4] = 47494.1512680647 # {⁴√[kg.m/s² / (m³ / s².kg) / (kg²/m⁶]) = m}
# Schwarzschild radius (Rs) of mass 'm'

From the above formulas:
G = g.A / 4.π.m = 6.67359232004334E-11 {m³ / s².kg}
4.π.G = g.A / m = 8.38628344228057E-10 {m³ / s².kg}
This is strong evidence that Fᴺ = 4.π.G.m₁.m₂ / 4.π.R²
In other words; Newton’s gravitational force equation should read: Fᴺ = G.m₁.m₂ / A
in which his gravitational constant would be: G = 8.38628344228057E-10 m³/s²/kg
I.e. gravitational force is constant irrespective of distance from the centre of its source, but a point force at any distance (R) will vary according to its distribution over the spherical area (at R).
This argument upholds the conservation of energy law.

Two aspects of this discovery require further explanation:
1) aₒ = 1·5/A.V (V = 1m³) where A = 2.83458918818674E+10 m²; what is 'A'? > TBA
2) Fᴾ/Fᴺ = 4 > TBA
Neither of which alter the final result

If the relationship between Planck's and Newton's sub-atomic particles is correct:
Vₑ.Vᵨ [Planck] = Vₑ.Vᵨ [Newton] must hold true
where Vₑ & Vᵨ are the force-centre and satellite (e.g. proton & electron) volumes respectively
If G & ρᵤ are as defined above:
Vₑ.Vᵨ [Newton] = Vₑ.Vᵨ [Planck] = 3E-91 {m⁶} (exactly)
If G & ρᵤ are not as defined above:
Vₑ.Vᵨ [Newton] ≠ Vₑ.Vᵨ [Planck] ≠ 3E-91 {m⁶}
Because Vₑ [Planck] = Vᵨ [Planck]; R = ⁶√[9 x 3E-91 ÷ 16π²] = 5.07563837996471E-16 m is
the radius of the Planck particle, it may be inferred that all the above relationships are correct.

And finally; according to Planck: G = 2π.λ².c³ / h = λ².c³ / ħ {m³ / s².kg}
confirming the units for 'G'

In his solution for 'G', Keith Dixon-Roche has fully analysed and linked Planck's atom, Rydberg's atom and Newton's theories all of which complement each other perfectly.
The discovery of so many sub-atomic associations with G means that Newton actually did anticipate both Poincaré and Planck, which probably means his theories can be applied throughout all science, from the largest to the smallest ...
i.e. there is probably no need for a unification theory

### E = m.c²

Combining the theories from Newton, Planck and Poincaré:
Newton: G = aₒ.c² / m {see G above}
Planck: F = c⁴/G
F = E/R = m.c⁴ / aₒ.c²
E = m.c² (Poincaré)

Kristian Huygens gave us the relationship between acceleration and velocity; v² = 2.a.R
and Henri Poincaré showed us that E = m.c² ⁽¹⁰⁾, which today is generally believed to represent kinetic relativism
but now we know that gravitational energy is twice kinetic energy in circular orbits (e.g. in atoms): PE = 2.KE = m.v²

If we assume a limiting gravitational energy that will trap light, it is probably equivalent to that defined by Henri Poincaré, i.e. for any specified mass; m.c² = m.2.g.R
where 'g' is the gravitational acceleration at its outer surface (at radius 'R')

Therefore, for any specific gravitational energy, according to: E = m.c²
we should be able to find the associated limiting mass with respect to its ability to emit light:
c² = 2.g/R → g = c² / 2.R
E = m.g.R → m.R.c² / 2.R → ½.m.c² (kinetic energy at light speed)
i.e. if 2.g.R ≥ c² for a given force centre, light will have insufficient energy to escape its surface.
if g = G.m/R² then 2.G.m/R = c² represents the limiting mass

c² in this famous equation therefore represents a limiting gravitational acceleration that may be used to define the gravitational energy required to trap light, and the formula becomes:
E = m.g.R
where the term 'm.g' refers to the gravitational force on light.

'E' in this formula is not kinetic energy, it is gravitational, i.e. Henri Poincaré's famous formula probably wasn't showing us what has euphemistically become relativism;
between them, Isaac Newton and Henri Poincaré were showing us how to size a black hole!

E = m.c² = m.2.R.g
c² = 2.R.g
g = G.m/R²
c² = 2.R.G.m/R² = 2.G.m/R
R = 2.G.m/c²
If 'm' is the mass of a proton:
R = 2.G.m/c² = 2 x 6.67359232004334E-11 x 1.67262164E-27 ÷ 299792459²
R = 2.48396784934951E-54 m
Which is the Scwarzschild radius of a proton confirming that it cannot trap light; i.e. it is insufficiently dense for its size.

If the above is a correct interpretation of E = m.c², it is probable that mass does not vary with velocity and therefore, the speed of light is not necessarily a limiting velocity for matter.

### What is Gravity?

In his studies of the atom, Keith Dixon-Roche has come to the conclusion that gravity is magnetism.
The basis for this claim will be presented in the atom when published later this year.

### Epilogue

Newton's Principia covered a good deal more than the above, but we have limited our selection to those theories necessary to calculate the behaviour of a single orbiting body.

What makes the above all the more startling is that Newton achieved it with his simple diagram (Fig 8), a new and (as yet) untried mathematical process (calculus) and no computational assistance {none of which can be claimed by CalQlata}
We {@ CalQlata} are suitably impressed

### Notes

1. Giordano Bruno (friar, mathematician and astrologer) was burnt to death in 1600 by the religious authorities for suggesting that the stars were other suns with orbiting planets, and Galileo was forced to recant his findings in order to escape the same fate
2. The projectile is fired from the surface at 67° with an initial velocity of 10.06761m/s. The half-parameter (p) of the parabolic curve with the best parallel match is 1.58m and the half-parameter (p) for the curve that passes through the same co-ordinates at the latus rectum is 1.7m
3. Gottfried Leibniz (1646 to 1716) created his own version of calculus after meeting with Isaac Newton in 1676, during which he may have been given sight of Newton’s work on the subject from about 1666
4. Earth-Sky fact sheet
5. The sun's mass (m₁) has been optimised against the gravitational constant (G) to give a zero error margin for all calculations
6. Figures for March 2016 were selected for this calculation as they represent the lowest error margin (0.033389%) when compared with the expected (theoretical) distance at the apogee
7. Halley's comet is not spherical, it is roughly cylindrical/oblong (like a peanut, potato, etc.). This radius is calculated by converting an equivalent cylinder into a sphere of the same volume. Minor variations in this variable will have no effect on the above results.
8. Neptune's orbit is about 4.5E+012m
9. The calculation for each planetary mass should include the masses of their moons, but because moons tend to be so much smaller than their planets (except Pluto), ignoring them will make little difference to the final result.
10. Henri Poincaré's original formula was: c = √(e/m)

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

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