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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

Isaac Newton's gretest discoveries

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.

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).

The theory of swept area proportional to orbital time

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)

The parabolic curve of a falling body

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.

The formula for gravitational force

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)

Earth's orbital path around the sun

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 = 1.49577429E+11m (R + R)/2)
b = 1.49556914E+11 {√[a².(1-e²)]}
e = 0.016561679 {a.e² + R.e + R - a = 0}
p = 1.49536402E+11m {a.(1-e²)}
ƒ = 1.47100176E+11m {a.(1-e)}
x’ = 2.47725342E+09m {a-ƒ}
F = 3.65792137E+22N to 3.42342763E+22N
v = 30278.03698m/s to 29291.46597m/s
R = 1.47100176E+11m to 1.52054683E+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

Proof on an elliptical orbit

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)

From: Sin²(ω.t) + Cos²(ω.t) = 1

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

Euclidean geometry diagrams for equal areas

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 (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):

Newton's diagram for the proof of orbital motion

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.

Method for finding unknown variables

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

Newtons diagram in complete form

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°:

LX(Y)
50119.8970215(-6.310487628)
25179.800507(-11.66860541)
10229.5181732(-9.486414926)
1257.0676077(-0.349331185)
0.1259.7089445(-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:

Lhh
502201.2147485.049
25580.08371196.551
1095.12179123.8112
10.9635160.992077
0.10.09700.0974

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

Newtons Diagram for his Proposition VI

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 = 6.67128190396304E-11 (gravitational constant)
m₁ = 1.9885E+30 (sun mass)
m₂ = 5.9665854161E+24 (earth mass)
R = 1.52054683E+11 (distance between mass centres)
F = G.m₁.m₂ / R² = 3.4234276256E+22 (centripetal)

h = 4.45390456835E+15 (constant of motion - see Corollary 1 above)
m = 5.9665854161E+24 (earth mass)
p = 1.49536402E+11 (ellipse parameter)
R = 1.52054683E+11 (distance between mass centres)
F = m.h² / p.R² = 3.4234276256E+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.

CalQlata's Contribution {© 05/03/16}

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 1.49577429421E+11m
therefore T² / a³ = 2.975944645E-19 {s²/m³}

G = 6.67128190396304E-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 = 7.5381558461E-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 = 6.2831853071796² ÷ 6.67128190396304E-11 ÷ 1.9885E+30
   = 2.975944645E-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
and is one of the reasons why CalQlata automatically defaults to:
G = 6.67128190396304E-11 N.m²/kg²
for the gravitational constant

Alternative Velocity Calculation

An easier way to calculate orbital velocity

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 7.02784920E+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 = 1.47100176E+11 m (@ A)
v = (2 x 7.02784920E+22) ÷ (31558149 x 1.47100176E+11) = 30278.0369776399 m/s
see Isaac Newton (1642 to 1727) above 30278.0369776401 m/s (calculated using; h=v.R)

The earth's minimum velocity occurs when R = 1.52054683E+11 m (@ B)
v = (2 x 7.02784920E+22) ÷ (31558149 x 1.52103776E+11) = 29291.4659718081 m/s
see Isaac Newton (1642 to 1727) above 29291.4659718082 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

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}

StepVariableFormulaUnitsDescription
Force Centre
0GInputN.m²/kg²gravitational constant
0m₁Inputkgmass
Orbiting Body
0m₂Inputkgmass
0R₂Inputmradius
0tInputsspin period
1J2.m₂.R₂²/5kg.m²polar moment of inertia
Orbit Shape
0TInputsorbit period
1a³√[G.m₁ / (2.π/T)²]mmajor semi-axis
3b√[a².(1-e²)]mminor semi-axis
2e[-R̂ + √(R̂² - 4.a.{R̂-a})] / 2.aeccentricity
3pa.(1-e²)mhalf-parameter
3ƒa.(1-e)mfocus distance from Peri-helion/gee
4x'a-ƒmfocus distance from ellipse centre (Fig 5)
4Lπ . √[ 2.(a²+b²) - (a-b)² / 2.2 ]mellipse circumference
1K(2.π)² / G.m₁s²/m³factor
4Aπ.a.borbit total area
Body Properties at Perihelion or Perigee
0Inputmdistance between force centre and body
1G.m₁.m₂ / R̂²Ncentripetal force on orbiting body
1g-G.m₁ / R̂²m/s²gravitational acceleration on body
5h / R̂m/sbody velocity
4h√[F.p.R̂² / m₂]m²/sNewton's motion constant
2PEm₂.g.R̂N.mpotential energy
6KE½.m₂.v̌² + ½.J.(2π/t)²N.mkinetic energy
7EPE+KEN.mtotal energy
Body Properties at Aphelion or Apogee
11[-b - √(b² - 4.a.c)] / 2.a #mdistance between force centre and body
12G.m₁.m₂ / Ř²Ncentripetal force on orbiting body
12g-G.m₁ / Ř²m/s²gravitational acceleration on body
12h / Řm/sbody velocity
8hhm²/sNewton's motion constant
{conservation of angular momentum}
13PEm₂.g.ŘN.mpotential energy
14KEE-PEN.mkinetic energy
8EEN.mtotal energy
{conservation of energy}
# variables for {from: E = ½.m₂.v̌² + ½.I.(G.m₁/R₂³)² + m₂.g.Ř}
9A½.m₂.h²part formula
9B½.I.G².m₁²part formula (see ## below)
9CG.m₁.m₂part formula
## B = Ř⁴.(E.Ř² + C.Ř - A)
Solve for Ř using: E.Ř² + C.Ř - A = 0
10aEvariable
10bCvariable
10c-Avariable
Notes:
1) variables with the same step (sequence) number can be calculated in any order within that step
2) if e is negative, the values for and/or T will be incorrect

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:

VariableCalculation ResultsUnitsNASA Fact Sheet
Force Centre
G6.67128190396304E-11N.m²/kg²-
m₁1.9885E+30⁽⁵⁾kg1.9885E+30
Orbiting Body
m₂5.966585416E+24kg5.9723E+24
R₂6371000.685m
t86164.1s
J3.6392486951E+36kg.m²-
Orbit Shape
T31558118.4s31558118.4
a1.495773327E+11m149.6E+09
b1.495567334E+11m-
e0.01659564778560.0167
p1.495361368E+11m-
ƒ1.47095E+11m147.09E+09
x'2.482332731E+09m-
L9.397573857E+11m-
K2.975944645E-19s²/m³-
A7.027836155E+22-
Body Properties at Perihelion
1.47095E+11m147.09E+09
3.658178805E+22N-
g-0.006131109419m/s²-
30279.07556m/s30290
h4.45390062E+15m²/s-
PE-5.380998113E+33N.m-
KE2.7351496307E+33N.m-
E-2.645838806E+33N.m-
Body Properties at Aphelion
1.520596655E+11m152.1E+09
3.422438995E+22N-
g-0.00573600939m/s²-
29287.21025333m/s29290
h4.45390062E+15m²/s-
PE-5.204730335E+33N.m-
KE2.558891529E+33N.m-
E-2.645838806E+33N.m-
variables for
A5.918026579E+55-
B3.202222579E+76-
C7.91517917E+44-
a-2.645838806E+33-
b7.91517917E+44-
c-5.918026579E+55-

The Moon's Orbit

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

VariableCalculation ResultsUnitsNASA Fact Sheet
Force Centre
G6.67128190396304E-11N.m²/kg²-
m₁5.9665854161E+24kg-
Orbiting Body
m₂7.34892163E+22kg7.346E+22
R₂1.73749451E+06m
t2360620.8s
J8.87422613E+34kg.m²-
Orbit Shape
T2.36059142E+06s2360594.88
a3.83006099E+08m3.844E+08
b3.82284594E+08m-
e0.06135176120.0549
p3.81564449E+08m-
ƒ3.59508E+08m3.633E+08
x'2.34980987E+07m-
L2.4042322E+09m-
K9.91801091E-14s²/m³-
A4.59983612E+17-
Body Properties at Perigee
3.59508E+08⁽⁶⁾m3.633E+08
2.26329755E+20N-
g-0.00307976825m/s²-
1084.03417m/s1076
h3.89718955E+11m²/s-
PE-8.13673575E+28N.m-
KE4.31796941E+28N.m-
E-3.81876634E+28N.m-
Body Properties at Apogee
4.06504197E+08m4.055E+08
1.77022595E+20N-
g-0.00240882409m/s²-
958.708317m/s964
h3.89718955E+11m²/s-
PE-7.19604278E+28N.m-
KE3.37727644E+28N.m-
E-3.81876634E+28N.m-
variables for
A5.58080283E+45-
B7.03025059E+63-
C2.92522160E+37-
a-3.81876634E+28-
b2.9252216E+37-
c-5.58080283E+45-

Jupiter's Orbit

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

VariableCalculation ResultsUnitsNASA Fact Sheet
Force Centre
G6.67128190396304E-11N.m²/kg²-
m₁1.9885E+30⁽⁵⁾kg1.9885E+30
Orbiting Body
m₂1.89819E+27kg1.89819E+27
R₂69911000m71492000
t35729.856s
J4.15964786E+37kg.m²-
Orbit Shape
T3.74335690E+08s
a7.77975417E+11m7.7857E+11
b7.77073253E+11m-
e0.04814473050.0489
p7.76172136E+11m-
ƒ7.4052E+11m7.4052E+11
x'3.74554168E+10m-
L4.88532993E+12m-
K2.97594464E-19s²/m³-
A1.89923064E+24-
Body Properties at Perihelion
7.4052E+11m7.4052E+11
4.59199054E+23N-
g-2.41914167E-04m/s²-
13702.81125m/s13720
h1.01472058E+16m²/s-
PE-3.40046083E+35N.m-
KE1.78208755E+35N.m-
E-1.61837328E+35N.m-
Body Properties at Aphelion
8.15430834E+11m8.1662E+11
3.78704365E+23N-
g-1.99508145E-04m/s²-
12443.981m/s12440
h1.01472058E+16m²/s-
PE-3.08807216E+35N.m-
KE1.46969888E+35N.m-
E-1.61837328E+35N.m-
variables for
A9.77243119E+58-
B3.66012862E+77-
C2.51810925E+47-
a-1.61837328E+35-
b2.51810925E+47-
c-9.77243119E+58-

Halley's Comet Orbit

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

VariableCalculation ResultsUnitsNASA Fact Sheet
Force Centre
G6.67128190396304E-11N.m²/kg²-
m₁1.9885E+30⁽⁵⁾kg1.9885E+30
Orbiting Body
m₂2.2E+14kg2.2E+14
R₂5646.216173⁽⁷⁾m-
t0snot known
J2.80541862E+21kg.m²-
Orbit Shape
T2.3998896E+09s2399889600
a2.68484505E+12m2.68378579E+012 (17.94 AU)
b6.80474366E+11m-
e0.967348450.967
p1.72466327E+11m-
ƒ8.76643518E+10m8.76643518E+10 (0.587 AU)
x'2.59718070E+12m-
L1.15501026E+13m-
K2.97594464E-19s²/m³-
A5.73958998E+24-
Body Properties at Perihelion
8.76643518E+10m8.76643518E+10 (0.587 AU)
3.7976201E+12N-
g-0.01726191m/s²-
54562.79069m/s-
h4.78321168E+15m²/s-
PE-3.32915904196E+23N.m-
KE3.27480794085E+23N.m-
E-5.43511011092E+21N.m-
Body Properties at Aphelion
5.28202575E+12⁽⁸⁾m-
1.04605987E+09N-
g-4.75481759E-06m/s²-
905.5638704m/s-
h4.78321168E+15m²/s-
PE-5.52531516249E+21N.m-
KE9.02050515674E+19N.m-
E-5.43511011092E+21N.m-
variables for
A2.516702536E+45-
B2.468524580E+61-
C2.91848569E+34-
a-5.43511011092E+21-
b2.9184856945E+34-
c-2.51670253561E+45-

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.

Gravitational Constant

Newton's gravitational constant (G) is variously quoted and still not completely settled.
It is claimed by some, that objects travelling faster than the speed of light can escape from a black hole.
If this claim is correct, gravitational acceleration must be related to the speed of light.
So if, for example, we multiply the gravitational constant (G) by the speed of light (c) we get:
G.c = 6.67128190396304E-11 x 299792459 = 0.020000000067 (almost exactly 1/50)
as either of the two above values (G and/or c) could be slightly incorrect;
for now, we can claim that G.c = 1/50

Moreover, the units of the gravitational constant are N.m²/kg² or m³/s²/kg and the units for the speed of light are m/s

Therefore; m³/s²/kg . m/s also equals; m/s² / kg/m³ . ?/s
that should perhaps be; m/s² / kg/m³ . rad/s

Dirac defined his constant (ħ) as Planck's constant (h) thus: ħ = h / 2π
and Planck's principal values are defined thus:
Planck's time: t = (ħ.G / c⁵)⁰˙⁵
Planck's length; ℓ = (ħ.G / c³)⁰˙⁵
Planck's mass; m = (ħ.c / G)⁰˙⁵
The implication being that 'G' in the above formulas includes rotary motion (radians)
We may therefore conclude that the units for Newton's gravitational constant (G) are in fact: rad/s . m/s² / kg/m³
or; G = ω.g / ρ

We now need to know ...
What does 'ω' (rotational {angular} velocity) represent?
Perhaps, for example, it has something to do with a planet's spin ...
The solution to which will allow us to identify an exact value for Newton's gravitational constant (G) because we already have accurate values for 'g' and 'ρ' for the earth and same formula will apply to all heavenly bodies.

Whilst CalQlata cannot, yet, fully explain this value or the relationship between gravity and the speed of light, it is such a coincidence that CalQlata has settled upon the value; 6.67128190396304E-11 N.m²/kg² for 'G' until/unless we disprove it.

Moreover, if this assumption turns out to be correct, Newton actually anticipated Poincaré's formula 200 years earlier than Poincaré

An explanation for the above will follow as soon as we have established one (see CalQlata's work on the Atom).

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. Henri Poincaré's original formula was: c = √(e/m)
  10. Question: why doesn’t E = ½.m.c² ?
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