iSoul In the beginning is reality

Circular orbits

With circular motion there is a radius and circumference that may be measured as distance or duration. Call the spatial circumference S, and the temporal circumference T, which is known as the period. Distinguish the spatial and temporal versions of the radius, R, by using R and Q. Then S = 2πR and T = 2πQ. Also, R = Qv, and Q = Ru, with velocity, v, and legerity, u.

Circular orbits are a convenient entry into Kepler’s and Newton’s laws of planetary motion. Copernicus thought the orbits were circular, and most planetary orbits are in fact nearly circular. We have then three propositions from the perspective of the Sun toward each orbiting planet:

1. Each planet orbits the Sun in a circular path with radius R.
2. The Sun is at the center of mass of each planet’s orbit.
3. The speed of each planet is constant.

Let’s follow the exposition given in Elements of Newtonian Mechanics by J.M. Knudsen and P.G. Hjorth (Springer, 1995), starting on page 6. Because the speed is constant, the acceleration follows the equation for uniform circular motion derived previously:

a = / R,

in which v is the speed and a is the acceleration. We have from the definition of speed:

v = S / T = 2πR / T.

Elimination of v from these equations leads to

aR / ,

or a ∝ S / .

Kepler’s third law states that the orbital period, T, and the semi-major axis, A, are related as . For circular orbits this becomes

,

or .

Combining this with the equation for acceleration yields

a ∝ 1/,

or a ∝ 1/,

or a ∝ 1/T4/3,

or a ∝ 1/Q4/3.

Inserting the first acceleration into Newton’s second law leads to:

Fm / ,

with force, F, and mass, m. The force is directed toward the Sun, with a magnitude inversely proportional to the square of the distance from the Sun.

Because of Newton’s third law, there is an equal and opposite force from each planet toward the Sun. Which is to say:

FM / .

for mass, M. The combined law of gravitation is thus:

F = GmM/,

for some constant G.

Now consider gravity at the surface of the Earth. Newton showed that the gravitational force of a body would be the same if the mass were concentrated at its center, so the above equation can be used with the radius of the Earth, ρ²:

F = GmM/ρ² = mg,

with g as the acceleration of gravity on Earth. Then

g = GM/ρ².

If the known values of G, M, and ρ are inserted into this equation, the result is g = 9.8 m/s².

Let us now modify this derivation so that distance is the independent variable instead of time. This is better called a law of levitation since it is naturally directed toward the smaller mass. We have then three propositions from the perspective of the Earth, toward each transiting celestial body:

1. Each planet orbits the Sun in a circular path with period T.
2. Each planet is at the center of vass of the Sun’s orbit in 3D time (see here).
3. The pace of each planet is constant.

Because the pace is constant, the expedience follows the equation for uniform circular motion derived previously:

b = / Q,

in which u is the pace and b is the expedience. Again, distinguish the spatial and temporal versions of the radius, R, by using R and Q. Then S = 2πR and T = 2πQ. Also, R = Qv, and Q = Ru, with velocity, v, and legerity, u.

We have from the definition of pace:

u = T / S = 2πQ / S.

Elimination of u from these equations leads to

bQ / ,

or bT / ,

or bT / .

Kepler’s third law states that the orbital period, T, and the semi-major axis, A, are related as . For circular orbits this becomes

,

or ,

or .

Combining this with the equation for expedience yields

b ∝ 1/Q1/3,

or b ∝ 1/T1/3,

or b ∝ 1/S1/2,

or b ∝ 1/R1/2.

Inserting the latter expedience into Newton’s second law in the form of rush, H, gives:

Hn / R1/2,

with rush, H, and vass, n. The rush is directed away from the Sun, with a magnitude inversely proportional to the square root of the distance from the Sun.

Because of Newton’s third law, there is an equal and opposite rush toward each planet from the Sun. Which is to say:

HN / R1/2,

for vass, N. The combined law of levitation is thus:

Γ = LnN/R1/2,

for some constant L. Since this is a function of the inverse square root instead of the inverse square, it is more sensitive to changes in R, i.e., distance above the surface of the Earth.

Consider levity at the surface of the Earth, using the above equation with the radius of the Earth, ρ:

H = LnN/ρ1/2 = nh,

with h as the expedience of levity on Earth. Then

h = LN/ρ1/2.

The values for L, N, and ρ would be inserted into this equation to determine the value of h.