This was derived for circular orbits *here*. In this post it is derived directly from Newtonian gravitation.

The law of gravitation states that the gravitational force is proportional to the product of the masses divided by the square of the semi-major axis, *A*_{s}, of the orbit:

*F = G mM / A _{s}*².

Because of Newton’s second law, the gravitational acceleration is

*a = Gm / A _{s}*².

The object is to transpose this into time-space by expressing it in terms of *modulation* and *vass*). The first step is to use Kepler’s third law, which is an estimate that states:

*T*² ∝ *A*_{s}³, or

*T*^{4/3} ∝ *A*_{s}²,

where *T* is the sidereal period. This may be expressed in terms of the radial time, i.e., the corresponding time to traverse the semi-major axis, *A*_{t} = *t*, estimated as a proportion of the period:

*A _{t}* =

*t*∝

*T*,

so that

*t*^{4/3} ∝ *A _{s}*².

This allows the gravitational acceleration to be expressed in terms of radial time:

*a* ∝ *m**t*^{–4/3}.

The next step is to integrate, with *a* = d*v*/d*t*:

∫ d*v* = ∫ C_{1} *mt*^{–4/3} d*t* = –3 C_{1} *mt*^{–1/3} + C_{2} = *v*.

Then integrate again, with *v* = d*r*/d*t*:

*r* = ∫ d*r* = ∫ –3 C_{1} *mt*^{–1/3} + C_{2} d*t* = –(9/2) C_{1} *mt*^{2/3} + C_{2}*t* + C_{3},

where the radial distance *r* = *A*_{s}. For the purposes of stating a law, set C_{2} = C_{3} = 0, and then solve for *t*:

*t* = –(2/9*m*C_{1})^{3/2} *r*^{3/2} = –C_{4} *ℓr*^{3/2},

where ℓ is the vass. Then take the derivative:

d*t*/d*r* = –(3/2) C_{4} *ℓr*^{1/2} = *u*,

where *u* is the legerity. Take the derivative again to get:

d*u*/d*r* = –(3/4) C_{4} ℓ/*r*^{1/2} = *b*,

where *b* is the *modulation* of levitation, since the sign is the opposite of gravitation. This leads to

*Γ* = *H ℓN*/*r*^{1/2},

for vasses ℓ and *L*, the levitational constant *H*, and the levitational *surge **Γ*.