# Newtonian computation of Lagrange point L1

This post is a copy of the computation here. I made the equation pretty and added some comments to accompany my calculation without Newtonian branding. As per Otto’s comment, I now realize that my calculation in that post is wrong and needs to be amended.

1. Derivation of Kepler’s Law

[Comment: This "derivation" is really branding of Kepler's Rule, which is geometric ,with ideological units named after Newton.]

$\\r = \textrm{Earth-Sun distance}\\ R = \textrm{Earth-L1 distance}\\ v = \textrm{Earth's velocity around the Sun}\\ G = \textrm{Gravitational constant}\\ M = \textrm{Mass of the Sun}\\ m = \textrm{Mass of the Earth}\\$

From Newtonian gravity considerations (Section 20 in that site):

$\frac{GMm}{r^2}=\frac{mv^2}{r}$

Eliminate superfluous terms:

$\frac{GM}{r}=v^2$

$\\T = \textrm{Orbital period of the Earth}\\ 2\pi r = \textrm{distance covered by Earth in each orbit}\\$

$\\vT = 2\pi r\\ \\ v = \frac{2\pi r}{T}\\ \\ v^2 = \frac{4\pi^2r^2}{T^2}$

And

$v^2 = \frac{GM}{r} = \frac{4\pi^2r^2}{T^2}$

Rearrange:

$\frac{GM}{r^3} = \frac{4\pi^2}{T^2}$

Here the author says, “from which Kepler’s Third Law for circular orbits follows (just multiply by r3T2)”

$\\r^3T^2\times \frac{GM}{r^3} = r^3T^2\times\frac{4\pi^2}{T^2}\\ \\ \frac{GM}{4\pi^2} = \frac{r^3}{T^2}$

***

So far the author went through elaborate algebraic gymnastics to brand Kepler’s geometric rule written as

$\frac{R_0^3}{T_0^2}=\frac{R^3}{T^2}\\$

with

$\frac{GM}{4\pi^2}=\frac{R^3}{T^2}\\$

where

$\frac{R_0^3}{T_0^2}=\frac{GM}{4\pi^2}$

2. Computation of Langrange point

For reference here’s the diagram again:

There is a spacecraft of mass msc located on the line connecting the Earth and the Sun and it is located at a distance R from the Earth and (r-R) from the Sun. Then the author introduces Force and he bases his argument of bodies acting on each other by this Newtonian force. But eventually, he eliminates all force terms and mass terms and the last operational equation contains no force (and mass enters only as a ratio). His argument goes like this: The force F pulling the spacecraft to the Sun is decreased by the pull of the Earth in the opposite direction:

$F =\frac{GM \: m_\textrm{sc}}{r-R}-\frac{Gm\: m_\textrm{sc}}{R^2}$

Assume that the spacecraft msc represented with its mass in the formulas moves in a circle around the Sun with velocity vsc: It is assumed that msc has centrifugal force balanced by the gravitational attraction of the Earth. Then, if the centrifugal force that arises as the msc orbits around the Sun balances the pull of the Sun. (The author also gives this alternative phrasing: “the attraction supplies the centripetal force”) Of course, this discussion involving the Newtonian force is purely academic and ideological because the operational final formula contains no force term. Basically he replaces, F (the force of the Sun) with

$\frac{m_\textup{sc}\, v_\textup{sc}^2}{r-R}$

to get

$\frac{GM\, m_\textup{sc}}{(r-R)^2}-\frac{Gm\, m_\textup{sc}}{R^2}=\frac{m_\textup{sc}\, v_\textup{sc}^2}{r-R}$

Multiply through by

$\frac{(r-R)}{m_\textup{sc}}$

to get

$\frac{GM}{(r-R)}-\frac{Gm(r-R)}{R^2}=v_\textup{sc}^2$

Let’s call this “Equation A.”

This equation contains only GM which is nothing more than a Newtonian branded unit. There is no force or mass in this formula. Orbits are geometric not dynamical; orbits are geometric, not forceful.

* * *

Now assume that the spacecraft msc also moves in a circle around the Sun or radius (r-R). This requires the Earth to be always in place to pull in the opposite direction from the Sun… (the author is still talking about forces even though he eliminated the force terms from the equations.) The orbital period Tsc of the satellite then satisfies, as before,

$\\v_\textup{sc}\, T_\textup{sc} = 2\pi(r-R)\\ \\v_\textup{sc} = \frac{2\pi(r-R)}{T_\textup{sc}} \\ \\ \\v_\textup{sc}^2=\frac{4\pi^2(r-R)^2}{T_\textup{sc}^2}$

Substitute the value of vsc2 in Equation A:

$\frac{GM}{(r-R)}-\frac{Gm(r-R)}{R^2}=\frac{4\pi^2(r-R)^2}{T_\textup{sc}^2}$

Divide by (r-R)2

$\frac{GM}{(r-R)^3}-\frac{Gm}{R^2(r-R)}=\frac{4\pi^2}{T_\textup{sc}^2}$

which resembles the earlier “Third Law” equation (Equation A above) except that the Earth’s opposing pull (Gm/R2(r-R)) is now added.

* * *

But will the Earth always located where its pull on the satellite is exactly opposite to the Sun? No. Unless the two orbital periods are the same:

$T_\textup{sc}= T$

Only then does the spacecraft’s motion match that of the Earth and the distance between the spacecraft and the Earth stays constant. That in general only happens at one value of R, that is, only at one distance from the Earth and that distance R is now the unknown number we need to calculate. If Tsc = T, the two relations with 4pi2/T2 on the right side are equal to each other, thus providing an equation from which R can be derived. That equation is

$\frac{GM}{(r-R)^3}-\frac{Gm}{R^2(r-R)}=\frac{GM}{r^3}$

Divide by GM to get rid of the superfluous term G

$\frac{1}{(r-R)^3}-\frac{m}{M}\frac{1}{R^2(r-R)}=\frac{1}{r^3}$

Let’s call this “Equation B.” If  we write what Newtonians call “the ratio of masses” in Keplerian notation we see that “mass” involves nothing more than period and radius:

$\frac{m}{M}\equiv \frac{r_0^3}{t_0^2}\times \frac{T_0^2}{R_0^3}$

Zero subscripts indicate that these are constant terms. * * * From here on, he applies algebraic manipulations to rearrange the final equation to compute the value z = R/r.

$\\y = \frac{m}{M}=\frac{3}{1,000,000} = 3\times 10^\textup{-6}\\ \\z = \frac{R}{r}$

Substitute y and Divide Equation B by r3 and then divide both denominator and numerator by r3 and substitute z to get

$\frac{1}{(1-z)^3}-\frac{y}{z^2(1-z)}=1$

He then solves this equation for z = R/r to get

$\frac{\textup{Distance between the Earth and L1}}{\textup{Distance between Earth and Sun}}= \frac{R}{r}= z \approx \frac{1}{100}=0.01$

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