7. Using Kepler’s Rule Newton compares the mean motion of an earth satellite near the earth’s surface and a satellite (the moon) at 60 earth radii.
   
8. Newton uses a pendulum to obtain the orbit of an earth satellite but let’s use the modern value of 5054.75 seconds. By Kepler’s Rule,
   

\[\frac{r^3}{t^2}=\frac{R^3}{T^2}=\frac{1}{(5054.75)^2}=\frac{60^3}{T^2}\]

where, \(r\) is the radius of the earth and it is unity, \(t\) is the mean motion of a satellite at \(r\), \(R\) is the earth-moon distance which is taken as \(60r\), \(T\) is the mean motion at \(R\) which is \[T=27.19\;\text{days}\] and Kepler’s Rule is confirmed.

9. This is it. This proposition proves that Kepler’s Rule is valid for the earth-moon system. No law other than Kepler’s Rule is used in the computations.
   
10. In Newton’s calculations I see no terms representing Newton’s supernatural cause he calls force.
    
11. Next we’ll check Newton’s actual text to expose his sophistry and lies pretending to do calculations with his supernatural cause he calls force.
    
12. Similar calculations existed in Thomas Streete’s Astronomia Carolina where Newton first saw Kepler’s Rule. Newton copied Streete’s calculations but pretended to use his force laws to compute orbits. This is deception and fraud.
    

1. In prop. 4, Newton states the first of the major results which indicate that the system of the world proposed in the Principia is radically different from any in existence.
   
2. This proposition declares that the moon “gravitates” toward the earth and that it is the “force of gravity” which keeps the moon “in its orbit” by always drawing it “back from rectilinear motion.”
   
3. Here Newton is not making a supposition about the nature or mode of action of “gravity”, merely assuming that every reader is familiar with gravity, or weight.
   
   1. This is not true. In fact this is a lie first told by Newton and still repeated by Newton’s apologists. Newton assumes that this “force of gravity” sets distant objects to motion by traveling the distance in between without time passing. This is by definition. So, Newton defines the “nature” and “mode of action” of the supernatural cause he calls “force” or “gravity” or “force of gravity.”
      
   2. How can Cohen say that the mode of action of gravity is not known. It is known. Gravity as a Newtonian jargon is a supernatural agent that has the defining property of timeless travel. If so, what are we talking about? How can Cohen seriously talk about this absurd force holding planets in orbit? This force does not exist in this world. Newton’s explanation of orbits is null and void because it explain a natural phenomenon with a supernatural cause. It is also deception and fraud because Newton hides the fact that he is explaining orbits with a supernatural cause.
      
   3. Are we familiar with “gravity or weight”? I don’t think so. We observe the stone fall and we also know that the stone does not fall because a supernatural cause stationed at the center of the earth (supposedly, Newton never explicitly says so) extends to the stone without time passing and sets it into motion towards the center. If I were to read this story in a fairy tale I would still question it. Gravity that we observe is pure acceleration. There is no force involved. It’s exactly the same thing you experience in an airplane accelerating to take off. This acceleration “pulls” you toward the back of the airplane. Do you conclude from this observation that there is a mass located at the back of the plane and a force emanating from that mass pulls you toward it? No. This is the reason tha acceleration looks instantaneous because there is no attraction.
      
4. The purpose of prop. 4, is twofold.
   
5. Newton wants to show, first, that gravity –terrestrial gravity, the gravity with which we on earth are all familiar– extends out to the moon and does so according to the law of the inverse square; and second, that it is this force of gravity which causes the moon to fall inward from its inertial linear path and so keep in its orbit.
   

[page 205, continued]

1. In this moon test, Newton first computes how far the moon (or an object placed in the moons’s orbit, 60 earth-radii from the earth’s center) would actually fall in one minute if deprived of all forward motion.
   
2. The result is \(15\tfrac{1}{12}\) Paris feet.
   
3. If gravity diminishes by the inverse-square law and if the earth’s gravity extends to the moon, then it follows that a leavy body on the earth’s surface should fall freely in one minute through \(60\times 60\) or \(3,600\) of the above \(15\tfrac{1}{12}\) Paris feet in one second (more accuratly, 15 feet, 1 inch, \(1\tfrac{4}{9}\) lines).
   
4. This result agrees so closely with actual terrestrial experiments that Newton can conclude that the force keeping the moon in its orbit is that very force which “we generally call gravity.”
   
   1. This thing we generally call gravity is not caused by Newton’s supernatural cause he calls force.
      
5. An alternative proof, given in the scholium, is based on consideration of a hypothetical terrastrial satellite.
   
6. Later on, in book 3, prop. 37, corol. 7 (added in the second edition), Newton will present a second such moon test.
   

Newton knew the acceleration due to gravity at the Earth’s surface:
\[ g = 9.8 \, \text{m/s}^2 \]

The Moon orbits the Earth with an approximately circular orbit. The centripetal acceleration \(a_{\text{moon}}\) required to keep the Moon in its orbit is:
\[ a_{\text{moon}} = \frac{v^2}{r} \]

However, instead of velocity \(v\), we can use the Moon’s orbital period \(T\) and orbital radius \(r\) to express \(a_{\text{moon}}\):
\[ a_{\text{moon}} = \frac{4\pi^2 r}{T^2} \]

Newton observed that the gravitational force diminishes with the square of the distance (\( F \propto \frac{1}{r^2} \)). He compared the radius of the Moon’s orbit (\( r_{\text{moon}} \)) to the radius of the Earth (\( r_{\text{earth}} \)):
\[ \frac{r_{\text{moon}}}{r_{\text{earth}}} \approx 60 \]

Thus, the acceleration due to gravity at the Moon’s distance should be reduced by a factor of \( 60^2 = 3600 \):
\[ a_{\text{predicted}} = \frac{g}{3600} \]

The Moon’s orbital period is about \( T = 27.3 \, \text{days} \), and the mean distance from the Earth is \( r_{\text{moon}} = 3.84 \times 10^8 \, \text{m} \). Plugging into the formula for \( a_{\text{moon}} \):
\[ a_{\text{moon}} = \frac{4\pi^2 (3.84 \times 10^8)}{(27.3 \times 24 \times 3600)^2} \approx 0.00272 \, \text{m/s}^2 \]

Newton’s predicted value for \( a_{\text{predicted}} = \frac{9.8}{3600} \approx 0.00272 \, \text{m/s}^2 \) matches the actual centripetal acceleration \( a_{\text{moon}} \). This agreement confirmed that the same gravitational force that causes objects to fall on Earth also governs the Moon’s orbit.

—

Would you like me to expand this into a more step-by-step problem with exercises?

I’m confused about this notation, can you clarify? $a_\text{moon}:a_\text{moon}

Ah, great catch! It looks like the notation I used got a little scrambled. Let me explain it clearly:

The goal is to express the Moon’s centripetal acceleration \( a_{\text{moon}} \) using modern formulas. We can calculate it in two equivalent ways:

1. Using velocity \( v \) and the radius of the Moon’s orbit \( r \):
   \[ a_{\text{moon}} = \frac{v^2}{r} \]
   
2. Using the Moon’s orbital period \( T \) (the time it takes to complete one orbit) and the radius \( r \):
   The formula for centripetal acceleration in terms of \( T \) is:
   \[ a_{\text{moon}} = \frac{4\pi^2 r}{T^2} \]
   

Here:

• \( r \) = radius of the Moon’s orbit (\( 3.84 \times 10^8 \, \text{m} \))
  
• \( T \) = orbital period of the Moon (\( 27.3 \times 24 \times 3600 \, \text{seconds} \))
  

This formula is derived from the relationship between circular motion and the Moon’s angular velocity. Does this version help clarify it? I can also walk you through how the formula \( a_{\text{moon}} = \frac{4\pi^2 r}{T^2} \) is derived if you want!

For any two circular orbits around the same central body (here, the Earth):
\[ \frac{T_1^2}{r_1^3} = \frac{T_2^2}{r_2^3} \]
where:

• \( T \) is the orbital period,
  
• \( r \) is the orbital radius.
  

For objects orbiting the Earth, this constant ratio can be related to the mass \( M_E \) of the Earth and the gravitational constant \( G \):
\[ T^2 = \frac{4\pi^2}{G M_E} r^3 \]

—

A freely falling stone (or an earth-skimming satellite) would have an orbital radius equal to the Earth’s radius \( r_{\text{earth}} \approx 6.37 \times 10^6 \, \text{m} \).

We’ll assume a circular orbit at this radius and aim to find the period \( T_{\text{earth}} \). Once we have the period, we can derive \( g \) by recognizing it as the centripetal acceleration for this “orbit.”

—

The Moon’s orbital radius and period are approximately:

• \( r_{\text{moon}} = 3.84 \times 10^8 \, \text{m} \)
  
• \( T_{\text{moon}} = 27.3 \, \text{days} = 2.36 \times 10^6 \, \text{seconds} \)
  

We’ll apply Kepler’s Rule for the two orbits:
\[ \frac{T_{\text{earth}}^2}{r_{\text{earth}}^3} = \frac{T_{\text{moon}}^2}{r_{\text{moon}}^3} \]

—

Rearranging and substituting the values:
\[ T_{\text{earth}} = T_{\text{moon}} \left( \frac{r_{\text{earth}}}{r_{\text{moon}}} \right)^{3/2} \]

Substitute:
\[ T_{\text{earth}} = (2.36 \times 10^6) \left( \frac{6.37 \times 10^6}{3.84 \times 10^8} \right)^{3/2} \]

Carrying out the calculation:
\[ T_{\text{earth}} \approx 5060 \, \text{seconds} \, (\approx 84.3 \, \text{minutes}) \]

—

The centripetal acceleration for an object in circular orbit is:
\[ a = \frac{4\pi^2 r}{T^2} \]

Substitute \( r = r_{\text{earth}} \) and \( T = T_{\text{earth}} \):
\[ g = \frac{4\pi^2 (6.37 \times 10^6)}{(5060)^2} \]

Calculating:
\[ g \approx 9.8 \, \text{m/s}^2 \]

—

So yes, by applying Kepler’s Rule, you can derive the familiar \( g = 9.8 \, \text{m/s}^2 \). This shows that a freely falling stone is essentially in an extremely short-period orbit around the Earth, just like a satellite that skims the Earth’s surface. Newton’s insight into connecting the motion of terrestrial objects with celestial mechanics was revolutionary!