• Geometric Origin: The \(1/r^2\) dependence isn’t pulled out of thin air. It naturally arises from how a force spreads out in three-dimensional space. Imagine a point source radiating influence equally in all directions—the influence (or flux) passes through spherical surfaces centered on the source. Since the surface area of a sphere grows as \(4\pi r^2\), the intensity of the effect (force per unit area) decreases as \(1/r^2\).
• Empirical Evidence: This \(1/r^2\) law is supported by extensive observations and experiments—from planetary motions described by Kepler’s laws to precise measurements of gravitational forces on Earth (such as in Cavendish’s experiment). It consistently matches what we observe in nature.

• Constants as Scaling Factors: In the expression \(a = \frac{GM}{r^2}\), \(G\) (the gravitational constant) and \(M\) (the mass of the central body) are indeed constants for a given situation. They set the scale of the gravitational field, while the \(1/r^2\) factor tells us how that field’s strength diminishes with distance.
• Physical Interpretation: This means that if you double the distance \(r\) from the mass \(M\), the gravitational acceleration is reduced by a factor of four. It’s a precise quantitative prediction, not an arbitrary choice.

• Describing Forces: The expression \(a = \frac{GM}{r^2}\) is a statement about how the gravitational influence manifests as an acceleration. It tells us that the acceleration is not constant but varies with the distance \(r\) in a very specific way—namely, inversely as the square of \(r\).
• Underlying Physics: This relationship provides a physical framework for understanding gravity as a force that acts at a distance and decreases predictably with distance. It is far from an imaginative embellishment; it’s a core result that emerges from both theoretical reasoning (through the conservation of flux in three dimensions) and experimental validation.

When we say "Newton's inverse-square law," we mean that gravitational acceleration \(a\) is given by

\[ a = \frac{GM}{r^2}, \]

which expresses a very specific, experimentally validated dependence on distance. The \(1/r^2\) term isn’t just an arbitrary function—it’s a direct consequence of the geometry of space and the empirical behavior of gravitational forces. Far from "saving Newton's authority," this law provides a deep insight into how gravity operates in our universe.

Great. But since F is a placeholder and we need to replace it with R/T², I calculated that if we double R, F=R/T² varies as 1/8 not as 1/4 because there is an R in R/T². Does this make sense?

The key point is that in orbital motion the orbital period \(T\) is not independent of the radius \(R\); they are linked by Kepler’s third law. Let’s break it down:

For circular motion, the centripetal acceleration is given (up to the factor \(4\pi^2\)) by

\[ a \sim \frac{R}{T^2}. \]

However, this expression is only complete once you recognize that \(T\) depends on \(R\).

Kepler’s third law tells us that

\[ T^2 \propto R^3. \]

We can write this as

\[ T^2 = k\, R^3, \]

where \(k\) is a constant that depends on the mass of the central object and other factors.

Substitute \(T^2 = kR^3\) into the expression:

\[ \frac{R}{T^2} = \frac{R}{k R^3} = \frac{1}{k\, R^2}. \]

This shows that the acceleration (or the force per unit mass, when multiplied by the appropriate constants) actually varies as \(1/R^2\), not \(1/R\) or \(R\).

Now, if you double \(R\):

• The new radius is \(2R\).
• The new period squared is \(T^2_{\text{new}} = k\,(2R)^3 = 8kR^3\).

Then the new expression becomes

\[ \frac{2R}{T^2_{\text{new}}} = \frac{2R}{8kR^3} = \frac{1}{4kR^2}. \]

Thus, compared to the original

\[ \frac{R}{kR^3} = \frac{1}{kR^2}, \]

the acceleration is reduced by a factor of 4, consistent with the inverse-square law.

If one mistakenly assumed that \(T\) stays fixed when \(R\) is doubled, then \(R/T^2\) would indeed double. However, in orbital dynamics, doubling \(R\) causes \(T^2\) to increase by a factor of 8 (since \(T^2 \propto R^3\)). When you combine this with the doubling of \(R\) in the numerator, you get

\[ \frac{2R}{8T^2_{\text{initial}}} = \frac{R}{4T^2_{\text{initial}}}, \]

which is a quarter of the original value, not one-eighth.

The apparent discrepancy arises if one neglects the relationship between \(T\) and \(R\). Once you account for Kepler’s third law, the expression \(R/T^2\) transforms into a term proportional to \(1/R^2\). Therefore, if you double \(R\), the acceleration (or force per unit mass) indeed decreases by a factor of 4, not 8.

I didn't understand this substitution. We have T² = kR³. And we have R/T² = 1/R². We substitute T² and we get R/kR³ = 1/R². How did you get R/kR³ = 1/kR²?

You're right to ask for clarification. Let's go through the substitution step-by-step carefully.

We start with the given relationships:

1. Kepler's Third Law \[ T^2 = k R^3 \] where \( k \) is some constant.
2. Expression to Evaluate \[ \frac{R}{T^2} \]

\[ \frac{R}{T^2} = \frac{R}{k R^3} \]

Now, let's simplify the right-hand side:

\[ \frac{R}{k R^3} = \frac{1}{k R^2}. \]

This follows from dividing the numerator and denominator by \( R \):

\[ \frac{R}{k R^3} = \frac{1}{k R^2}. \]

Thus, we have:

\[ \frac{R}{T^2} = \frac{1}{k R^2}. \]

—

If you mistakenly assumed that:

\[ \frac{R}{T^2} = \frac{1}{R^2} \]

before substitution, that would be incorrect. The correct result is:

\[ \frac{R}{T^2} = \frac{1}{k R^2}. \]

The extra factor of \( k \) is there because Kepler’s law introduces this proportionality constant. The key takeaway is that \( R/T^2 \) does indeed follow an inverse-square relationship with \( R \), but with an extra constant \( k \).

1. Explicit Definition (\(\equiv\)): Using the identity symbol (\(\equiv\)) can clarify that we are defining a new form of the expression rather than stating a fundamental equality: \[ \frac{R}{T^2} \equiv \frac{R}{k R^3}. \] This makes it clear that the right-hand side is a rewritten version of the left-hand side under a specific assumption.
2. Transformation Notation (\(\rightarrow\)): If we want to indicate that we are performing a step in a derivation rather than stating an equality, we can use the right arrow: \[ \frac{R}{T^2} \rightarrow \frac{R}{k R^3}. \] This suggests that we are rewriting the left-hand side in a new form rather than asserting that both sides are strictly equal in all contexts.

• If you're in a mathematical context where the expression is an explicit algebraic identity, using \(\equiv\) might be clearer.
• If you're showing a logical progression or a transformation, the arrow notation (\(\rightarrow\)) is preferable.

In formal derivations, most physicists and mathematicians still use the standard equals sign, but I agree that using \(\equiv\) or \(\rightarrow\) in explanatory contexts can help prevent confusion.

Great, this makes sense. But I still have a question. First, as with Newton, I write Kepler's Rule with a constant term as R₀³/T₀² = R³/T². So, R₀³/T₀² would be your k. Then, I write R₀³/T₀² = R/T² = 1/R². From this it is clear that as T and R vary R/T² = 1/R² will stay constant. But you and Newtonians are reading this same proportionality as "R/T² varies as 1/R²". Can you clarify the difference between these two ways of reading the same proportionality? So, what is the difference between, "R/T² = 1/R² stays constant as T and R varies" and "R/T² varies as 1/R²". To me they don't seem to be saying the same thing.

You are treating the equation \[ \frac{R}{T^2} = \frac{1}{R^2} \] as a strict equality that holds for all values of \( R \) and \( T \). That is, you are asserting that as \( R \) and \( T \) change, the numerical value of \( R/T^2 \) remains fixed—it does not change. This means that, at every instant, \( R/T^2 \) always has the same value, which is \( 1/R^2 \).

This suggests that instead of saying "it varies," we should say that it is constant and equal to a function of \( R \).

The Newtonian way of reading the same relationship is different. When Newtonians say, \[ \frac{R}{T^2} \propto \frac{1}{R^2}, \] they are emphasizing how \( R/T^2 \) changes as \( R \) changes, rather than treating it as something constant.

In proportionality statements like this, "varies as" means that if \( R \) increases or decreases, then \( R/T^2 \) will change in the same way as \( 1/R^2 \). This does not necessarily mean that the value of \( R/T^2 \) itself is fixed at every instant—it just means that whatever happens to \( R \), the quantity \( R/T^2 \) responds according to \( 1/R^2 \).

—

• Your interpretation ("stays constant"): You are reading the equation as a strict equality, meaning \( R/T^2 \) has a fixed numerical value that does not change.
• The Newtonian interpretation ("varies as"): They are emphasizing the relationship—how \( R/T^2 \) responds to changes in \( R \)—rather than treating it as a fixed quantity.

—

Both interpretations can be valid, but they emphasize different things:

• If \( R/T^2 \) is truly constant, then its numerical value does not change.
• If we are just describing the dependency, then saying "varies as" is a way of capturing how it scales.

So, the question is whether \( R/T^2 \) remains literally the same value for all \( R \) and \( T \) or whether we are simply asserting a proportionality.

Would you say that your perspective assumes a deeper structural necessity to this relationship, rather than just an empirical pattern?

You wrote Kepler’s rule in the form:

\[ \frac{R}{T^2} = \frac{1}{R^2} \]

This suggests that \( R/T^2 \) is proportional to \( 1/R^2 \), meaning there exists some constant \( k \) such that:

\[ \frac{R}{T^2} = k \cdot \frac{1}{R^2} \]

which can be rewritten as:

\[ k = \frac{R}{T^2} \cdot R^2. \]

If \( k \) is a constant, this would mean:

\[ k = \frac{R \cdot R^2}{T^2} = \frac{R^3}{T^2}. \]

This actually aligns with the traditional formulation of Kepler’s Rule:

\[ \frac{R^3}{T^2} = k. \]

So, algebraically, your expression should be written as:

\[ \frac{R}{T^2} = \frac{1}{R^2} \quad \Rightarrow \quad k = \frac{R^3}{T^2}. \]

But your original formulation \( k = R/T^2 = 1/R^2 \) is incorrect because it incorrectly sets \( R/T^2 \) equal to \( 1/R^2 \) without the necessary proportionality constant.

Would you agree that \( k = R^3/T^2 \) is the correct way to write it?

So, we agree that k=R³/T² is Kepler's Rule and it is correct to write it this way. But when we transform R³/T² algebraically and write it as R/T² = 1/R², we can no longer write k=R/T² = 1/R² because this way of writing implies that k=R/T² and k=1/R² which is incorrect. Any comments on this?

You're absolutely right. The correct expression from Kepler’s Rule is

\[ k = \frac{R^3}{T^2}, \]

which means that

\[ \frac{R}{T^2} = \frac{k}{R^2}. \]

This tells us that \(R/T^2\) is proportional to \(1/R^2\) with the constant of proportionality \(k\).

If one were to write

\[ \frac{R}{T^2} = \frac{1}{R^2}, \]

that would implicitly assume \(k = 1\), which isn’t generally true. Therefore, writing

\[ k = \frac{R}{T^2} = \frac{1}{R^2} \]

is misleading because it suggests that \(k\) equals both \(R/T^2\) and \(1/R^2\) independently, which conflates the proportionality factor with the variables.

The correct interpretation is:

• Kepler’s Law in full: \[ k = \frac{R^3}{T^2}, \]
• Expressing the dependency: \[ \frac{R}{T^2} = \frac{k}{R^2}. \]

This shows that as \(R\) and \(T\) vary, the quantity \(R/T^2\) scales with \(1/R^2\), but with the constant \(k\) explicitly present.

So, your observation is spot on: while it’s acceptable to say “\(R/T^2\) varies as \(1/R^2\)” (meaning that the scaling behavior is \(1/R^2\)), it is incorrect to then equate \(k\) directly to both \(R/T^2\) and \(1/R^2\) unless \(k\) is specifically normalized to 1.

This distinction is important to maintain proper dimensional consistency and clarity in the relationship.

Great. This makes it clearer the function of k. But I still don't understand how R/T² varies as k/R². Because I read this proportionality as "as R and T varies the proportionality R³/T² = k stays constant." Can you make the distinction between my reading that the proportionality stays constant and you reading of R/T² varying as k/R²?

You're noticing a subtle but important point. Let's break it down:

1. Kepler’s Third Law: Kepler’s law is usually expressed as \[ \frac{R^3}{T^2} = k, \] where \( k \) is a constant. This means that for any orbit, regardless of its size, the combination \( R^3/T^2 \) remains fixed.
2. Deriving \( R/T^2 \): From Kepler’s law, if you solve for \( R/T^2 \), you get \[ \frac{R^3}{T^2} = k \quad \Longrightarrow \quad \frac{R}{T^2} = \frac{k}{R^2}. \] This tells us that \( R/T^2 \) is not a constant by itself; rather, it changes inversely with \( R^2 \).
3. Two Readings:
   • Your Reading: You’re focusing on the full proportionality \( R^3/T^2 = k \). This expression stays constant as \( R \) and \( T \) vary. In other words, for different orbits, if you calculate \( R^3/T^2 \), you always get the same number \( k \).
   • The "Varies as" Reading: When I say \( R/T^2 \) varies as \( k/R^2 \), I mean that if you isolate the ratio \( R/T^2 \), it is not constant; its value depends on the radius \( R \). Specifically, as \( R \) increases, \( R/T^2 \) decreases following the rule \[ \frac{R}{T^2} = \frac{k}{R^2}. \] This is a statement about how one part of the full relationship behaves on its own.

4. Reconciling the Two Views: Both views are correct but they focus on different aspects:

   • The full combination \( R^3/T^2 \) is constant (by Kepler's law).
   • However, the ratio \( R/T^2 \) depends on \( R \) as \( k/R^2 \).

   In other words, while the product \( R^3/T^2 \) remains the same for all orbits, the term \( R/T^2 \) must change with \( R \) to ensure that when multiplied by \( R^2 \) (i.e., \( R/T^2 \times R^2 \)), you recover the constant \( k \).

5. Summary:

   • "The proportionality stays constant": This refers to \( R^3/T^2 = k \), which does not change with \( R \) or \( T \).
   • "R/T² varies as k/R²": This indicates that if you look at the ratio \( R/T^2 \) by itself, its value changes as \( 1/R^2 \) (with the constant \( k \) scaling it).

   They are two sides of the same coin: one is the complete, constant relationship; the other is a derived, variable-dependent expression.

I hope this clarifies the distinction between the two interpretations!

1. Kepler's Law: We have \[ \frac{R^3}{T^2} = k, \] which implies \[ T^2 = \frac{R^3}{k}. \] This tells us that if you change \(R\), \(T\) will adjust accordingly.
2. Expressing \(R/T^2\): Substituting \(T^2 = R^3/k\) into the ratio \(R/T^2\) gives: \[ \frac{R}{T^2} = \frac{R}{R^3/k} = \frac{k}{R^2}. \] This shows that the ratio \(R/T^2\) is completely determined by \(R\) (with \(k\) as a constant).
3. Interpreting "Varies as": When we say “\(R/T^2\) varies as \(1/R^2\) (or \(k/R^2\)),” we mean that if you consider a family of orbits with different radii \(R\), the value of \(R/T^2\) will change in proportion to \(1/R^2\).
   • For example, if you double \(R\), then \[ \frac{k}{(2R)^2} = \frac{k}{4R^2}, \] which is one quarter of the value you had before.

4. Constant vs. Varying Quantities:

   • \(R^3/T^2 = k\) is constant: No matter which orbit you consider (provided the central mass remains the same), the combination \(R^3/T^2\) is fixed.
   • \(R/T^2\) is not constant by itself: It changes with \(R\) according to the relationship \(R/T^2 = k/R^2\).

   Although \(T\) changes with \(R\), once you substitute for \(T\) using Kepler’s law, the dependency of \(R/T^2\) on \(R\) becomes explicit.

• Your Reading: "The proportionality \(R^3/T^2 = k\) stays constant." This means that for any valid orbit, when you compute \(R^3/T^2\), you'll always get the constant \(k\).
• The "Varies as" Reading: "The quantity \(R/T^2\) varies as \(k/R^2\)." This is a statement about how \(R/T^2\) depends on \(R\): if you compare orbits with different radii, then \(R/T^2\) will take on values proportional to \(1/R^2\).

Even though \(R\) and \(T\) are not independent (because Kepler’s law ties them together), once you choose an orbit with a given \(R\), \(T\) is fixed and so is \(R/T^2\). When considering different orbits, the change in \(R/T^2\) with \(R\) follows the rule \(R/T^2 = k/R^2\).

I hope this helps clarify the distinction between saying “the combination \(R^3/T^2\) stays constant” and “\(R/T^2\) varies as \(1/R^2\)”—they’re simply two different ways of describing the same interdependent relationship.

We can also start from the relation R/T² = k/R² to obtain that as we double R, T varies 8 fold, right?: R/T² = k/R² 2R/T² = k/4R² 8R/T² = k/R² or 8R³/T² = k This seems to say that as we double R the satellite moves 8 times slower. Is this correct? Can you give an actual example with real numbers?

Let's clarify the derivation and meaning step by step.

Kepler's third law is usually written as

\[ \frac{R^3}{T^2} = k, \]

where \(k\) is a constant (for a given central mass). This means that for any orbit,

\[ T^2 = \frac{R^3}{k}. \]

From the above law, if you isolate \(R/T^2\) you get

\[ \frac{R}{T^2} = \frac{R}{R^3/k} = \frac{k}{R^2}. \]

This tells us that \(R/T^2\) is not itself constant; it changes with \(R\) according to the relation

\[ \frac{R}{T^2} \propto \frac{1}{R^2}. \]

Let’s examine what happens when we double \(R\).

1. Original Orbit: Let the original radius be \(R\) and period be \(T\). Then

   \[ k = \frac{R^3}{T^2}. \]

2. New Orbit with \(R' = 2R\): Let the new period be \(T'\). Then Kepler's law gives

   \[ k = \frac{(2R)^3}{T'^2} = \frac{8R^3}{T'^2}. \]

   Equate this with the original expression for \(k\):

   \[ \frac{R^3}{T^2} = \frac{8R^3}{T'^2}. \]

   Cancel \(R^3\) (assuming \(R \neq 0\)):

   \[ \frac{1}{T^2} = \frac{8}{T'^2}. \]

   Solve for \(T'^2\):

   \[ T'^2 = 8\, T^2 \quad \Longrightarrow \quad T' = \sqrt{8}\, T = 2\sqrt{2}\, T \approx 2.83\, T. \]

This tells us that when \(R\) doubles, the period increases by a factor of about 2.83—not by 8.

The orbital speed \(v\) is given by

\[ v = \frac{2\pi R}{T}. \]

• Original Speed: \(v = \frac{2\pi R}{T}\).
• New Speed: \(v' = \frac{2\pi (2R)}{T'} = \frac{4\pi R}{2\sqrt{2}\, T} = \frac{2\pi R}{\sqrt{2}\, T} = \frac{v}{\sqrt{2}}.\)

So, when \(R\) doubles, the orbital speed decreases by a factor of \(1/\sqrt{2} \approx 0.707\).

Suppose we choose:

• Original orbit: \(R = 1\) unit, \(T = 1\) unit of time. Then \[ k = \frac{1^3}{1^2} = 1. \] The orbital speed is \[ v = \frac{2\pi \cdot 1}{1} = 2\pi \approx 6.28\, \text{units/time}. \]
• New orbit: \(R' = 2\) units. Then the new period is: \[ T' = \sqrt{\frac{(2)^3}{1}} = \sqrt{8} \approx 2.83\, \text{units of time}. \] The new orbital speed is: \[ v' = \frac{2\pi \cdot 2}{2.83} \approx \frac{12.57}{2.83} \approx 4.44\, \text{units/time}. \]

Notice:

• The period increased from 1 to about 2.83 (a factor of \(2\sqrt{2}\)).
• The orbital speed decreased from about 6.28 to 4.44 (a reduction factor of roughly \(1/\sqrt{2} \approx 0.707\)).

When you write

\[ \frac{R}{T^2} = \frac{k}{R^2}, \]

this is simply a rearrangement of Kepler's law. It tells us how the combination \(R/T^2\) depends on \(R\). It does not mean that if you double \(R\) the quantity \(R/T^2\) itself automatically increases or decreases by a factor of 8 in an isolated sense. The key is that \(T\) is also changing with \(R\).

• Your Reading: You see \(R^3/T^2 = k\) as a constant relation that holds for all orbits. That’s correct.
• The "Varies as" Reading: When we say \(R/T^2\) varies as \(k/R^2\), we mean that if you examine different orbits, the numerical value of \(R/T^2\) will be determined by \(R\) according to \(k/R^2\).

There is no situation in which doubling \(R\) directly causes \(R/T^2\) to change by a factor of 8 because \(T\) adjusts in such a way (by increasing by a factor of \(\sqrt{8}\)) that the overall relation \(R^3/T^2\) remains constant.

• No, the satellite does not move 8 times slower. Doubling \(R\) leads to the period \(T\) increasing by a factor of \(2\sqrt{2} \approx 2.83\) (so the orbit takes 2.83 times as long), and the orbital speed decreases by a factor of \(1/\sqrt{2} \approx 0.707\).
• Actual Example: With \(R=1\) and \(T=1\), doubling \(R\) to 2 gives \(T \approx 2.83\) and \(v\) decreases from about 6.28 to about 4.44.

Your derived algebraic steps need to account for the fact that \(T\) is not independent of \(R\); its change ensures that \(R^3/T^2\) stays constant.

I know you don't draw pictures, but can you recommend a source where I can find a geometric representation of the acceleration R/T²?

I understand you're seeking a visual representation of the acceleration \( R/T^2 \) in the context of orbital motion. While I cannot provide images directly, I can guide you to resources that offer geometric illustrations of centripetal acceleration, which is mathematically expressed as \( a_c = \frac{v^2}{r} \) and is equivalent to \( R/T^2 \) in uniform circular motion.

Recommended Resource:

• Wikipedia's "Centripetal Force" Article: This entry provides detailed explanations and diagrams illustrating centripetal acceleration. The section titled "Derivation" includes vector diagrams that depict how the velocity vector changes direction, leading to centripetal acceleration directed toward the center of the circular path. citeturn0search12

Key Concepts Illustrated:

• Velocity and Acceleration Vectors: Diagrams show that, in uniform circular motion, the velocity vector is tangent to the path, while the acceleration vector points radially inward, perpendicular to the velocity.
• Derivation of Centripetal Acceleration: Geometric derivations demonstrate how the change in the velocity vector's direction, while maintaining constant speed, results in an inward acceleration of magnitude \( a_c = \frac{v^2}{r} \).

These visual aids can help you understand how the acceleration \( R/T^2 \) manifests in circular motion, providing a clearer geometric perspective on the concept.

Can you think of any other examples of a proportionality where the same term is on both sides? Despite all your explanations this does not make sense to me. It seems to be there is a play on words here. If these two are proportional the proportionality will stay constant as R changes.

One way to see this is to consider a purely algebraic example. Suppose we have a relationship

\[ \frac{x^3}{y^2} = C, \]

with \(C\) a constant. Then we can rearrange this to

\[ \frac{x}{y^2} = \frac{C}{x^2}. \]

Here, the variable \(x\) appears on both sides. Notice that:

• The original relation, \(x^3/y^2 = C\), tells you that as \(x\) and \(y\) change, the combination \(x^3/y^2\) stays constant.
• But if you isolate \(x/y^2\), you get an expression that depends on \(x\) as \(C/x^2\). That means the value of \(x/y^2\) will vary with \(x\) even though the whole combination \(x^3/y^2\) remains fixed.

This is exactly what happens with Kepler’s law:

• We start with \[ \frac{R^3}{T^2} = k, \] which stays constant as \(R\) and \(T\) vary together.
• Then, isolating \(R/T^2\) gives \[ \frac{R}{T^2} = \frac{k}{R^2}. \] Here, even though \(k\) is constant, the expression \(R/T^2\) is not constant—it depends on \(R\) through the factor \(1/R^2\).

So, while the overall proportionality \(R^3/T^2 = k\) remains constant, the sub-expression \(R/T^2\) “varies as” \(k/R^2\). In other words, if you compare two orbits with different radii, the value of \(R/T^2\) in each orbit will be different, and the way it differs is exactly by a factor of \(1/R^2\).

This isn’t a play on words but simply a result of algebraic manipulation: the fact that the same variable \(R\) appears on both sides of the derived relation shows how a part of the relationship (here, \(R/T^2\)) scales with \(R\) when the whole combination \(R^3/T^2\) is fixed.

If you’d like to see more visual or geometric illustrations of such relationships, I recommend looking into textbooks or online resources on orbital mechanics and proportional reasoning. For instance, the Wikipedia article on [Kepler's laws](https://en.wikipedia.org/wiki/Kepler%27s_laws) and [centripetal force](https://en.wikipedia.org/wiki/Centripetal_force) often include diagrams that show how these scaling laws come about.

Does this help clarify the difference between saying the overall proportionality is constant and the fact that a derived term can still vary with a variable like \(R\)?

I understand, R/T² is not constant. I don't understand how it change as 1/R² since R/T² also has an R in it. I mean, R/T² cannot change as 1/R² because it has an R in it.

The key is to remember that \(T\) is not independent of \(R\); they are linked by Kepler's third law. Let me walk through the steps:

1. Start with Kepler’s Third Law: \[ \frac{R^3}{T^2} = k \quad \Longrightarrow \quad T^2 = \frac{R^3}{k}. \]
2. Substitute into \(R/T^2\): Replace \(T^2\) in the expression: \[ \frac{R}{T^2} = \frac{R}{\frac{R^3}{k}}. \]
3. Simplify the Expression: This simplifies as follows: \[ \frac{R}{\frac{R^3}{k}} = \frac{R \cdot k}{R^3} = \frac{k}{R^2}. \] Notice that the \(R\) in the numerator cancels with one factor of \(R\) in \(R^3\) (which is \(R \times R^2\)), leaving you with \(1/R^2\).
4. Interpretation: Even though the original term \(R/T^2\) appears to have an \(R\) in it, the fact that \(T^2\) scales as \(R^3\) means that when you substitute, the overall dependency on \(R\) becomes \(\frac{k}{R^2}\). That is why we say \(R/T^2\) varies as \(1/R^2\) (with \(k\) as the constant of proportionality).

In summary: The presence of \(R\) in the numerator of \(R/T^2\) does not contradict the fact that, once you substitute \(T^2 = R^3/k\), the expression simplifies to \(k/R^2\). This shows that as \(R\) increases, \(R/T^2\) decreases proportionally to \(1/R^2\).

Ok, I'll read this now. But before reading, I know T is not independent of R. This is why it is silly to say that R/T² varies is 1/R². They are not independent of each others. But let me read your answer first, it looks promising.

Yes, I see your concern. The phrase "varies as 1/R²" might sound misleading if taken to imply independent variation. What it really means in this context is that as R changes, R/T² changes in a way that follows the 1/R² proportionality, given the underlying constraint \( T^2 \sim R^3 \). The dependence is baked into the math, and we're not treating \( R \) and \( T \) as independent variables. Let me know if my explanation helps clarify this!