Table of Contents
Principia - I.4 Cor. 6
If the periodic times be in the sesquiplicate ratio of the radii, and therefore the velocities inversely in the subduplicate ratio of the radii, the centripetal forces will be inversely as the squares of the radii, and conversely.
Why are velocities inversely proportional to the square root of radii
[chatgpt]
Take a look at the Corollary 6 of the 4. Proposition of Book 1 in Newton’s Principia:
If the periodic times be in the sesquiplicate ratio of the radii, and therefore the velocities inversely in the subduplicate ratio of the radii, the centripetal forces will be inversely as the squares of the radii, and conversely.
Here, “If the periodic times be in the sesquiplicate ratio of the radii,” simply means Period T is proportional to the 3/2 power of the radius R.
And, “the velocities inversely in the subduplicate ratio of the radii,” means that velocity V is proportional to the square root of the radius.”
Correct?
Now, explain why velocities are proportional to the square root of the radii.
Let’s start with the relationship between the period \(T\) and the radius \(R\). According to Kepler’s third law (as expressed in Newton’s Principia), the period of an orbit is proportional to the \(3/2\) power of the radius:
\[
T \propto R^{3/2}
\]
For a circular orbit, the distance traveled in one complete revolution is the circumference \(2\pi R\). Since the orbital speed \(V\) is the distance traveled per unit time, we have:
\[
T = \frac{2\pi R}{V}
\]
Now, rearrange this equation to solve for \(V\):
\[
V = \frac{2\pi R}{T}
\]
Substitute the proportionality \(T \propto R^{3/2}\) into this expression:
\[
V \propto \frac{R}{R^{3/2}} = \frac{1}{R^{1/2}}
\]
This shows that:
\[
V \propto \frac{1}{\sqrt{R}}
\]
In words, the orbital speed is inversely proportional to the square root of the radius. This is what is meant by “velocities inversely in the subduplicate ratio of the radii” — the term “subduplicate” refers to the square root.
Summary
- Sesquiplicate Ratio: \(T \propto R^{3/2}\)
- Circumference: \(2\pi R\)
- Orbital Speed: \(V = \dfrac{2\pi R}{T} \implies V \propto \frac{R}{R^{3/2}} = \frac{1}{\sqrt{R}}\)
Thus, if you increase the radius, the velocity decreases as the inverse square root of the radius. This is why the velocities are said to be in the inverse subduplicate (i.e., square root) ratio to the radii.
—
Velocity and Radius are linked
Just to be clear, when we say V is proportional to the inverse square root R, does this make it clear that when we change V, the radius R must also change, according to Kepler’s Rule? We cannot have an object move on the same R with different velocities. Can you clarify the confusion?
Yes, that’s exactly right. In a Keplerian (or Newtonian gravitational) system, for a circular orbit, the orbital velocity and the orbital radius are not independent—they are linked by the dynamics of the system. Here’s a clear breakdown:
- Unique Relation:
For an object in a stable circular orbit, the gravitational force provides the necessary centripetal force. This balance is given by:
\[ \frac{GMm}{R^2} = \frac{mV^2}{R} \]
Simplifying (and canceling the mass \(m\)) gives:
\[ V^2 = \frac{GM}{R} \quad \Rightarrow \quad V = \sqrt{\frac{GM}{R}} \]
This shows that the velocity \(V\) is uniquely determined by the radius \(R\) (and the mass \(M\) of the central body). - Implication of the Relation:
When we say “\(V\) is proportional to the inverse square root of \(R\)” (i.e., \(V \propto 1/\sqrt{R}\)), it means that for a given central mass:
- If \(R\) increases: \(V\) decreases following the \(\frac{1}{\sqrt{R}}\) law.
- If \(R\) decreases: \(V\) increases accordingly.
- If \(R\) increases: \(V\) decreases following the \(\frac{1}{\sqrt{R}}\) law.
- No Freedom to Vary \(V\) Independently:
Because of this relation, you cannot have two different orbital velocities at the same radius in a stable orbit. If you try to change \(V\) while keeping \(R\) fixed, the balance of forces would be disturbed, and the object would not remain in a circular orbit. It might move into an elliptical orbit or escape altogether, depending on the change. - Connection to Kepler’s Rule:
Kepler’s third law, which states that the orbital period \(T\) is proportional to \(R^{3/2}\) (\(T \propto R^{3/2}\)), is derived from the same dynamics. Since for a circular orbit the velocity is \(V = \frac{2\pi R}{T}\), substituting the relationship for \(T\) reinforces that \(V \propto 1/\sqrt{R}\).
Summary
- Circular Orbit Dynamics: \(V = \sqrt{\frac{GM}{R}}\) ensures that for any given \(R\), there is only one value of \(V\) that maintains a stable orbit.
- Implication of the Proportionality:
The statement \(V \propto \frac{1}{\sqrt{R}}\) means that any change in orbital speed must be accompanied by a corresponding change in the orbital radius if the system is to obey Kepler’s and Newton’s laws. - No Arbitrary Variation:
You cannot arbitrarily change \(V\) without affecting \(R\); they are inherently linked by the gravitational dynamics.
So, in summary, when we state that \(V\) is proportional to the inverse square root of \(R\), it indeed implies that a change in one must result in a corresponding change in the other in a Keplerian orbit. There is no freedom to have an object moving at two different velocities while remaining at the same orbital radius under these conditions.
—
Velocity as geometry and in Kepler’s Rule
But this is not the case for our initial relation V=2pi R/T, correct? Here, I can keep the period constant and increase R and velocity will increase. This looks confusing to me. Maybe you can go in more details
You’re right to notice this apparent contradiction! Let’s break it down carefully and see why the equation \(V = \frac{2\pi R}{T}\) does not contradict Kepler’s Rule.
—
1. Understanding \( V = \frac{2\pi R}{T} \)
This equation comes from basic kinematics. It simply states that for a circular motion:
\[
V = \frac{\text{Circumference}}{\text{Period}} = \frac{2\pi R}{T}
\]
This equation, by itself, does not tell us how \(T\) and \(R\) are related. It simply states that if you know the period \(T\) and the radius \(R\), you can compute the velocity.
—
2. The Role of Kepler’s Third Law
Kepler’s Third Law states:
\[
T \propto R^{3/2}
\]
This is a physical constraint that applies in a gravitational system where the motion is governed by a central force. If we impose this condition on \(T\), we can substitute it into \(V = \frac{2\pi R}{T}\) to obtain the correct relationship between \(V\) and \(R\).
Substituting \(T \propto R^{3/2}\):
Since \(T = k R^{3/2}\) for some constant \(k\), we substitute into the velocity equation:
\[
V = \frac{2\pi R}{k R^{3/2}}
\]
Simplifying:
\[
V = \frac{2\pi}{k} R^{1 - 3/2} = \frac{2\pi}{k} R^{-1/2}
\]
Thus, we recover:
\[
V \propto \frac{1}{\sqrt{R}}
\]
which is consistent with Newton’s formulation and Kepler’s Rule.
—
3. What If We Keep \(T\) Constant?
If you assume \(T\) is independent of \(R\) (that is, you artificially force \(T\) to stay the same while changing \(R\)), then:
\[
V = \frac{2\pi R}{T}
\]
suggests that as \(R\) increases, \(V\) increases proportionally. However, this scenario does not apply to gravitational orbits. It only applies to an artificial system where you are setting the period by some external means (like an engine-driven system, not gravity).
In a gravitational orbit, \(T\) and \(R\) cannot be independent because Kepler’s Rule enforces \(T \propto R^{3/2}\). So, in reality, you cannot change \(R\) while keeping \(T\) constant in a natural gravitational system.
—
4. Why Is This Confusing?
- The equation \( V = \frac{2\pi R}{T} \) is always mathematically true, but it does not tell you how \(V\), \(R\), and \(T\) are related in a physical system.
- In a gravitational orbit, \(T\) and \(R\) are linked by Kepler’s Rule, so we must substitute \(T \propto R^{3/2}\), which gives \(V \propto 1/\sqrt{R}\).
- If you assume \(T\) is constant (ignoring Kepler), you get a different result (\(V \propto R\)), but this is not how gravity works.
—
5. Final Clarification
- In a gravitational orbit: You cannot independently change \(V\) without changing \(R\), because \(T \propto R^{3/2}\).
- If \(T\) were constant: This would not describe a gravitational orbit, but rather a hypothetical system where something else (like an engine) forces an object to complete its orbit in the same time regardless of \(R\).
So, Kepler’s Rule ensures that for gravitational orbits, you cannot have an object moving at two different velocities at the same radius. The correct velocity-radius relation comes from the combination of \(V = \frac{2\pi R}{T}\) and \(T \propto R^{3/2}\), leading to \(V \propto 1/\sqrt{R}\).
Does this help clarify the confusion?