<p>The relationship between <span class="math-container">$a$</span>, <span class="math-container">$b$</span> and the eccentricity of an orbit <span class="math-container">$e$</span> is
<span class="math-container">$$ a = b\left(\frac{1+e}{1-e}\right)\ .$$</span></p>
<p>It is clear therefore from a mathematical point of view that you cannot get a similar constant relationship by replacing <span class="math-container">$a^3$</span> by <span class="math-container">$b^3$</span> in Kepler's third law. You would have to use an expression involving both <span class="math-container">$b$</span> and <span class="math-container">$e$</span>.</p>
<p>Another way of saying this is that you can have (infinitely) many orbits that have the same <span class="math-container">$a$</span> and <span class="math-container">$T$</span>, but which have different <span class="math-container">$b, e$</span> combinations.</p>
<p>I guess from a fundamental point of view it boils down to being able to express the total energy of an orbit as
<span class="math-container">$$ E = -\frac{GMm}{2a}$$</span>
which is independent of the shape of the ellipse.</p>

CC BY-SA 4.0
2021-04-08T19:28:25.017Z
"Another way of saying this is that you can have (infinitely) many orbits that have the same a and T, but which have different b,e combinations." How come? If for one orbit b is greater then the circumference will be greater => bigger T. Am I missing something obvious? I dont understand why T wouldn't change
xenfoulis
CC BY-SA 4.0
2021-04-08T20:09:37.207Z
@xenfoulis The period just depends on the mean radius of the orbit. And $a$ happens to be the mean radius. It's easy to see that it's the mean of the minimum & maximum radius, but you need calculus to show that it truly is the mean radius. And you need calculus to prove that $$\frac{a^3}{T^2}=\frac{GM}{4\pi^2}$$ That's one of the main things Newton did when he developed calculus and his theory of gravitation.
PM 2Ring
CC BY-SA 4.0
2021-04-08T23:09:30.933Z
@PM2Ring I think the OP completely accepts Newton as given and does not challenge it, but are still looking for more than a "because the equation is..." answer to "why not"?
uhoh
CC BY-SA 4.0
2021-04-08T23:32:51.870Z
@uhoh I just posted a link to WP's article on vis-viva. That might help, and it doesn't use calculus in its derivation.
PM 2Ring
CC BY-SA 4.0
2021-04-09T06:54:05.477Z
@uhoh exactly, my physics teacher showed us the Kepler's laws today and asked the class "why can't we use the smi-minor axis in this formula?", I'm pretty sure I was the only one that didn't know why... It's still kind of hard to understand but I think will just accept that only _a_ works for my own good
xenfoulis
CC BY-SA 4.0
2021-04-09T07:32:01.120Z
@xenfoulis No I'm pretty sure you were among the majority, but you might simply be the only person who *realized* that they didn't know why. Most people are satisfied with "because the equation..." as understanding, but sometimes we can gain further insight by asking "Yes, but *why* is that equation the 'reason'?"
uhoh
CC BY-SA 4.0
2021-04-08T17:56:23.430Z
2021-04-08T17:56:23.430Z
ProfRob
<p>Consider this gif from <a href="https://en.wikipedia.org/wiki/Orbital_eccentricity" rel="noreferrer">wikipedia</a>.  All the orbits in the animation have the same orbital period <span class="math-container">$T$</span> and the same semi-major axis <span class="math-container">$a$</span>, but different semi-minor axes <span class="math-container">$b_1,b_2...b_5$</span>.  This shows that the orbital period is independent of the semi-minor axis.</p>
<p><a href="https://i.sstatic.net/VD8nz.gif" rel="noreferrer"><img src="https://i.sstatic.net/VD8nz.gif" alt="enter image description here" /></a></p>
<p>As an analogy, imagine twin figure skaters spinning with the same angular momentum.  The red figure skater holds a weight halfway away from her body for a full revolution.  The pink figure skater holds the same weight far away for part of a revolution, but close in for the other part of the same revolution, in a way that the average distance of the weight from her body is halfway.  Both figure skaters will have the same period of revolution, but the length of the path of the weight will be larger for the red figure skater!</p>
<p>The time for a revolution for each skater is a function of the average distance they hold out the weight, rather than the difference between the closest and furthest distance they hold the weight!</p>
<p>Similarly, given fixed masses, the period of an orbit is ONLY dependent on the mean distance between the bodies: <span class="math-container">$a$</span>. It is NOT dependent on any variables relating only to variations in distance between the bodies (i.e. eccentricity, semi-minor axis).</p>

CC BY-SA 4.0
2021-04-08T20:17:06.373Z
2021-04-08T23:56:50.757Z
2021-04-08T23:56:50.757Z
Connor Garcia
<p>A derivation of Kepler's laws using vector calculus can be found in David M. Bressoud's delightful little textbook <em>Second Year Calculus</em> (Springer-Verlag, New York, 1991).  Kepler's third law is proved on page 72, demonstrating why the third law is what it is and not something different.  As to whether some simple physical intuition can lead you to Kepler's laws, I rather doubt it, considering the historical missteps preceding the discovery of Kepler's laws.</p>

CC BY-SA 4.0
2021-04-09T20:43:57.087Z
2021-04-09T20:43:57.087Z
Greg Marks
<p>Why can't we use the semi-minor axis in Kepler's third law? The formula for the third law is:</p>
<p><span class="math-container">$$\dfrac{a^3}{T^2} = \text{const.} $$</span></p>
<p>Instead of semi-major axis (a) why can't we use the semi-minor axis (b)?</p>
<p><span class="math-container">$$\dfrac{b^3}{T^2} \neq \text{const.} $$</span></p>
<p>I just checked it (I calculated the results for some planets using the semi-minor axis) and I really don't know why it's not constant.</p>

CC BY-SA 4.0
2021-04-08T15:46:04.230Z
Well, those equations can't both be correct at the same time. If we have a bunch of ellipses with identical $a$ but different eccentricity, they all have the same period, but different $b$s.
PM 2Ring
CC BY-SA 4.0
2021-04-08T19:47:55.363Z
How come orbits with different bs and same as can have the same period?
xenfoulis
CC BY-SA 4.0
2021-04-08T20:17:09.460Z
Here's a Python program I wrote a few years ago that shows that orbital period only depends on $a$ and not on eccentricity. It plots 7 bodies that all start at the same position & speed, but heading in different angles. It only calculates the gravity between the sun & each body, it ignores the gravitational attraction between the bodies, and bodies can pass through each other. https://gist.github.com/PM2Ring/d7878c904df8da838f76dc4a15c6c746
PM 2Ring
CC BY-SA 4.0
2021-04-08T21:01:29.283Z
Great program I must say, I have a question though, isn't the eccentricity for each of these orbits the same?
xenfoulis
CC BY-SA 4.0
2021-04-08T23:29:17.933Z
You may find this helpful: https://en.wikipedia.org/wiki/Vis-viva_equation
PM 2Ring
CC BY-SA 4.0
2021-04-09T06:50:34.610Z
@PM2Ring thank you, I'll give it a read this evening
xenfoulis
CC BY-SA 4.0
2021-04-09T07:14:37.590Z
Would a derivation of Kepler third law be useful? I think that would make it clear why it doesn't work with $b$. If yes, I can write an answer on that.
zabop
CC BY-SA 4.0
2021-04-09T12:04:16.167Z
I don't want to waste anymore of anyones time. Ill make do with what I have already, thank you very much for the offer :)
xenfoulis
CC BY-SA 4.0
2021-04-09T15:55:09.520Z
Kepler's third law also implies that $T^2\cdot E^3=\text{const}$, where $E$ is orbital energy. Thus $a$ and $b$ are not quite equivalent. $a$ is the mean radius of the orbit and is inversely proportional to orbital energy. $b$ *alone* is not related to orbital energy: the $(b, e)$ pair is needed for that.
Kuba hasn't forgotten Monica
CC BY-SA 4.0
2021-04-10T18:24:45.050Z
@xenfoulis Sorry, I made a very bad typo in my previous (now deleted) comment. I *meant* to say: each orbit has different eccentricity, listed under `ecc` in the printed information. eccentricity = (focal distance) / (semi-major axis).
PM 2Ring
<mathematics>
Why can't we use the semi-minor axis in Kepler's third law?
xenfoulis
# Why can't we use the semi-minor axis in Kepler's third law?

(Asked by: username_0 on 2021-04-08)

Why can't we use the semi-minor axis in Kepler's third law? The formula for the third law is:

$$\dfrac{a^3}{T^2} = \text{const.} $$

Instead of semi-major axis (a) why can't we use the semi-minor axis (b)?

$$\dfrac{b^3}{T^2} \neq \text{const.} $$

I just checked it (I calculated the results for some planets using the semi-minor axis) and I really don't know why it's not constant.

**Tags:** mathematics
> **Comments:**
> **username_1**: Well, those equations can't both be correct at the same time. If we have a bunch of ellipses with identical $a$ but different eccentricity, they all have the same period, but different $b$s.  
> **username_0**: How come orbits with different bs and same as can have the same period?  
> **username_1**: Here's a Python program I wrote a few years ago that shows that orbital period only depends on $a$ and not on eccentricity. It plots 7 bodies that all start at the same position & speed, but heading in different angles. It only calculates the gravity between the sun & each body, it ignores the gravitational attraction between the bodies, and bodies can pass through each other. https://gist.github.com/PM2Ring/d7878c904df8da838f76dc4a15c6c746  
> **username_0**: Great program I must say, I have a question though, isn't the eccentricity for each of these orbits the same?  
> **username_1**: You may find this helpful: https://en.wikipedia.org/wiki/Vis-viva_equation  
> **username_0**: @PM2Ring thank you, I'll give it a read this evening  
> **username_2**: Would a derivation of Kepler third law be useful? I think that would make it clear why it doesn't work with $b$. If yes, I can write an answer on that.  
> **username_0**: I don't want to waste anymore of anyones time. Ill make do with what I have already, thank you very much for the offer :)  
> **username_3**: Kepler's third law also implies that $T^2\cdot E^3=\text{const}$, where $E$ is orbital energy. Thus $a$ and $b$ are not quite equivalent. $a$ is the mean radius of the orbit and is inversely proportional to orbital energy. $b$ *alone* is not related to orbital energy: the $(b, e)$ pair is needed for that.  
> **username_1**: @xenfoulis Sorry, I made a very bad typo in my previous (now deleted) comment. I *meant* to say: each orbit has different eccentricity, listed under `ecc` in the printed information. eccentricity = (focal distance) / (semi-major axis).  

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## 3 answers

### Answer by username_4 on 2021-04-08. Score: 14

The relationship between $a$, $b$ and the eccentricity of an orbit $e$ is
$$ a = b\left(\frac{1+e}{1-e}\right)\ .$$

It is clear therefore from a mathematical point of view that you cannot get a similar constant relationship by replacing $a^3$ by $b^3$ in Kepler's third law. You would have to use an expression involving both $b$ and $e$.

Another way of saying this is that you can have (infinitely) many orbits that have the same $a$ and $T$, but which have different $b, e$ combinations.

I guess from a fundamental point of view it boils down to being able to express the total energy of an orbit as
$$ E = -\frac{GMm}{2a}$$
which is independent of the shape of the ellipse.

> **Comments:**
> **username_0**: "Another way of saying this is that you can have (infinitely) many orbits that have the same a and T, but which have different b,e combinations." How come? If for one orbit b is greater then the circumference will be greater => bigger T. Am I missing something obvious? I dont understand why T wouldn't change  
> **username_1**: @xenfoulis The period just depends on the mean radius of the orbit. And $a$ happens to be the mean radius. It's easy to see that it's the mean of the minimum & maximum radius, but you need calculus to show that it truly is the mean radius. And you need calculus to prove that $$\frac{a^3}{T^2}=\frac{GM}{4\pi^2}$$ That's one of the main things Newton did when he developed calculus and his theory of gravitation.  
> **username_5**: @PM2Ring I think the OP completely accepts Newton as given and does not challenge it, but are still looking for more than a "because the equation is..." answer to "why not"?  
> **username_1**: @uhoh I just posted a link to WP's article on vis-viva. That might help, and it doesn't use calculus in its derivation.  
> **username_0**: @uhoh exactly, my physics teacher showed us the Kepler's laws today and asked the class "why can't we use the smi-minor axis in this formula?", I'm pretty sure I was the only one that didn't know why... It's still kind of hard to understand but I think will just accept that only _a_ works for my own good  
> **username_5**: @xenfoulis No I'm pretty sure you were among the majority, but you might simply be the only person who *realized* that they didn't know why. Most people are satisfied with "because the equation..." as understanding, but sometimes we can gain further insight by asking "Yes, but *why* is that equation the 'reason'?"  

---

### Answer by username_6 on 2021-04-08. Score: 9

Consider this gif from [wikipedia](https://en.wikipedia.org/wiki/Orbital_eccentricity). All the orbits in the animation have the same orbital period $T$ and the same semi-major axis $a$, but different semi-minor axes $b\_1,b\_2...b\_5$. This shows that the orbital period is independent of the semi-minor axis.

[![enter image description here](https://i.sstatic.net/VD8nz.gif)](https://i.sstatic.net/VD8nz.gif)

As an analogy, imagine twin figure skaters spinning with the same angular momentum. The red figure skater holds a weight halfway away from her body for a full revolution. The pink figure skater holds the same weight far away for part of a revolution, but close in for the other part of the same revolution, in a way that the average distance of the weight from her body is halfway. Both figure skaters will have the same period of revolution, but the length of the path of the weight will be larger for the red figure skater!

The time for a revolution for each skater is a function of the average distance they hold out the weight, rather than the difference between the closest and furthest distance they hold the weight!

Similarly, given fixed masses, the period of an orbit is ONLY dependent on the mean distance between the bodies: $a$. It is NOT dependent on any variables relating only to variations in distance between the bodies (i.e. eccentricity, semi-minor axis).


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### Answer by username_7 on 2021-04-09. Score: 0

A derivation of Kepler's laws using vector calculus can be found in David M. Bressoud's delightful little textbook *Second Year Calculus* (Springer-Verlag, New York, 1991). Kepler's third law is proved on page 72, demonstrating why the third law is what it is and not something different. As to whether some simple physical intuition can lead you to Kepler's laws, I rather doubt it, considering the historical missteps preceding the discovery of Kepler's laws.


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