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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.

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    $\begingroup$ 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. $\endgroup$
    – PM 2Ring
    Apr 8, 2021 at 15:46
  • $\begingroup$ How come orbits with different bs and same as can have the same period? $\endgroup$
    – xenfoulis
    Apr 8, 2021 at 19:47
  • $\begingroup$ 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. gist.github.com/PM2Ring/d7878c904df8da838f76dc4a15c6c746 $\endgroup$
    – PM 2Ring
    Apr 8, 2021 at 20:17
  • $\begingroup$ Great program I must say, I have a question though, isn't the eccentricity for each of these orbits the same? $\endgroup$
    – xenfoulis
    Apr 8, 2021 at 21:01
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    $\begingroup$ 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. $\endgroup$
    – Kuba hasn't forgotten Monica
    Apr 9, 2021 at 15:55

4 Answers 4

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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.

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    $\begingroup$ @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. $\endgroup$
    – PM 2Ring
    Apr 8, 2021 at 20:09
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    $\begingroup$ @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"? $\endgroup$
    – uhoh
    Apr 8, 2021 at 23:09
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    $\begingroup$ @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. $\endgroup$
    – PM 2Ring
    Apr 8, 2021 at 23:32
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    $\begingroup$ @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 $\endgroup$
    – xenfoulis
    Apr 9, 2021 at 6:54
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    $\begingroup$ @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'?" $\endgroup$
    – uhoh
    Apr 9, 2021 at 7:32
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Consider this gif from wikipedia. 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

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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When the planet is at one end of the major axis, that's an extremum with respect to time of distance from the sun (either aphelion or perihelion, depending which end of the major axis). That means that, at that point, the radial component of the planet's velocity is zero, and therefore there is no contribution of radial motion to the kinetic energy. Hence, the total energy is just the sum of the contribution of azimuthal motion to the kinetic energy (which depends only on distance from the sun and some constants like angular momentum and the mass of the planet) and the gravitational potential energy (which depends only on distance from the sun and some constants like the universal gravitational constant, the mass of the sun, and the mass of the planet). Conservation of energy means the total energy is constant, so equating the total energy at aphelion and perihelion gives an equation that rearranges to show that the angular momentum is proportional to the quotient of the geometric mean of perihelion and aphelion distances by the square root of the semi-major axis. The equal-areas-in-equal-times principle means that the period is proportional to the quotient of the product of the semi-major axis and the geometric mean of perihelion and aphelion distances by the angular momentum, so one ends up with a formula for the period in terms of the semi-major axis, independent of any other feature of the orbital geometry.

One can't do the same trick with the two ends of the minor axis, because there are not extrema with respect to time of distance from the sun there, so the radial component of velocity is non-zero, and there's an extra, complicating term in the energy equation representing the contribution of radial motion to the kinetic energy.

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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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