0
$\begingroup$

While I was trying to derive Kepler's third law of planetary motion, I tried the gravitational force for the Earth method which goes something like this:

$$\frac{mv^2}{r}=\frac{GMm}{r^2}$$ $$\Rightarrow\frac{4\pi r^2}{T^2}=\frac{GM}{r}$$ From further rearrangement of the equation, we can figure out that $T^2\propto r^3$.

Now, I am trying to derive the same thing using angular momentum (assuming no external force disrupts the Sun-Earth system). So, what I did goes as follows: $$L=mvr$$ $$\Rightarrow L=m\times \frac{2\pi r}{T}\times r$$ Now since, $\frac{dL}{dt}=0$ (angular momentum is conserved), we can treat $L$ like a constant (from what I think). $$\Rightarrow T=\frac{2\pi r^2 m}{L}$$ $$\therefore T\propto r^2$$ But this is not what Kepler's third law of planetary motion says. Now I am sure my angular momentum method has gone wrong somewhere, I am just not sure where. Can someone please point out my error?

$\endgroup$
0

2 Answers 2

3
$\begingroup$

Clearly, $L$ varies with $r$ somehow. So, let's work from first principles; in fact, let's not restrict ourselves to circular orbits. The Binet equation proves closed orbits are elliptical with semi-latus rectum $\ell=\frac{h^2}{G\mu}$, with $h$ the orbiting body's specific angular momentum and $\mu$ the orbiting and central bodies' reduced mass. This implies $L\propto h$ is proportional to lengths' square roots. Indeed, $L\propto r^{1/2}$ is just what you need to fix your problem.

As a bonus, I'll give the full proof of Kepler's third law. Area is swiped out at rate $\frac12r^2\dot{\theta}=\frac12h$. Let $a,\,b$ denote the major and minor semi-axes, so the area is $\pi ab$, and the orbital period is $T=\frac{2\pi ab}{h}$. If $e$ is the orbit's eccentricity then$$a=\frac{\ell}{1-a^2},\,b=\frac{\ell}{\sqrt{1-a^2}}\implies b^2=a\ell\implies T^2=\frac{4\pi a^2b^2}{G\mu\ell}=\frac{4\pi a^3}{G\mu}.$$

$\endgroup$
1
$\begingroup$

Even though the argument is completely general, let me confine to the simple case of circular motion.

The first derivation mentioned in the question is valid. The starting formula, equating the expression of the modulus of acceleration in a uniform circular motion to the acceleration due to Newton's force, is valid for all the circular motions, each corresponding to a different pair $r$,$T$.

The second cannot work. The angular momentum is a constant in a circular motion of radius $r$ and period $T$, but it is a different constant in a circular motion of radius $r'$ and period $T'$. Therefore, it is not possible to use the conservation of angular momentum to derive a relation between $r$ and $T$.

Said in another way, the third Kepler's law is not a consequence of the angular momentum conservation, valid for all the central forces, but depends on the $1/r^2$ character of the force law.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.