1
$\begingroup$

Kepler's third law states $$T^2\propto a^3$$

When $T$ is in years and $a$ is in AU, the proportionality constant becomes $1$. This can't be a coincidence; I would like to know the physical reason for it.

$\endgroup$
  • $\begingroup$ This follows from the $1/r^2$ nature of the force. It’s in every 1st year textbook... What exactly are you asking? $\endgroup$ – ZeroTheHero Oct 12 '19 at 23:10
  • 2
    $\begingroup$ @ZeroTheHero The law follows from the inverse-square nature of the force but the fact that the proportionality constant for the relation $T^2\propto a^3$ is simply a matter of definition--it doesn't even need to invoke the nature of the force. If the Kepler's law had been $T^2\propto a^5$ (i.e., if the force hadn't had the inverse square nature) then also the proportionality constant would have been unity given the definition of the said units. $\endgroup$ – Dvij Mankad Oct 12 '19 at 23:18
  • $\begingroup$ @DvijMankad yes of course but there’s always a system of units where the proportionality constant is $1$ so I’m not sure what the OP is asking but he seems to have found what he was looking for... $\endgroup$ – ZeroTheHero Oct 13 '19 at 0:47
  • $\begingroup$ @ZeroTheHero Yes, exactly, and I think what the OP wants to understand is that why that system of units turns out to be the system of the units "year" and "AU". As I said, it is trivial, follows directly from the definitions of those units. $\endgroup$ – Dvij Mankad Oct 13 '19 at 0:58
1
$\begingroup$

The physical reason is that 1 year is the period of the Earth and 1 A.U is the radius of the Earth's orbit.

Never noticed that before.

Looking at the force balance version of the law:

$$ \frac{GMm}{R^2} = \frac{mv^2}{R}$$

with

$$ v = \frac{2\pi R}T$$

gives:

$$ \frac{GM}{R} = v^2 = \frac{4\pi^2R^2}{T^2}$$

so that

$$ \frac{R^3}{T^2} = \frac{GM}{4\pi^2} = \frac{1\, (AU)^3}{(\rm year)^2}$$

$\endgroup$
  • $\begingroup$ Your derivation is limited to circular orbits but this law applies to elliptic orbits as well. $\endgroup$ – ZeroTheHero Oct 12 '19 at 23:11
  • $\begingroup$ The point of the derivation was to show that mass of the sun as $GM$ is $4\pi^2$ cubic astronomical units per year squared. $\endgroup$ – JEB Oct 13 '19 at 16:38
  • $\begingroup$ @ZeroTheHero it turns out the OP's observation is more than a curiosity: en.wikipedia.org/wiki/Solar_mass $\endgroup$ – JEB Oct 14 '19 at 17:39
2
$\begingroup$

It is almost trivial! The law in SI units: $$\tag{1} T^2 = \frac{4 \pi^2}{G M} \, a^3. $$ Now write this, for $T_0 = 1~\mathrm{year}$ and $a_0 = 1~\mathrm{AU}$: $$\tag{2} \frac{T^2}{T_0^2} = \frac{\displaystyle{\frac{4 \pi^2}{G M} \, a^3}}{\displaystyle{\frac{4 \pi^2}{G M} \, a_0^3}} \equiv \frac{a^3}{a_0^3}. $$ Now, define $a' = a / a_0$ and $T' = T / T_0$, so $a'$ is now measured in UA and $T'$ in years: $$\tag{3} T'^{2} = a'^3. $$

$\endgroup$
2
$\begingroup$

It's not really a physical reason but you are correct in suspecting that this shouldn't be a coincidence. The fact that the proportionality constant becomes unity follows from the definition of a year and an astronomical unit. A year is defined precisely as the period of time it takes for the Earth to complete a revolution around the Sun. An astronomical unit is defined precisely as the distance from the Earth to the Sun. Since the orbit of the Earth is nearly circular, one can take the semi-major axis to be approximately the same as the radius of the Earth's orbit, which is the same as the distance from the Earth to the Sun in this approximation. Thus, if $T^2=ka^3$ then in the units of a year and $\text{AU}$ for $T$ and $a$ respectively, for the case of the Earth, by definition, both $T$ and $a$ are $1$. This determines $k$ to be unity.


Edit

Since other responses to the post invoke the precise form of the gravitational force, I would like to point out that while Kepler's law obviously follows from the inverse-square nature of the gravitational force, the fact that the proportionality constant for the relation $T^2\propto a^3$ is unity is simply a matter of definition--explaining it doesn't even need to invoke the nature of the force. If Kepler's law had been $T^2\propto a^5$ (i.e., if the force hadn't had the inverse-square nature) then also the proportionality constant would have been unity given the definition of the said units.

$\endgroup$
  • $\begingroup$ Can the downvoter elaborate? $\endgroup$ – Dvij Mankad Oct 15 '19 at 13:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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