Physical reason for $T^2=a^3$ when $T$ is in years and $a$ is in AU 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.
 A: 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.
A: 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:
$$\tag{3}
T'^{2} = a'^3.
$$
A: 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 
$$ \frac{R^3}{T^2} = \frac{GM}{4\pi^2} = \frac{1\, (AU)^3}{(\rm year)^2}$$
