Angular momentum of earth from Lagrangian, am I correct?

[genericusrnme]

Hey, I was just playing about with some lagrangian mechanics and tried to work out the angular momentum of the earth;
Starting with the Lagrangian
$$\mathcal{L} = \left(\frac{1}{2}m (r')^2 + \frac{1}{2}m r^2 (\theta ')^2\right)+\frac{G m M}{r}$$
Applying the Euler Lagrange eqn to prove conservation of momentum conjugate to angle
$$0 = m r^2\theta '=L$$
And solving for angular velocity
$$\theta '=\frac{L}{m r^2}$$
Then applying the eqn to the radial component
$$m r (\theta ')^2-\frac{G m M}{r^2}= m a$$
Assuming on a stable orbit net force on r should be zero and substituting in what I found for angular velocity
$$\frac{L^2}{m r^3}-\frac{G m M}{r^2}=0$$
Then solving for L
$$L = \sqrt{G m^2M r}$$
Plugging in values from wiki
$$L = \sqrt{\left(6.67\times 10^{-11}\right)\left(6\times 10^{24}\right)^2\left(2\times 10^{30}\right)\left(15\times 10^6\right)}$$
$$L = 2.68395*10^38$$

But I think I'm two orders of magnitude off, I remember it being to the power 40 not 38
Have I done something wrong here or made any incorrect assumptions?

 

[genericusrnme]

Further more I get
$$\theta '=\frac{L}{m r^2}$$
$$\theta '\approx \frac{2.7\times 10^{39}}{\left(6\times 10^{24}\right)\left(1.5\times 10^9\right)^2}$$
$$\theta '\approx 0.0002$$
Which proves I'm 2 orders of magnitude off because I get
$$0.0002\ 60\ 60\ 24\ 365 = 6307.2$$
which is about three orders of magnitude off of 2pi

where did my 1000 go? :(

 

[TheDragon]

1 Astronomical Unit = 149 597 870 700 meters=15 *10^10 and not 15*10^6.