Angular momentum of earth from Lagrangian, am I correct?

  • #1

Main Question or Discussion Point

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
[tex]\mathcal{L} = \left(\frac{1}{2}m (r')^2 + \frac{1}{2}m r^2 (\theta ')^2\right)+\frac{G m M}{r}[/tex]
Applying the Euler Lagrange eqn to prove conservation of momentum conjugate to angle
[tex]0 = m r^2\theta '=L[/tex]
And solving for angular velocity
[tex]\theta '=\frac{L}{m r^2}[/tex]
Then applying the eqn to the radial component
[tex]m r (\theta ')^2-\frac{G m M}{r^2}= m a[/tex]
Assuming on a stable orbit net force on r should be zero and substituting in what I found for angular velocity
[tex]\frac{L^2}{m r^3}-\frac{G m M}{r^2}=0[/tex]
Then solving for L
[tex]L = \sqrt{G m^2M r}[/tex]
Plugging in values from wiki
[tex]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)}[/tex]
[tex]L = 2.68395*10^38[/tex]

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?
 

Answers and Replies

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

where did my 1000 go? :(
 
Last edited:
  • #3
10
0
1 Astronomical Unit = 149 597 870 700 meters=15 *10^10 and not 15*10^6.
 

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