The moon orbits the Earth so that the same side always faces the Earth. Determine the ratio of its spin angular momentum (about its own axis) and its orbital angular momentum. (In this case, treat the Moon as a particle orbiting the Earth). Distance between the Earth and the Moon is 3.85 ｘ10⁸ m. Radius of the Moon is 1.74 ｘ10⁶ m. (Ans: 8.2 ｘ10⁻⁶)



Data Given:

Distance between the Earth and the Moon = Radius of the moon's orbit = r = 3.85 ｘ10⁸ m

Radius of the Moon  = Rm = 1.74 ｘ10⁶ m

To Find:

Ratio of spin angular momentum and orbital angular momentum = →Ls→Lo  = ?


Solution:


Using formula for Moon Angular momentum (spin) →Ls in terms of angular velocity  ധ and Moment of inertia I

→Ls = Iധ  

Where for sphere (moon), I = 2MnR2m5, Thus

→Ls = 2MnR2mധ5T -------(1)



Now using formula for Moon orbital momentum (spin) →Lo in terms of angular velocity  ധ and Moment of inertia I


→Lo = Iധ  


Where for orbit (ring) I = Mnr2, Thus
→Lo = Mnr2ധ  ---------(2)


Now dividing equation (1) by (2) we get


→Ls→Lo  = 2MnR2mധ5TMnr2mധ


Or (by Simplifying)

→Ls→Lo  = 2R2m5r2

Putting the corresponding values

→Ls→Lo  = 2(1.74ｘ10⁶m)25(3.85ｘ10⁸m)2

→Ls→Lo  = 2(3.0276ｘ10¹²m2)5(14.8225ｘ10¹⁶m2)

→Ls→Lo = 6.0552ｘ10¹²m274.1125ｘ10¹⁶m2

→Ls→Lo  = 0.08170 ｘ10⁻⁴

or

→Ls→Lo  = 8.170 ｘ10⁻⁶

or

→Ls→Lo  = 8.170 ｘ10⁻⁶ -----Ans.



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