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 ï½˜10⁶ 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  = `\R_m` = 1.74 ｘ10⁶ m

To Find:

Ratio of spin angular momentum and orbital angular momentum = `\frac {vec {L_s}}{vec {L_o}}`  = ?


Solution:


Using formula for Moon Angular momentum (spin) `\vec {L_s}` in terms of angular velocity  à´§ and Moment of inertia `\I`

`\vec {L_s}` = `\I`à´§  

Where for sphere (moon), `\I` = `\frac {2M_nR_m^2}{5}`, Thus

`\vec {L_s}` = `\frac {2M_nR_m^2à´§}{5T}` -------(1)



Now using formula for Moon orbital momentum (spin) `\vec {L_o}` in terms of angular velocity  à´§ and Moment of inertia `\I`


`\vec {L_o}` = `\I`à´§  


Where for orbit (ring) `\I` = `\M_nr^2`, Thus
`\vec {L_o}` = `\M_nr^2`à´§  ---------(2)


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


`\frac {vec {L_s}}{vec {L_o}}`  = `\frac {frac {2M_nR_m^2à´§}{5T}}{M_nr_m^2à´§}`


Or (by Simplifying)

`\frac {vec {L_s}}{vec {L_o}}`  = `\frac {2R_m^2}{5r^2}`

Putting the corresponding values

`\frac {vec {L_s}}{vec {L_o}}`  = `\frac {2(1.74 ｘ10⁶ m)^2}{5 (3.85 ｘ10⁸ m)^2}`

`\frac {vec {L_s}}{vec {L_o}}`  = `\frac {2(3.0276 ｘ10¹² m^2)}{5 (14.8225 ｘ10¹⁶ m^2)}`

`\frac {vec {L_s}}{vec {L_o}}` = `\frac {6.0552 ｘ10¹² m^2}{74.1125 ｘ10¹⁶ m^2}`

`\frac {vec {L_s}}{vec {L_o}}`  = 0.08170 ï½˜10⁻⁴

or

`\frac {vec {L_s}}{vec {L_o}}`  = 8.170 ï½˜10⁻⁶

or

`\frac {vec {L_s}}{vec {L_o}}`  = 8.170 ï½˜10⁻⁶ -----Ans.



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