Answer :
Angular momentum of moon = 2.88 × [tex]10^{34}[/tex]kg.[tex]\frac{m^{2} }{sec}[/tex]
The orbital angular momentum of the Moon is 1.22 × [tex]10^{5}[/tex] times more than its rotating angular momentum.
Mass of moon = 7.35 × 10²²kg
radius (r) = 1.74 × [tex]10^{6}[/tex]m
Time period (T) = 27.3days = 27.3 × 24 × 3600sec
distance from earth(R) = 3.84 × [tex]10^{8}[/tex] (= orbiting radius)
a) Angular momentum of moon in its orbit around earth is
[tex]L_{orb}[/tex] = [tex]I_{w}[/tex]
in where I = moment of inertia = m[tex]R^{2}[/tex] = 7.35 × 10²² ( 3.84 × [tex]10^{8}[/tex])²
= 1.084 × [tex]10^{40}[/tex] kg.m²
and ω = 2π/T
= 2π/27.3 × 24 × 3600 = 2.66 × [tex]10^{-6}[/tex][tex]\frac{rad}{sec}[/tex]
[tex]L_{orb}[/tex] = [tex]I_{w}[/tex] = 1.084 × [tex]10^{40}[/tex] kg.m² × 2.66 × [tex]10^{-6}[/tex]
[tex]L_{orb}[/tex] = 2.88 × [tex]10^{34}[/tex]kg.[tex]\frac{m^{2} }{sec}[/tex]
b) Angular momentum of moon around it's axis is (due to rotation),
[tex]L_{m} = I_{m} W_{m}[/tex] = [tex]\frac{2}{5}[/tex] mr² × 2π/T = [tex]\frac{2}{5}[/tex] × 7.35 × [tex]10^{22}[/tex] × (1.74 × [tex]10^{6}[/tex])² × 2.66 × [tex]10^{-6}[/tex]kg.[tex]\frac{m^{2} }{sec}[/tex]
[tex]L_{orb}[/tex] = 2.37 × [tex]10^{29}[/tex] kg.[tex]\frac{m^{2} }{sec}[/tex]
c) The required fraction is:
[tex]\frac{L_{orb} }{L_{rot} }[/tex] = [tex]\frac{2.88(10^{34} )}{2.37(10^{29} )}[/tex] = 1.22 × [tex]10^{5}[/tex]
[tex]\frac{L_{orb} }{L_{rot} }[/tex] = 1.22 × [tex]10^{5}[/tex]
What in physics do you mean by orbital angular momentum?
An aspect of an electron's rotational motion called orbital angular momentum is connected to the orbital's geometry. If detection is attempted, the electron will be located in the orbital, which is the area surrounding the nucleus. The orbital around the nucleus has been been referred to as a "electron cloud."
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