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Q42PE

Expert-verifiedFound in: Page 358

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**Consider the Earth-Moon system. Construct a problem in which you calculate the total angular momentum of the system including the spins of the Earth and the Moon on their axes and the orbital angular momentum of the Earth-Moon system in its nearly monthly rotation. Calculate what happens to the Moon’s orbital radius if the Earth’s rotation decreases due to tidal drag. Among the things to be considered are the amount by which the Earth’s rotation slows and the fact that the Moon will continue to have one side always facing the Earth.**

** **If the rotation speed of the earth is decreased by 10%, then the orbital radius of the moon will increase by 1.88 %.

- The radius of the moon: R
_{m}= 1.737×10^{6}m. - Mass of the moon: M
_{m}= 7.3477×10^{22}kg. - The radius of the earth: R
_{e}= 6.378×10^{6}m. - Mass of the earth: M
_{e}= 5.972×10^{24}kg. - The radius of the moon's orbit: R
_{m}= 3.85×10^{8}m.

**The spin angular momentum of earth, the spin angular moment of the moon, and the orbital angular momentum of the moon point in the same direction. Hence, the total angular momentum of the system will be the vector sum of all these angular momenta.**

L=L_{e}+L_{m}+L_{o}

**The angular momentum can be calculated using the equation **

L=Iw

**Here, I is the moment of inertia; for spherical solid bodies, it is 2MR ^{2}/5, and ω is the angular frequency of the rotation. The angular frequency of the earth can be calculated by diving 2π by a period of the earth. Similarly, the angular frequency of the moon is calculated by diving 2π by a period of the moon. As the moon is locked, the orbital and spin angular frequencies are the same.**

**Once the total angular momentum of the system is known, we can use the conservation of angular momentum to calculate the percentage change in the radius of the orbit when the angular frequency of earth rotation is decreased.**

The total angular momentum of the moon-earth system is,

\(L = {L_e} + {L_m} + {L_o}\)

Here, L_{e} is the angular momentum of the earth, L_{m} is the angular momentum of the moon, and L_{o} is the angular momentum due to the orbital motion of the moon.

\(L=\dfrac{2}{5}{M_e}{R_e}^2{\omega_e}+\dfrac{2}{5}{M_m}{R_m}^2{\omega_m}+{M_m}{R_o}^2{\omega _m}\)

Here, ω_{e }and ω_{m }are the angular frequencies of the rotation of the earth and moon.

Earth has a period of 24 hours and the period of the moon is 27.32 days. Hence,

\(\begin{array}{c}{\omega _e}=\frac{{2\pi }}{{24\times60\times60}}\\=7.25\times{10^{-5}}\;{\rm{rad/s}}\end{array}\)

\(\begin{array}{c}{\omega_m}=\frac{{2\pi}}{{27.32\times24\times60\times60}}\\=2.662\times {10^{ - 6}}\;{\rm{rad/s}}\end{array}\)

Substituting all the known and calculated values in equation 1, we get,

\(\begin{array}{c}L=\left({\frac{2}{5}\times5.972\times{{10}^{24}}\times{{\left({6.378\times {{10}^6}}\right)}^2}\times7.27\times{{10}^{-5}}}\right)+\\\left({\frac{2}{5}\times7.348\times{{10}^{22}}\times{{\left({1.737\times{{10}^6}}\right)}^2}\times2.662 \times{{10}^{-6}}}\right)+\\\left({7.348\times{{10}^{22}}\times{{\left( {3.85\times{{10}^8}}\right)}^2}\times2.662\times{{10}^{-6}}}\right)\\=0.706.\times{10^{34}}+ 1.359\times{10^{29}}+2.89\times{10^{34}}\\L=3.650\times{10^{34}}\;{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}\end{array}\)

^{ }

This is the total angular momentum of the earth-moon system.

For convenience, let's assume that the angular frequency of the earth has decreased by 10%. If we decrease 10% of the angular frequency of the earth, we will have 90% of the remaining angular frequency of the earth. According to the conservation of momentum, this decrease will lead to an increase in the radius of the moon's orbit. Let us call this changed radius R_{mc }and calculate it.

\(\begin{array}{c}3.650 \times {10^{34}} = \left( {\frac{2}{5} \times 5.972 \times {{10}^{24}} \times {{\left( {6.378 \times {{10}^6}} \right)}^2} \times 7.27 \times {{10}^{ - 5}} \times \left( {90/100} \right)} \right)\\ + \left( {\frac{2}{5} \times 7.348 \times {{10}^{22}} \times {{\left( {1.737 \times {{10}^6}} \right)}^2} \times 2.662 \times {{10}^{ - 6}}} \right)\\ + \left( {7.348 \times {{10}^{22}} \times {R_{mc}}^2 \times 2.662 \times {{10}^{ - 6}}} \right)\\3.650 \times {10^{34}} = \left( {\frac{2}{5} \times 5.972 \times {{10}^{24}} \times {{(6.378 \times {{10}^6})}^2} \times 7.27 \times {{10}^{ - 5}} \times \left( {90/100} \right)} \right)\\ + \left[ {\left( {7.348 \times {{10}^{22}} \times 2.662 \times {{10}^{ - 6}}} \right) \times \left( {\frac{2}{5} \times {{\left( {1.737 \times {{10}^6}} \right)}^2} + {R_{mc}}^2} \right)} \right]\\3.650 \times {10^{34}} = 6.358 \times {10^{33}} + \left( {1.959 \times {{10}^{17}}} \right) \times \left( {1.2068 \times {{10}^{12}} + {R_{mc}}^2} \right)\end{array}\)

Solving further as,

\(\begin{array}{c}3.0142 \times {10^{34}} = \left( {1.959 \times {{10}^{17}}} \right) \times \left( {1.2068 \times {{10}^{12}} + {R_{mc}}^2} \right)\\1.53 \times {10^{17}} = \left( {1.2068 \times {{10}^{12}} + {R_{mc}}^2} \right)\\{R_{mc}}^2 \sim 1.53 \times {10^{17}}\\{R_{mc}} = 3.9225 \times {10^8}\;{\rm{m}}\end{array}\)^{ }

The new radius of the moon's orbit will be \(3.9225 \times {10^8}\;{\rm{m}}\) .

^{ }

Now, the percent change in the radius of the moon's orbit is,

\(\begin{array}{l}\% change = \left( {\frac{{3.9225 \times {{10}^8} - 3.85 \times {{10}^8}}}{{3.85 \times {{10}^8}}}} \right) \times 100\\\% change = 1.88\;\% \end{array}\)

Thus, if the rotation speed of the earth is decreased by 10%, then the orbital radius of the moon will increase by 1.88 %.

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