(A) Compare the current rotational angular momentum of Earth (use I = 0.33M R2 )

Question

(A) Compare the current rotational angular momentum of Earth (use I = 0.33M R2 ) to the orbital angular momentum of the Moon as it orbits Earth (assume the Moon's orbit is circular) and the rotational angular momentum of the Moon (assume the Moon is a rigid uniform sphere). Note the Moon orbits synchronously, that is, its orbital period equals its rotation period.

(B) The Moon's orbit is observed to be getting larger at a rate of about 4 cm per year. Comparing the age of the Earth-Moon system (4.5 Gyr) to the current orbit size divided by the recession rate suggests the Moon originally was much closer to Earth than it is now. Assuming the Moon has always rotated synchronously and the total angular momentum of the Earth+Moon system has remained a constant, what was the rotation period of Earth when the Moon's orbit was1/6 as large as it is now?

Explanation / Answer

solving first question

orbital angular momentum of earth=Iw

I=0.33MR^2

w=2pi/T

Earth completes one rotation in 24 hours

so w=2pi/(24*60*60)=7.27*10^(-5) rad/sec

angular momentum=0.33*5.972*10^24*6371000^2=5.8*10^33

now for moon

time=30 days =30*24*60*60 sec

I=0.4MR^2

M is mass of moon,R=radius of orbit = 384,400 km

Angular momentum=Iw=0.4*7.34*10^22*384400000^2*2p/(30*24*3600)=1.05*10^34

ratio=58/105=0.55

for orbital angular momentum

=Iw

I=0.4*M*R^2

R=1737000 m

we got momenum=2.14*10^29 kgm^2

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