Question

Today, the Moons orbit around Earth has a semi-major axis of a=384,400 km and an orbital period of 27.32166 days. a. The MooToday, the Moon’s orbit around Earth has a semi-major axis of a=384,400 km and an orbital period of 27.32166 days.

a. The Moon slowly moves outward due to tidal braking of the Earth’s rotation, and at some future date the Moon will have an orbital period of 47 days. Compute the semi-major axis of the Moon’s orbit at this future date (express your answer in kilometers).

semi-major axis = 5.5*10^5 km

b. Today, the Moon has an angular diameter of about 0.5° as seen from Earth. When the Moon has an orbital period of 47 days, what will its angular size be? Can there be total solar eclipses at this future time?

c. When the Moon’s orbital period slows to 47 days, tidal braking will cause the Earth’s rotation period to also be 47 days – a condition known as “tidal locking” (1 day = 86400 SI seconds). The current rate of tidal braking of Earth’s rotation is dProt/dt=1.6 milliseconds per century. At this braking rate, how many years will it take for the Earth’s rotation to slow to 47 days?

d. What is the Hill Radius of the Earth in the tidal field of the Sun? Use this equation: Compare rH with the size of the Moon’s orbit in the future when the Earth and Moon are tidally locked that you found in part a. When the Earth and Moon are tidally locked, is the Moon in any danger of wandering away from its orbit around the Earth? Explain your reasoning.

I have already done part a (answer above). I am having particular trouble with part b but would like answers to all parts of the question. You can also check my answer for part a if you wish! Please write legibly and explain your reasoning clearly as you go.

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Answer #1

Ans. Given date: Current moons semi major axis a= 384,400 km. orbited period op: 27.32166 days. a) In future semi major axissince, ase diameter of moon. - ae = diameter of moon. QIB = ae - 01- Geo F inance C0.50) (1.43569 a) oi = 0.3482 64597 (i= 0.YH ( + mom a mp = 5.9736 x 1024 kg M sun = 1.489) x1030 kg. a= I AU. : The 5.9736x1024) cool 3 1489147030 1/3 in 8H = 0.0/000

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