Subfield: Mechanics

EVOLUTION OF THE EARTH-MOON SYSTEM 

Scientists can determine the distance Earth-Moon with great precision. They achieve this by bouncing a laser beam on special mirrors deposited on the Moon's surface by astronauts in 1969, and measuring the round travel time of the light (see Figure 1).

<img_4459>

Figure 1. A laser beam sent from an observatory is used to measure accurately the distance between the Earth and the Moon.

With these observations, they have directly measured that the Moon is slowly receding from the Earth. That is, the Earth-Moon distance is increasing with time. This is happening because due to tidal torques the Earth is transferring angular momentum to the Moon, see Figure 2. In this problem you will derive the basic parameters of the phenomenon.



<img_4430>

Figure 2. The Moon's gravity produces tidal deformations or "bulges" in the Earth. Because of the Earth's rotation, the line that goes through the bulges is not aligned with the line between the Earth and the Moon. This misalignment produces a torque that transfers angular momentum from the Earth's rotation to the Moon's translation. The drawing is not to scale.

1. Conservation of Angular Momentum. 

Let $L_{1}$ be the present total angular momentum of the Earth-Moon system. Now, make the following assumptions: i) $L_{1}$ is the sum of the rotation of the Earth around its axis and the translation of the Moon in its orbit around the Earth only. ii) The Moon's orbit is circular and the Moon can be taken as a point. iii) The Earth's axis of rotation and the Moon's axis of revolution are parallel. iv) To simplify the calculations, we take the motion to be around the center of the Earth and not the center of mass. Throughout the problem, all moments of inertia, torques and angular momenta are defined around the axis of the Earth. v) Ignore the influence of the Sun.
Context question:
1a Write down the equation for the present total angular momentum of the Earth-Moon system. Set this equation in terms of $I_{E}$, the moment of inertia of the Earth; $a_{E 1}$, the present angular frequency of the Earth's rotation; $I_{M 1}$, the present moment of inertia of the Moon with respect to the Earth's axis; and $a_{M 1}$, the present angular frequency of the Moon's orbit.
Context answer:
\boxed{$L_{1}=I_{E} \omega_{E 1}+I_{M 1} \omega_{M 1}$}


Extra Supplementary Reading Materials:

This process of transfer of angular momentum will end when the period of rotation of the Earth and the period of revolution of the Moon around the Earth have the same duration. At this point the tidal bulges produced by the Moon on the Earth will be aligned with the line between the Moon and the Earth and the torque will disappear.
Context question:
1b Write down the equation for the final total angular momentum $L_{2}$ of the Earth-Moon system. Make the same assumptions as in Question 1a. Set this equation in terms of $I_{E}$, the moment of inertia of the Earth; $a_{2}$, the final angular frequency of the Earth's rotation and Moon's translation; and $I_{M 2}$, the final moment of inertia of the Moon.
Context answer:
\boxed{$L_{2}=I_{E} \omega_{2}+I_{M 2} \omega_{2}$}


Context question:
1c Neglecting the contribution of the Earth's rotation to the final total angular momentum, write down the equation that expresses the angular momentum conservation for this problem.
Context answer:
$I_{E} \omega_{E 1}+I_{M 1} \omega_{M 1}=I_{M 2} \omega_{2}=L_{1}$


Extra Supplementary Reading Materials:

2. Final Separation and Final Angular Frequency of the Earth-Moon System. 

Assume that the gravitational equation for a circular orbit (of the Moon around the Earth) is always valid. Neglect the contribution of the Earth's rotation to the final total angular momentum.

Question: 2a Write down the gravitational equation for the circular orbit of the Moon around the Earth, at the final state, in terms of $M_{E}, a_{2}, G$ and the final separation $D_{2}$ between the Earth and the Moon. $M_{E}$ is the mass of the Earth and $G$ is the gravitational constant.

Solution:
$\omega_{2}^{2} D_{2}^{3}=G M_{E}$

Final answer: $\omega_{2}^{2} D_{2}^{3}=G M_{E}$