IPhO 2009 · Theory problem 1

Evolution of the Earth-Moon System

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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).

Figure 1

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

With this 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.

Figure 2

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.

Part Question Points
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; \(\omega_{E1}\), the present angular frequency of the Earth’s rotation; \(I_{M1}\), the present moment of inertia of the Moon with respect to the Earth’s axis; and \(\omega_{M1}\), the present angular frequency of the Moon’s orbit. 0.2

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.

Part Question Points
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; \(\omega_2\), the final angular frequency of the Earth’s rotation and Moon’s translation; and \(I_{M2}\), the final moment of inertia of the Moon. 0.2
Part Question Points
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. 0.3

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.

Part Question Points
2a Write down the gravitational equation for the circular orbit of the Moon around the Earth, at the final state, in terms of \(M_E\), \(\omega_2\), \(G\) 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. 0.2
Part Question Points
2b Write down the equation for the final separation \(D_2\) between the Earth and the Moon in terms of the known parameters, \(L_1\), the total angular momentum of the system, \(M_E\) and \(M_M\), the masses of the Earth and Moon, respectively, and \(G\). 0.5
Part Question Points
2c Write down the equation for the final angular frequency \(\omega_2\) of the Earth-Moon system in terms of the known parameters \(L_1\), \(M_E\), \(M_M\) and \(G\). 0.5

Below you will be asked to find the numerical values of \(D_2\) and \(\omega_2\). For this you need to know the moment of inertia of the Earth.

Part Question Points
2d Write down the equation for the moment of inertia of the Earth \(I_E\) assuming it is a sphere with inner density \(\rho_i\) from the center to a radius \(r_i\), and with outer density \(\rho_o\) from the radius \(r_i\) to the surface at a radius \(r_o\) (see Figure 3). 0.5
Figure 3

Figure 3. The Earth as a sphere with two densities, \(\rho_i\) and \(\rho_o\).

Determine the numerical values requested in this problem always to two significant digits.

Part Question Points
2e Evaluate the moment of inertia of the Earth \(I_E\), using \(\rho_i=1.3\times10^4\ \mathrm{kg\,m^{-3}}\), \(r_i=3.5\times10^6\ \mathrm{m}\), \(\rho_o=4.0\times10^3\ \mathrm{kg\,m^{-3}}\), and \(r_o=6.4\times10^6\ \mathrm{m}\). 0.2

The masses of the Earth and Moon are \(M_E=6.0\times10^{24}\ \mathrm{kg}\) and \(M_M=7.3\times10^{22}\ \mathrm{kg}\), respectively. The present separation between the Earth and the Moon is \(D_1=3.8\times10^8\ \mathrm{m}\). The present angular frequency of the Earth’s rotation is \(\omega_{E1}=7.3\times10^{-5}\ \mathrm{s^{-1}}\). The present angular frequency of the Moon’s translation around the Earth is \(\omega_{M1}=2.7\times10^{-6}\ \mathrm{s^{-1}}\), and the gravitational constant is \(G=6.7\times10^{-11}\ \mathrm{m^3\,kg^{-1}\,s^{-2}}\).

Part Question Points
2f Evaluate the numerical value of the total angular momentum of the system, \(L_1\). 0.2
Part Question Points
2g Find the final separation \(D_2\) in meters and in units of the present separation \(D_1\). 0.3
Part Question Points
2h Find the final angular frequency \(\omega_2\) in \(\mathrm{s^{-1}}\), as well as the final duration of the day in units of present days. 0.3

Verify that the assumption of neglecting the contribution of the Earth’s rotation to the final total angular momentum is justified by finding the ratio of the final angular momentum of the Earth to that of the Moon. This should be a small quantity.

Part Question Points
2i Find the ratio of the final angular momentum of the Earth to that of the Moon. 0.2

3. How much is the Moon receding per year?

Now, you will find how much the Moon is receding from the Earth each year. For this, you will need to know the equation for the torque acting at present on the Moon. Assume that the tidal bulges can be approximated by two point masses, each of mass \(m\), located on the surface of the Earth, see Fig. 4. Let \(\theta\) be the angle between the line that goes through the bulges and the line that joins the centers of the Earth and the Moon.

Figure 4

Figure 4. Schematic diagram to estimate the torque produced on the Moon by the bulges on the Earth. The drawing is not to scale.

Part Question Points
3a Find \(F_c\), the magnitude of the force produced on the Moon by the closest point mass. 0.4
Part Question Points
3b Find \(F_f\), the magnitude of the force produced on the Moon by the farthest point mass. 0.4

You may now evaluate the torques produced by the point masses.

Part Question Points
3c Find the magnitude of \(\tau_c\), the torque produced by the closest point mass. 0.4
Part Question Points
3d Find the magnitude of \(\tau_f\), the torque produced by the farthest point mass. 0.4
Part Question Points
3e Find the magnitude of the total torque \(\tau\) produced by the two masses. Since \(r_o\ll D_1\) you should approximate your expression to lowest significant order in \(r_o/D_1\). You may use that \((1+x)^a\approx1+ax\), if \(x\ll1\). 1.0
Part Question Points
3f Calculate the numerical value of the total torque \(\tau\), taking into account that \(\theta=3^\circ\) and that \(m=3.6\times10^{16}\ \mathrm{kg}\) (note that this mass is of the order of \(10^{-8}\) times the mass of the Earth). 0.5

Since the torque is the rate of change of angular momentum with time, find the increase in the distance Earth-Moon at present, per year. For this step, express the angular momentum of the Moon in terms of \(M_M\), \(M_E\), \(D_1\) and \(G\) only.

Part Question Points
3g Find the increase in the distance Earth-Moon at present, per year. 1.0

Finally, estimate how much the length of the day is increasing each year.

Part Question Points
3h Find the decrease of \(\omega_{E1}\) per year and how much is the length of the day at present increasing each year. 1.0

4. Where is the energy going?

In contrast to the angular momentum, that is conserved, the total (rotational plus gravitational) energy of the system is not. We will look into this in this last section.

Part Question Points
4a Write down an equation for the total (rotational plus gravitational) energy of the Earth-Moon system at the present, \(E\). Put this equation in terms of \(I_E\), \(\omega_{E1}\), \(M_M\), \(M_E\), \(D_1\) and \(G\) only. 0.4
Part Question Points
4b Write down an equation for the change in \(E\), \(\Delta E\), as a function of the changes in \(D_1\) and in \(\omega_{E1}\). Evaluate the numerical value of \(\Delta E\) for a year, using the values of changes in \(D_1\) and in \(\omega_{E1}\) found in questions 3g and 3h. 0.4

Verify that this loss of energy is consistent with an estimate for the energy dissipated as heat in the tides produced by the Moon on the Earth. Assume that the tides rise, on the average by 0.5 m, a layer of water \(h=0.5\ \mathrm{m}\) deep that covers the surface of the Earth (for simplicity assume that all the surface of the Earth is covered with water). This happens twice a day. Further assume that 10% of this gravitational energy is dissipated as heat due to viscosity when the water descends. Take the density of water to be \(\rho_{\mathrm{water}}=10^3\ \mathrm{kg\,m^{-3}}\) and the gravitational acceleration on the surface of the Earth to be \(g=9.8\ \mathrm{m\,s^{-2}}\).

Part Question Points
4c What is the mass of this surface layer of water? 0.2
Part Question Points
4d Calculate how much energy is dissipated in a year? How does this compare with the energy lost per year by the Earth-Moon system at present? 0.3

Official English text of the problem.

Answers

For the bulge forces and torques, define the distances \[ d_c=\sqrt{D_1^2+r_o^2-2D_1r_o\cos\theta}, \qquad d_f=\sqrt{D_1^2+r_o^2+2D_1r_o\cos\theta}. \] Let \(T_1\) and \(T_2\) denote the present and final day lengths. Numerical results are given to two significant figures.

  • 1a: \(\displaystyle L_1=I_E\omega_{E1}+I_{M1}\omega_{M1}\).
  • 1b: \(\displaystyle L_2=(I_E+I_{M2})\omega_2\).
  • 1c: \(\displaystyle I_E\omega_{E1}+I_{M1}\omega_{M1}=L_1\simeq I_{M2}\omega_2\).
  • 2a: \(\displaystyle \omega_2^2D_2^3=GM_E\).
  • 2b: \(\displaystyle D_2=\frac{L_1^2}{GM_EM_M^2}\).
  • 2c: \(\displaystyle \omega_2=\frac{G^2M_E^2M_M^3}{L_1^3}\).
  • 2d: \(\displaystyle I_E=\frac{8\pi}{15}\left[\rho_i r_i^5+\rho_o(r_o^5-r_i^5)\right]\).
  • 2e: \(\displaystyle I_E=8.0\times10^{37}\ \mathrm{kg\,m^2}\).
  • 2f: \(\displaystyle L_1=3.4\times10^{34}\ \mathrm{kg\,m^2\,s^{-1}}\).
  • 2g: \(\displaystyle D_2=5.5\times10^8\ \mathrm{m}=1.4D_1\).
  • 2h: \(\displaystyle \omega_2=1.6\times10^{-6}\ \mathrm{s^{-1}},\qquad T_2/T_1=47\).
  • 2i: \(\displaystyle \frac{L_{E2}}{L_{M2}}=\frac{I_E}{M_MD_2^2}=3.6\times10^{-3}\).
  • 3a: \(\displaystyle F_c=\frac{GM_Mm}{d_c^2}\).
  • 3b: \(\displaystyle F_f=\frac{GM_Mm}{d_f^2}\).
  • 3c: \(\displaystyle \tau_c=\frac{GM_MmD_1r_o\sin\theta}{d_c^3}\).
  • 3d: \(\displaystyle \tau_f=\frac{GM_MmD_1r_o\sin\theta}{d_f^3}\).
  • 3e: \(\displaystyle \tau=\tau_c-\tau_f\simeq\frac{3GM_Mmr_o^2}{D_1^3}\sin(2\theta)\), directed along the Moon’s orbital angular momentum.
  • 3f: \(\displaystyle \tau=4.1\times10^{16}\ \mathrm{N\,m}\).
  • 3g: \(\displaystyle \dot D_1=\frac{2\tau}{M_M}\sqrt{\frac{D_1}{GM_E}}=3.5\times10^{-2}\ \mathrm{m\,yr^{-1}}=3.5\ \mathrm{cm\,yr^{-1}}\).
  • 3h: In one year, \(\displaystyle \Delta\omega_{E1}=-1.6\times10^{-14}\ \mathrm{s^{-1}}\) and \(\displaystyle \Delta T_1=+1.9\times10^{-5}\ \mathrm{s}\).
  • 4a: \(\displaystyle E=\frac12I_E\omega_{E1}^2-\frac{GM_EM_M}{2D_1}\).
  • 4b: \(\displaystyle \Delta E\simeq I_E\omega_{E1}\Delta\omega_{E1}+\frac{GM_EM_M}{2D_1^2}\Delta D_1=-9.1\times10^{19}\ \mathrm{J}\) in one year.
  • 4c: \(\displaystyle M_{\mathrm{water}}=4\pi r_o^2h\rho_{\mathrm{water}}=2.6\times10^{17}\ \mathrm{kg}\).
  • 4d: \(\displaystyle E_{\mathrm{heat,yr}}=9.2\times10^{19}\ \mathrm{J}\), in close agreement with the mechanical energy loss of \(9.1\times10^{19}\ \mathrm{J}\) per year.

Solution

All calculations use the assumptions stated in the problem. Numerical results are rounded only at the end. For yearly changes, take \[ t_{\mathrm{yr}}=365\times24\times3600\ \mathrm{s}. \]

1a. Present angular momentum

The Earth’s spin and the Moon’s revolution are in the same sense about parallel axes, so their angular momenta add: \[ \boxed{L_1=I_E\omega_{E1}+I_{M1}\omega_{M1}.} \] Since the Moon is treated as a point at distance \(D_1\) from the Earth’s axis, \[ I_{M1}=M_MD_1^2. \]

1b. Final angular momentum

In the final synchronized state, both angular frequencies equal \(\omega_2\). Thus \[ \boxed{L_2=I_E\omega_2+I_{M2}\omega_2=(I_E+I_{M2})\omega_2,} \qquad I_{M2}=M_MD_2^2. \]

1c. Angular momentum conservation

There is no external torque in the stated model. The tidal torques transfer angular momentum internally, giving \(L_1=L_2\).

Neglecting only the final Earth spin contribution, \[ \boxed{I_E\omega_{E1}+I_{M1}\omega_{M1}\simeq I_{M2}\omega_2.} \] The present Earth spin must still be included in \(L_1\).

2a. Circular-orbit equation

The gravitational force supplies the Moon’s centripetal acceleration: \[ \frac{GM_EM_M}{D_2^2}=M_M\omega_2^2D_2. \] Cancelling \(M_M\) gives \[ \boxed{\omega_2^2D_2^3=GM_E.} \]

2b. Final separation

Within the approximation of part 1c, \[ L_1=M_MD_2^2\omega_2. \] Using \(\omega_2=\sqrt{GM_E/D_2^3}\), \[ L_1=M_M\sqrt{GM_ED_2}. \] Squaring and solving, \[ \boxed{D_2=\frac{L_1^2}{GM_EM_M^2}.} \]

2c. Final angular frequency

Substitution of the preceding result into the circular-orbit equation yields \[ \omega_2 =\sqrt{GM_E}\left(\frac{GM_EM_M^2}{L_1^2}\right)^{3/2}. \] Therefore \[ \boxed{\omega_2=\frac{G^2M_E^2M_M^3}{L_1^3}.} \]

2d. Earth’s moment of inertia

A thin spherical shell of radius \(r\), density \(\rho\), and thickness \(dr\) has \[ dm=4\pi\rho r^2\,dr, \qquad dI=\frac23r^2\,dm=\frac{8\pi}{3}\rho r^4\,dr. \] Integrating over the two density regions, \[ I_E=\frac{8\pi}{3} \left[ \rho_i\int_0^{r_i}r^4\,dr+ \rho_o\int_{r_i}^{r_o}r^4\,dr \right]. \] Thus \[ \boxed{ I_E=\frac{8\pi}{15} \left[\rho_i r_i^5+\rho_o(r_o^5-r_i^5)\right]. } \]

2e. Numerical moment of inertia

Substituting the given densities and radii, \[ I_E=\frac{8\pi}{15} \left[ (1.3\times10^4)(3.5\times10^6)^5 +(4.0\times10^3) \left((6.4\times10^6)^5-(3.5\times10^6)^5\right) \right]\mathrm{kg\,m^2}. \] Hence \[ \boxed{I_E=8.0\times10^{37}\ \mathrm{kg\,m^2}.} \]

2f. Present total angular momentum

Using the given present angular frequencies, \[ L_1=I_E\omega_{E1}+M_MD_1^2\omega_{M1}. \] The two contributions are \[ I_E\omega_{E1}=5.8\times10^{33}\ \mathrm{kg\,m^2\,s^{-1}}, \] \[ M_MD_1^2\omega_{M1}=2.8\times10^{34}\ \mathrm{kg\,m^2\,s^{-1}}. \] Adding them with unrounded intermediate values, \[ \boxed{L_1=3.4\times10^{34}\ \mathrm{kg\,m^2\,s^{-1}}.} \]

2g. Numerical final separation

Using part 2b and the unrounded value of \(L_1\), \[ \boxed{D_2=\frac{L_1^2}{GM_EM_M^2}=5.5\times10^8\ \mathrm{m}.} \] Relative to the present separation, \[ \boxed{\frac{D_2}{D_1}=1.4.} \]

2h. Final frequency and day length

From part 2c, \[ \boxed{\omega_2=1.6\times10^{-6}\ \mathrm{s^{-1}}.} \]

Since a rotation period is \(T=2\pi/\omega\), \[ \frac{T_2}{T_1} =\frac{2\pi/\omega_2}{2\pi/\omega_{E1}} =\frac{\omega_{E1}}{\omega_2}. \] Using unrounded intermediate values gives \[ \boxed{T_2=47T_1,} \] so the final day lasts 47 present days.

2i. Check of the neglected final spin

Both final angular momenta contain the same angular frequency: \[ \frac{L_{E2}}{L_{M2}} =\frac{I_E\omega_2}{M_MD_2^2\omega_2} =\frac{I_E}{M_MD_2^2}. \] Numerically, \[ \boxed{\frac{L_{E2}}{L_{M2}}=3.6\times10^{-3}.} \] The final Earth spin angular momentum is only \(0.36\%\) of the Moon’s orbital angular momentum, justifying its neglect.

3a. Force from the closest bulge

Choose the Earth’s center as the origin, put the Moon at \((D_1,0)\), and take positive \(y\) along the Moon’s instantaneous motion. The closest bulge is at \[ (r_o\cos\theta,r_o\sin\theta). \] Its distance from the Moon is therefore \[ d_c^2=(D_1-r_o\cos\theta)^2+(r_o\sin\theta)^2 =D_1^2+r_o^2-2D_1r_o\cos\theta. \] Newton’s gravitational law gives \[ \boxed{ F_c=\frac{GM_Mm}{D_1^2+r_o^2-2D_1r_o\cos\theta}. } \]

3b. Force from the farthest bulge

The farther bulge is at \[ (-r_o\cos\theta,-r_o\sin\theta). \] Consequently, \[ d_f^2=(D_1+r_o\cos\theta)^2+(r_o\sin\theta)^2 =D_1^2+r_o^2+2D_1r_o\cos\theta, \] and \[ \boxed{ F_f=\frac{GM_Mm}{D_1^2+r_o^2+2D_1r_o\cos\theta}. } \]

3c. Torque from the closest bulge

Only the force component perpendicular to the Earth-Moon line contributes to the torque about the Earth’s axis. For the closer bulge, \[ F_{c,y}=F_c\frac{r_o\sin\theta}{d_c}. \] Its torque is positive, in the direction of the Moon’s orbital angular momentum: \[ \tau_c=D_1F_{c,y}. \] Thus its magnitude is \[ \boxed{ \tau_c= \frac{GM_MmD_1r_o\sin\theta} {(D_1^2+r_o^2-2D_1r_o\cos\theta)^{3/2}}. } \]

3d. Torque from the farthest bulge

The farther bulge pulls the Moon in the negative \(y\) direction: \[ F_{f,y}=-F_f\frac{r_o\sin\theta}{d_f}. \] Its torque opposes the closer bulge’s torque. Its magnitude is \[ \boxed{ \tau_f= \frac{GM_MmD_1r_o\sin\theta} {(D_1^2+r_o^2+2D_1r_o\cos\theta)^{3/2}}. } \]

3e. Total torque to lowest nonzero order

Because \(d_c<d_f\), the closer bulge produces the larger torque. For the geometry shown, \[ \tau=\tau_c-\tau_f =GM_MmD_1r_o\sin\theta \left(\frac1{d_c^3}-\frac1{d_f^3}\right). \]

Define \(\epsilon=r_o/D_1\ll1\). Then \[ \frac1{d_c^3} =\frac1{D_1^3}(1+\epsilon^2-2\epsilon\cos\theta)^{-3/2} \simeq\frac1{D_1^3}(1+3\epsilon\cos\theta), \] and \[ \frac1{d_f^3} =\frac1{D_1^3}(1+\epsilon^2+2\epsilon\cos\theta)^{-3/2} \simeq\frac1{D_1^3}(1-3\epsilon\cos\theta). \] The zeroth-order terms cancel, leaving \[ \tau\simeq GM_MmD_1r_o\sin\theta \frac{6r_o\cos\theta}{D_1^4}. \] Therefore \[ \boxed{ \tau\simeq\frac{6GM_Mmr_o^2}{D_1^3}\sin\theta\cos\theta =\frac{3GM_Mmr_o^2}{D_1^3}\sin(2\theta). } \] This torque increases the Moon’s orbital angular momentum.

3f. Numerical torque

With \(\theta=3^\circ\), \[ \tau\simeq \frac{3(6.7\times10^{-11})(7.3\times10^{22}) (3.6\times10^{16})(6.4\times10^6)^2} {(3.8\times10^8)^3} \sin6^\circ. \] Hence \[ \boxed{\tau=4.1\times10^{16}\ \mathrm{N\,m}.} \]

3g. Present recession of the Moon

For a circular orbit at separation \(D\), \[ \omega_M=\sqrt{\frac{GM_E}{D^3}}, \] so the Moon’s angular momentum is \[ L_M=M_MD^2\omega_M=M_M\sqrt{GM_ED}. \] Differentiating with respect to time, \[ \tau=\frac{dL_M}{dt} =\frac{M_M}{2}\sqrt{\frac{GM_E}{D}}\,\dot D. \] Thus, at the present separation, \[ \boxed{\dot D_1=\frac{2\tau}{M_M}\sqrt{\frac{D_1}{GM_E}}.} \] Numerically, \[ \dot D_1=1.1\times10^{-9}\ \mathrm{m\,s^{-1}}. \] The increase over one year is \[ \Delta D_1=\dot D_1t_{\mathrm{yr}}, \] giving \[ \boxed{\Delta D_1=3.5\times10^{-2}\ \mathrm{m}=3.5\ \mathrm{cm} \quad\text{in one year}.} \]

3h. Earth’s slowing rotation and increasing day length

The torque on the Earth is opposite to the torque on the Moon. With constant \(I_E\), \[ I_E\dot\omega_E=-\tau. \] Therefore \[ \dot\omega_{E1}=-\frac{\tau}{I_E} =-5.2\times10^{-22}\ \mathrm{s^{-2}}, \] and the change in one year is \[ \boxed{ \Delta\omega_{E1}=-\frac{\tau t_{\mathrm{yr}}}{I_E} =-1.6\times10^{-14}\ \mathrm{s^{-1}}. } \]

Since \(T_E=2\pi/\omega_E\), a small change satisfies \[ \Delta T_E\simeq-\frac{2\pi}{\omega_E^2}\Delta\omega_E. \] Consequently, \[ \boxed{ \Delta T_1 =\frac{2\pi\tau t_{\mathrm{yr}}}{I_E\omega_{E1}^2} =1.9\times10^{-5}\ \mathrm{s} \quad\text{in one year}. } \] The day lengthens by approximately \(19\ \mathrm{\mu s}\) per year.

4a. Present mechanical energy

The mechanical energy consists of the Earth’s spin kinetic energy, the Moon’s orbital kinetic energy, and their gravitational potential energy: \[ E=\frac12I_E\omega_{E1}^2 +\frac12M_MD_1^2\omega_{M1}^2 -\frac{GM_EM_M}{D_1}. \] The circular-orbit relation gives \[ \frac12M_MD_1^2\omega_{M1}^2 =\frac{GM_EM_M}{2D_1}. \] Hence \[ \boxed{ E=\frac12I_E\omega_{E1}^2-\frac{GM_EM_M}{2D_1}. } \]

4b. Annual mechanical energy change

For changes \(\Delta\omega_{E1}\) and \(\Delta D_1\), the exact difference within this energy model is \[ \Delta E= \frac12I_E\left[(\omega_{E1}+\Delta\omega_{E1})^2-\omega_{E1}^2\right] -\frac{GM_EM_M}{2} \left[\frac1{D_1+\Delta D_1}-\frac1{D_1}\right]. \] The annual fractional changes are extremely small, so retaining first-order terms gives \[ \boxed{ \Delta E\simeq I_E\omega_{E1}\Delta\omega_{E1} +\frac{GM_EM_M}{2D_1^2}\Delta D_1. } \] The first term is negative: the Earth loses spin energy. The second is positive: the Moon’s orbital energy increases as its orbit expands.

Substituting the annual changes from parts 3g and 3h, retaining their unrounded values, \[ \boxed{\Delta E=-9.1\times10^{19}\ \mathrm{J} \quad\text{in one year}.} \]

The energy balance can also be expressed directly in terms of the torque: \[ \dot E_{\mathrm{spin}}=-\tau\omega_{E1}. \] Using the recession formula, \[ \dot E_{\mathrm{orbit}} =\frac{GM_EM_M}{2D_1^2}\dot D_1 =\tau\sqrt{\frac{GM_E}{D_1^3}}. \] Thus \[ \dot E=-\tau(\omega_{E1}-\omega_{M1}), \] where \(\omega_{M1}\) is understood through the circular-orbit relation. Since the Earth spins faster than the Moon revolves, the mechanical energy decreases; the difference becomes heat.

4c. Mass of the water layer

Since the layer is very thin compared with the Earth’s radius, its volume is its surface area times its thickness: \[ V\simeq4\pi r_o^2h. \] Therefore \[ M_{\mathrm{water}} =4\pi r_o^2h\rho_{\mathrm{water}} =4\pi(6.4\times10^6)^2(0.5)(10^3)\ \mathrm{kg}. \] Hence \[ \boxed{M_{\mathrm{water}}=2.6\times10^{17}\ \mathrm{kg}.} \]

4d. Annual heat dissipation

The stated average rise is \(H=0.5\ \mathrm{m}\). The gravitational energy gained by the layer in one rise is \[ E_{\mathrm{rise}}=M_{\mathrm{water}}gH. \] Ten percent is dissipated as heat during the subsequent descent. With two cycles per day, there are \(2\times365\) cycles per year: \[ E_{\mathrm{heat,yr}} =0.10(2\times365)M_{\mathrm{water}}gH. \] Substitution gives \[ \boxed{E_{\mathrm{heat,yr}}=9.2\times10^{19}\ \mathrm{J}.} \]

This agrees closely with the mechanical energy loss \[ |\Delta E|=9.1\times10^{19}\ \mathrm{J} \] found in part 4b. Within the simplified model, tidal dissipation accounts for the lost mechanical energy by converting it into heat.

The solution was checked against the official solution.