How do tides vary over a lunar cycle?

[Hak]

Homework Statement

You went to the equator near a coast and measured the height of the sea every minute of the day for a whole year.

You found that high tide, which is when the height of the water reaches a maximum and then begins to decrease, occurs every 12 hours. You also noticed that the height of the high tide fluctuates over time, following the lunar cycle.

1. For which phase of the Moon does the maximum height occur, and for which does the minimum?
2. Estimate how much the height of the high tide varies over the course of a lunar cycle, as a percentage.

Make as many approximations as you need and find the necessary data if you don't know it by heart (obviously without looking up explanations of tides...).

Relevant Equations

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1. All I know, from trite and rehashed information, is that the maximum height of the high tide occurs when the Sun, the Earth and the Moon are aligned, that is, in the phases of new moon and full moon, whereas the minimum height of the high tide occurs when the Sun, the Earth and the Moon form a right angle, that is, in the phases of first quarter and last quarter.

2. All I know is that the height of the high tide varies over the course of a lunar cycle depending on the distance between the Earth and the Moon, which is not constant.

What calculations and approximations should I sketch out? A calculative solution is called for, but I don't know how to get calculations behind such a problem. Any advice on this is appreciated, I hope you can help me in this regard. Thank you very much.

(If you feel that this problem is not "Introductory Physics", you are very free to move it, unfortunately I am not able to understand the real difficulty of the problem, but to me it does not seem easy...)

 

[jbriggs444]

Homework Statement: You went to the equator near a coast and measured the height of the sea every minute of the day for a whole year.

You found that high tide, which is when the height of the water reaches a maximum and then begins to decrease, occurs every 12 hours. You also noticed that the height of the high tide fluctuates over time, following the lunar cycle.

1. For which phase of the Moon does the maximum height occur, and for which does the minimum?
2. Estimate how much the height of the high tide varies over the course of a lunar cycle, as a percentage.

Make as many approximations as you need and find the necessary data if you don't know it by heart (obviously without looking up explanations of tides...).
Relevant Equations: /

1. All I know, from trite and rehashed information, is that the maximum height of the high tide occurs when the Sun, the Earth and the Moon are aligned, that is, in the phases of new moon and full moon, whereas the minimum height of the high tide occurs when the Sun, the Earth and the Moon form a right angle, that is, in the phases of first quarter and last quarter.

2. All I know is that the height of the high tide varies over the course of a lunar cycle depending on the distance between the Earth and the Moon, which is not constant.

What calculations and approximations should I sketch out? A calculative solution is called for, but I don't know how to get calculations behind such a problem. Any advice on this is appreciated, I hope you can help me in this regard. Thank you very much.

(If you feel that this problem is not "Introductory Physics", you are very free to move it, unfortunately I am not able to understand the real difficulty of the problem, but to me it does not seem easy...)

Reading the questioner's mind... You should ignore any variation in the distance to the moon. That would not follow the lunar cycle anyway. Can you see why?

Do you understand what causes the tides? What is your understanding?

 

[kuruman]

I think you should extend "all you know" by doing some preliminary research on the internet.
Read this for an introduction.

 

[Hak]

Reading the questioner's mind... You should ignore any variation in the distance to the moon. That would not follow the lunar cycle anyway. Can you see why?

Do you understand what causes the tides? What is your understanding?

My understanding is that the tides are mainly caused by the gravitational attraction of the Moon and the Sun on the Earth’s oceans. The Moon has a stronger influence than the Sun because it is much closer to the Earth. The gravitational force of the Moon and the Sun creates bulges of water on the side of the Earth facing them and on the opposite side: these bulges are the high tides, and the areas between them are the low tides. As the Earth rotates, different locations experience high and low tides at different times. So, do you mean that the distance to the Moon does vary slightly due to its elliptical orbit, but this variation does not follow the lunar cycle? Reasonable, because the lunar cycle is based on the phases of the Moon, which depend on its position relative to the Sun and the Earth: the distance to the Moon affects its apparent size and brightness, but not its phase. The variation in distance to the Moon does affect the tides, but in a different way than the lunar cycle. When the Moon is closer to the Earth (at perigee), it exerts a stronger gravitational force and causes higher high tides and lower low tides; when the Moon is farther from the Earth (at apogee), it exerts a weaker gravitational force and causes lower high tides and higher low tides. These variations occur about once a month, but not necessarily at the same time as the new or full moon.

I thought I could find the height by relating the variation of the height of high tide, the maximum height of high tide, the minimum height of high tide, the average tidal range at the location, and the angle between the Sun, Earth, and Moon. Something like this, that's all I thought of. I don't know how to formalize it from a physical point of view...

To express the variation as a percentage of the high tide height, we need to know what that value is. So, I'm not able to answer this.

Do you have some hints? Could you help me?

 

[jbriggs444]

My understanding is that the tides are mainly caused by the gravitational attraction of the Moon and the Sun on the Earth’s oceans. The Moon has a stronger influence than the Sun because it is much closer to the Earth.

But why should the sun attract the oceans more strongly than the rest of the Earth?

What would be the effect of the inverse square law for gravity?

 

[haruspex]

But why should the sun attract the oceans more strongly than the rest of the Earth?

… and why only on the side nearer the sun/moon?

These variations occur about once a month, but not necessarily at the same time as the new or full moon.

That would not rule out the two periods being exactly the same; could be just out of phase. But as you know, they have different periods. So does the question setter expect you to take into account the distance variation when calculating how the tide height varies over a lunar cycle?

What two parameters determine the Sun's and Moon's gravitational pulls?

 

[Hak]

But why should the sun attract the oceans more strongly than the rest of the Earth?

What would be the effect of the inverse square law for gravity?

I think the Sun attracts the oceans more strongly than the rest of the Earth because the oceans are closer to the Sun than the Earth’s center of mass. The gravitational force between two objects depends on the inverse square of the distance between them, so a small difference in distance can make a significant difference in force. So, the effect of the inverse square law for gravity would be that it makes gravity very weak at large distances, but very strong at small distances.

… and why only on the side nearer the sun/moon?

I think the oceans experience a net force toward the Sun because they are pulled more than the average. This net force creates a bulge or a high tide on the side of the Earth facing the Sun. A similar effect happens on the opposite side of the Earth, where the oceans are farther from the Sun than the Earth’s center of mass. The Sun pulls the Earth more than the oceans, creating a net force away from the Sun, and that force also creates another bulge or another high tide on the side of the Earth away from the Sun.

What two parameters determine the Sun's and Moon's gravitational pulls?

In my opinion, the two parameters that determine the Sun’s and Moon’s gravitational pulls are their masses and their distances from the Earth: the gravitational force between two objects is proportional to the product of their masses and inversely proportional to the square of their distance. This means that the more massive and closer an object is, the stronger its gravitational pull is, according to the equation for calculating the gravitational force:

$$F=G \frac{m_1m_2}{r^2}$$, where ##F## is the force, ##G## is the universal gravitational constant, ##m_1## and ##m_2## are the masses of the two objects, and ##r## is the distance between them.

Is it correct?

 

[haruspex]

the two parameters that determine the Sun’s and Moon’s gravitational pulls are their masses and their distances from the Earth: the gravitational force between two objects is proportional to the product of their masses and inversely proportional to the square of their distance

So can you calculate the relative strengths of the pulls?

 

[Hak]

So can you calculate the relative strengths of the pulls?

I don't know if I understand. I could use values taken from the net (I don't remember them by heart), so we can plug in the numbers for the Sun and the Earth:$$F_{S}=G \frac{m_{S}m_E}{r_S^2}$$where ##m_S## is the mass of the Sun, ##m_E## is the mass of the Earth, and ##r_S## is the average distance between them.$$F_S=(6.674 \times 10^{−11} N m^2/kg^2)\frac{(1.9885 \times 10^{30} kg)(5.9724 \times 10^{24} kg)}{(1.496 \times 10^{11} m)^2} =3.542 \times 10^{22} N$$Similarly, we can plug in the numbers for the Moon and the Earth:$$F_M=G \frac{m_M m_E}{r_M^2}$$where ##m_M## is the mass of the moon, and ##r_M## is the average distance between them.$$F_M= (6.674 \times 10^{−11} N m^2/kg^2) \frac{(7.342 \times 10^{22} kg)(5.9724 \times 10^{24} kg)}{(3.844×10^8 m)^2} = 1.981 \times 10^{20} N$$.

Is it correct?

 

[haruspex]

I don't know if I understand. I could use values taken from the net (I don't remember them by heart), so we can plug in the numbers for the Sun and the Earth:$$F_{S}=G \frac{m_{S}m_E}{r_S^2}$$where ##m_S## is the mass of the Sun, ##m_E## is the mass of the Earth, and ##r_S## is the average distance between them.$$F_S=(6.674 \times 10^{−11} N m^2/kg^2)\frac{(1.9885 \times 10^{30} kg)(5.9724 \times 10^{24} kg)}{(1.496 \times 10^{11} m)^2} =3.542 \times 10^{22} N$$Similarly, we can plug in the numbers for the Moon and the Earth:$$F_M=G \frac{m_M m_E}{r_M^2}$$where ##m_M## is the mass of the moon, and ##r_M## is the average distance between them.$$F_M= (6.674 \times 10^{−11} N m^2/kg^2) \frac{(7.342 \times 10^{22} kg)(5.9724 \times 10^{24} kg)}{(3.844×10^8 m)^2} = 1.981 \times 10^{20} N$$.

Is it correct?

I think I misled you by being careless with my question.
As you noted, each pulls more strongly on the side close to it than on the Earth CoM, and that more strongly than on the far side.
How much stronger is the Sun's field on the side of the Earth nearest it than on Earth's centre?
What about the moon?

 

[Hak]

I think I misled you by being careless with my question.
As you noted, each pulls more strongly on the side close to it than on the Earth CoM, and that more strongly than on the far side.
How much stronger is the Sun's field on the side of the Earth nearest it than on Earth's centre?
What about the moon?

Yes, you're right, I'm sorry for the misunderstanding. We can assume that the Earth’s radius is negligible compared to the distance from the sun, so we can use the average distance between them as ##r_S##. We can also assume that the mass of the Earth is evenly distributed, so we can use the same value for ##m_E## at both points. The only difference is that the point nearest to the sun is closer to it by half of Earth’s radius, or ##r_E/2##, where ##r_E## is Earth’s radius. Therefore, we can write:$$F_S=G\frac{m_Sm_E}{(r_S−r_E/2)^2}$$where ##F_S## is the force on the point nearest to the sun.$$F_c=G \frac{m_S m_E}{r_S^2}$$where ##F_c## is the force on Earth’s center.To find the ratio of these two forces, we can divide them:$$\frac{F_c}{F_S}=\frac{(r_s−r_e/2)^2}{r_S^2}$$Using the values from the net, plugging in the numbers:$$\frac{F_c}{F_S}=\frac{(1.496 \times 10^{11}−6.371 \times 10^6/2)^2}{(1.496 \times 10^{11})^2}=1.000000085$$This means that the Sun’s field is about ##0.0000085%## stronger on the side of the Earth nearest it than on Earth’s center.

To find how much stronger the moon’s field is on the side of the Earth nearest it than on Earth’s center, we can use the same approach. We can use the average distance between them as ##r_M##. We have:

$$ \frac{F_c}{F_M}= \frac{(3.844 \times 10^8 - 6.371 \times 10^6/2)^2}{(3.844 \times 10^8)^2} = 1.0000333 $$

This means that the Moon’s field is about ##0.0033%## stronger on the side of the Earth nearest it than on Earth’s center.

So, how is it going now?

 

[haruspex]

Haven’t forgotten about this, just having trouble finding enough time for it.

 

[Hak]

Haven’t forgotten about this, just having trouble finding enough time for it.

Don't worry, it's OK. I'll wait as long as you need...

 

[haruspex]

Seems to me it is all about equipotentials. At any instant (ignoring lag) the oceans' surface is all at the same potential.
We have four contributors to that: gravities of Earth, Sun and Moon, and, in a frame of reference rotating with Earth's orbit about the Sun, a potential associated with the centrifugal force. By integration, that last would be ##-\frac 12D_s^2\omega^2##, where ##D_s## is the distance from the Sun to the point.

Say a point on the surface nearest the moon is ##R+y## from the centre of the Earth, and a point a quarter of the way around the Earth from there is at ##R-y##. If we set the total potential at those two points equal, and take suitable approximations, we should get an expression for y. We can do this for each of the two cases: Sun aligned with Moon and 90° out.

 

[Hak]

Seems to me it is all about equipotentials. At any instant (ignoring lag) the oceans' surface is all at the same potential.
We have four contributors to that: gravities of Earth, Sun and Moon, and, in a frame of reference rotating with Earth's orbit about the Sun, a potential associated with the centrifugal force. By integration, that last would be ##-\frac 12D_s^2\omega^2##, where ##D_s## is the distance from the Sun to the point.

Say a point on the surface nearest the moon is ##R+y## from the centre of the Earth, and a point a quarter of the way around the Earth from there is at ##R-y##. If we set the total potential at those two points equal, and take suitable approximations, we should get an expression for y. We can do this for each of the two cases: Sun aligned with Moon and 90° out.

Thank you very much. I did not understand how you did the first integration, I will try to work on it. In general, I don't think I understand the physical situation very well, could you explain in more detail again? Anyway, I will try to follow your advice and find an expression for ##y##. Thank you.

 

[Hak]

Seems to me it is all about equipotentials. At any instant (ignoring lag) the oceans' surface is all at the same potential.
We have four contributors to that: gravities of Earth, Sun and Moon, and, in a frame of reference rotating with Earth's orbit about the Sun, a potential associated with the centrifugal force. By integration, that last would be ##-\frac 12D_s^2\omega^2##, where ##D_s## is the distance from the Sun to the point.

Say a point on the surface nearest the moon is ##R+y## from the centre of the Earth, and a point a quarter of the way around the Earth from there is at ##R-y##. If we set the total potential at those two points equal, and take suitable approximations, we should get an expression for y. We can do this for each of the two cases: Sun aligned with Moon and 90° out.

To find the potential associated with the centrifugal force, I integrated the expression for the centrifugal force with respect to the distance from the Sun. The centrifugal force is given by:
$$F_c=m \omega^2 D_s$$,

where ##m## is the mass of a point on the ocean’s surface, ##\omega## is the angular velocity of Earth’s orbit around the sun, and ##D_s## is the distance from the Sun to that point. The potential energy of a point due to this force is:

$$U_c=−\int F_c dD_s$$

Using the expression for ##F_c##, we can write:

$$U_c=−\int m\omega^2 D_s dD_s$$

Integrating, I get:

$$U_c=−\frac 12 m \omega^2 D_s^2+C$$

where ##C## is an arbitrary constant of integration. This isn't the same expression you gave in your answer, because there is an extra ##m##. Why?To find the expression for ##y##, I equated the total potential at two points on opposite sides of Earth. The total potential is given by:$$U=U_E+U_M+U_S+U_c$$,

where ##U_E## is the potential due to Earth’s gravity, ##U_M## is the potential due to the Moon’s gravity, ##U_S## is the potential due to the Sun’s gravity, and ##U_c## is the potential due to the centrifugal force. We can write:

$$U_E=−\frac{GM_Em}{R}$$

where ##G## is the universal gravitational constant, ##M_E## is Earth’s mass, and ##R## is Earth’s radius.

$$U_M=−\frac{GM_Mm}{R_M}(1+2y/R_M)$$where ##M_M## is the Moon’s mass, and ##R_M## is the distance from Earth’s center to the moon.$$U_S=−\frac{GM_Sm(1+2y/R_S)}{R_M}$$where ##M_S## is the sun’s mass, and ##R_S## is the distance from Earth’s center to the sun.$$U_c=-\frac 12 m \omega^2 D_s^2$$where ##\omega## is Earth’s orbital angular velocity, and ##D_s## is the distance from the Sun to a point on its orbit.For a point on Earth’s surface nearest to or farthest from the moon, we have ##y=R##. For a point on Earth’s surface perpendicular to that line, we have ##y=0##. Equating these two cases, we get:$$-\frac{GM_Em}{R}-\frac{GM_Mm}{R_M}(1+2R/R_M)−\frac{GM_Sm}{R_S}(1+2R/R_S)-\frac 12m\omega^2D_s^2 =-\frac{GM_Em}{R}-\frac{GM_Mm}{R_M}-\frac{GM_Sm}{R_S}- \frac 12 m\omega^2D_s^2$$Solving for ##R##, we get:$$R=0$$This means that there is no difference in height between these two points when the sun and moon are aligned. I don't believe it's right. Where is the mistake?When they are perpendicular, we have to use a different value for ##D_s##, which is given by:$$D_s=R_S \cos(\theta)+R \sin(\theta)$$where ##\theta## is the angle between Earth’s orbit and its axis of rotation. Using this value and equating the potentials again, I get:

$$-\frac{GM_Em}{R}-\frac{GM_Mm}{R_M}(1+2R/R_M)−\frac{GM_Sm}{R_S}(1+2R/R_S)-\frac 12m\omega^2(R_S \cos(\theta)+R \sin(\theta))^2 =-\frac{GM_Em}{R}-\frac{GM_Mm}{R_M}-\frac{GM_Sm}{R_S}- \frac 12 m\omega^2 (R_S \cos(\theta)+R \sin(\theta))^2$$Solving for ##R##, I get:$$R=\frac{GM_M(R_M−R)}{4(\omega^2(R_S \cos(\theta)+R \sin(\theta))+G M_S/R_S)}$$This is my expression for ##y## when the Sun and Moon are perpendicular.

Where am I going wrong? Thanks.

 

[haruspex]

This isn't the same expression you gave in your answer, because there is an extra m. Why?

You are confusing potential with potential energy. E.g. in electrostatics, if a charge q is at potential P then its potential energy is Pq. In the potentials in this problem, there should be no factor m.

I equated the total potential at two points on opposite sides of Earth.

Doesn't seem to me that's what you did.
##U_E## depends on y, but that will be the same on opposite sides. That's why I wrote you should equate the potentials for two points 90° apart, not 180°.

Also, I do not understand how you got those expressions for ##U_S## and ##U_M##.
There should be different expressions for the two points. If taking them on opposite sides, one should have +y where the other has -y.

But I made a serious omission earlier. We also need the centrifugal potential in respect of Earth's rotation about the Earth-Moon mass centre. In view of this, just start with the two body case (one being Earth, the other generic for now) and figure out the sum of the centrifugal and gravitational potentials. Don't include Earth's gravity at first.

 

[Hak]

You are confusing potential with potential energy. E.g. in electrostatics, if a charge q is at potential P then its potential energy is Pq. In the potentials in this problem, there should be no factor m.

Doesn't seem to me that's what you did.
##U_E## depends on y, but that will be the same on opposite sides. That's why I wrote you should equate the potentials for two points 90° apart, not 180°.

Also, I do not understand how you got those expressions for ##U_S## and ##U_M##.
There should be different expressions for the two points. If taking them on opposite sides, one should have +y where the other has -y.

But I made a serious omission earlier. We also need the centrifugal potential in respect of Earth's rotation about the Earth-Moon mass centre. In view of this, just start with the two body case (one being Earth, the other generic for now) and figure out the sum of the centrifugal and gravitational potentials. Don't include Earth's gravity at first.

You're right on everything.
I'm going to revise the expression for the centrifugal potential that I used before. So we have:

$$U_c=-\frac 12 \omega^2 D^2$$,

where ##\omega## is the angular velocity of the orbit, and ##D## is the distance from the center of mass of the system. If I understand your suggestion, I get that the gravitational potential is given by:$$U_g=- \frac{GM_1M_2}{r}$$,

where ##G## is the universal gravitational constant, ##M_1## and ##M_2## are the masses of the two bodies, and ##r## is the distance between them. The sum of these two potentials is:$$U=Uc+Ug=-\frac 12 \omega^2 D^2 - \frac{GM_1M_2}{r}$$.

Is it correct? How to continue?

 

[haruspex]

You're right on everything.
I'm going to revise the expression for the centrifugal potential that I used before. So we have:

$$U_c=-\frac 12 \omega^2 D^2$$,

where ##\omega## is the angular velocity of the orbit, and ##D## is the distance from the center of mass of the system. If I understand your suggestion, I get that the gravitational potential is given by:$$U_g=- \frac{GM_1M_2}{r}$$,

where ##G## is the universal gravitational constant, ##M_1## and ##M_2## are the masses of the two bodies, and ##r## is the distance between them. The sum of these two potentials is:$$U=Uc+Ug=-\frac 12 \omega^2 D^2 - \frac{GM_1M_2}{r}$$.

Is it correct? How to continue?

That's dimensionally wrong because you are still mixing potentials and potential energies.
The gravitational potential at R from a mass M is just ##-\frac{GM}R##.

 

[Hak]

That's dimensionally wrong because you are still mixing potentials and potential energies.
The gravitational potential at R from a mass M is just ##-\frac{GM}R##.

Yes yes, you are right, I already knew that. I made a mistake in copying from my previous post. Anyway, now what do you recommend to do?

 

[haruspex]

Yes yes, you are right, I already knew that. I made a mistake in copying from my previous post. Anyway, now what do you recommend to do?

Two things.
r needs to be not the distance between the two bodies but, rather, the distance from the non-Earth body to a point of the ocean surface.
There is a relationship between ##\omega##, M and the distance of the Earth from the common mass centre.

 

[Hak]

Two things.
r needs to be not the distance between the two bodies but, rather, the distance from the non-Earth body to a point of the ocean surface.
There is a relationship between ##\omega##, M and the distance of the Earth from the common mass centre.

Thanks.
To revise your first objection, I feel like I need to use a different value for ##r##. I could assume that this distance is approximately equal to the distance from the center of Earth to that point, plus or minus the radius of Earth, depending on whether the point is closer or farther from the non-Earth body. Therefore, I could write:

$$r=R_E \pm r_E$$where ##R_E## is the distance from Earth’s center to the non-Earth body, and ##r_E## is Earth’s radius. Using this value and substituting it in the expression for ##U_g##, we get:$$U_g=-\frac{GM}{R_E±r_E}$$.

I don't understand what this relationship between ##\omega##, ##M##, and the distance of Earth from the common mass center can be. I tried to apply the conservation of angular momentum to derive this relationship. The angular momentum of a two-body system is given by:$$L=m_1r_1^2\omega_1+m_2r_2^2\omega_2$$where ##m_1## and ##m_2## are the masses of Earth and the non-Earth body, ##r_1## and ##r_2## are their distances from the common mass center, and ##\omega_1## and ##\omega_2## are their angular velocities.

What do you think? What do you suggest?

 

[haruspex]

I don't understand what this relationship between ω, M, and the distance of Earth from the common mass center can be.

Centripetal force equals gravitational attraction.

 

[Hak]

So, is $$M_1 \omega^2 r = G\frac {M_1 M_2}{r^2}$$ correct?

 

[Hak]

So, is $$M_1 \omega^2 r = G\frac {M_1 M_2}{r^2}$$ correct?

Perhaps I have not yet grasped the point...

 

[haruspex]

So, is $$M_1 \omega^2 r = G\frac {M_1 M_2}{r^2}$$ correct?

Where r is the distance between the two major bodies, yes.

 

[Hak]

Where r is the distance between the two major bodies, yes.

OK, thanks, but what do you suggest now? How to continue? I am a bit confused.

 

[haruspex]

OK, thanks, but what do you suggest now? How to continue? I am a bit confused.

Sorry, only just got out of bed…
As noted, we need to consider the common mass centre, at least in the case of Earth+Moon. So I should have answered that the r on the left needs to be replaced by the distance from M1 to the common mass centre.

 

[Hak]

Sorry, only just got out of bed…
As noted, we need to consider the common mass centre, at least in the case of Earth+Moon. So I should have answered that the r on the left needs to be replaced by the distance from M1 to the common mass centre.

Don't worry, me too. Thanks for reply. What should we call this distance? However, I still do not understand why you have advanced this equality, nor why the distances are correct, nor whether my expression of potential in post #22 is correct, nor how this could lead to the correct procedure you have indicated. Sorry if I don't understand enough, thank you very much for any clarification.

 

[haruspex]

Don't worry, me too. Thanks for reply. What should we call this distance? However, I still do not understand why you have advanced this equality, nor why the distances are correct, nor whether my expression of potential in post #22 is correct, nor how this could lead to the correct procedure you have indicated. Sorry if I don't understand enough, thank you very much for any clarification.

In a two body system, M1 and M2 at distances R1, R2 from their common mass centre, the attractive force between them is ##\frac{GM_1M_2}{(R_1+R_2)^2}##. So ##\frac{GM_1M_2}{(R_1+R_2)^2}=M_1\omega^2 R_1=M_2\omega^2 R_2##.
Often we are dealing with one body very much more massive than the other, so we can treat it as fixed, but that does not work in this problem for Earth+Moon.

 

[Hak]

In a two body system, M1 and M2 at distances R1, R2 from their common mass centre, the attractive force between them is ##\frac{GM_1M_2}{(R_1+R_2)^2}##. So ##\frac{GM_1M_2}{(R_1+R_2)^2}=M_1\omega^2 R_1=M_2\omega^2 R_2##.
Often we are dealing with one body very much more massive than the other, so we can treat it as fixed, but that does not work in this problem for Earth+Moon.

Now I understand. Thank you very much. Regarding potential, what do you think? Most probably my expression in post #22 is not correct. Also you said not to consider (for the moment) Earth's gravity, how do you add it now? I am a bit confused because you had mentioned as many as 5 potential contributions (four initially, then you rightly added another). What advice do you give me to fix this?

 

[haruspex]

Now I understand. Thank you very much. Regarding potential, what do you think? Most probably my expression in post #22 is not correct. Also you said not to consider (for the moment) Earth's gravity, how do you add it now? I am a bit confused because you had mentioned as many as 5 potential contributions (four initially, then you rightly added another). What advice do you give me to fix this?

My suggestion is to work with the equation in post #30 to find the potentials at a point ##R_1+y## from the centre of ##M_1##, on the side nearest M2. Add together the centrifugal potential and that due to M2's gravitation (leave M1's out for now) and make an approximation for small y.
The reason for leaving M1's out is that we will need to add up the potentials for the two different M2s, and we don’t want to add M1's in twice.

 

[Hak]

My suggestion is to work with the equation in post #30 to find the potentials at a point ##R_1+y## from the centre of ##M_1##, on the side nearest M2. Add together the centrifugal potential and that due to M2's gravitation (leave M1's out for now) and make an approximation for small y.
The reason for leaving M1's out is that we will need to add up the potentials for the two different M2s, and we don’t want to add M1's in twice.

Thank you very much. I'll see what I can come up with. However, I still don't understand the precise meaning of ##M_1## and ##M_2##: which of these indicates the mass of the Sun? Which one that of the Moon? Which one that of the Earth? Thank you.

 

[haruspex]

Thank you very much. I'll see what I can come up with. However, I still don't understand the precise meaning of ##M_1## and ##M_2##: which of these indicates the mass of the Sun? Which one that of the Moon? Which one that of the Earth? Thank you.

M1 is always the Earth. M2 will be the Sun in one case and the Moon in the other. This way of doing things means we get a generic expression for the sum of the centrifugal and gravitational potentials then apply it in both M2 cases. Saves a bit of work.

 

[haruspex]

I set ##R_1, R_2## to be the distances to the common mass centre. I found the sum of the centrifugal and M2-based gravitational potentials for two points, one at ##y## from the centre of M1, towards M2, and one at 'moonrise', i.e such that the point and M2 subtend an angle of 90° at the centre of M1.
I then took the difference between these two sums and discarded terms order ##y^3## and higher. (Note that a trap with this sort of calculation is taking approximations too soon and ending up with 0.)
I ended up with ##\frac{GM_2y^2(R_2+3R_1)}{2(R_1+R_2)^3R_1}##. See if you get the same.

If this results in a tide height difference of h then that potential difference should equal ##gh##.

I haven't validated it numerically by plugging in the numbers for Earth and Moon.