How do tides vary over a lunar cycle?

[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.

 

[Hak]

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.

Thank you, @haruspex! I'll see if I can achieve the same, although it seems difficult. I will post my results even if they are wrong. However, I didn't understand one thing: from these expressions you got, how should you arrive at the answers to 1. and 2.? I didn't understand...

 

[haruspex]

Thank you, @haruspex! I'll see if I can achieve the same, although it seems difficult. I will post my results even if they are wrong. However, I didn't understand one thing: from these expressions you got, how should you arrive at the answers to 1. and 2.? I didn't understand...

You already answered 1 in post #1.
For 2., we need to estimate the max tide height due to the moon alone and that due to the sun alone. We can do both of those from the formula in post #35. Then we can see what tide heights we get when they reinforce and when they oppose.

But the first step is to validate that formula numerically for Earth-Moon. I leave that to you.

 

[Hak]

You already answered 1 in post #1.
For 2., we need to estimate the max tide height due to the moon alone and that due to the sun alone. We can do both of those from the formula in post #35. Then we can see what tide heights we get when they reinforce and when they oppose.

But the first step is to validate that formula numerically for Earth-Moon. I leave that to you.

As for the answer to point 1, I thought it was wrong, because you, @jbriggs444 and @kuruman had replied by asking me questions to get me to the correct answer. So, I didn't understand which point the answers to posts #2, #3, #5 and #6 were referring to. Is point #1 therefore correct in light of these replies? I did not understand.

As for point #2, I think I understand. I'll see if I can get the same result as you, then do the numerical verification. Thank you.

 

[haruspex]

As for the answer to point 1, I thought it was wrong, because you, @jbriggs444 and @kuruman had replied by asking me questions to get me to the correct answer. So, I didn't understand which point the answers to posts #2, #3, #5 and #6 were referring to. Is point #1 therefore correct in light of these replies? I did not understand.

As for point #2, I think I understand. I'll see if I can get the same result as you, then do the numerical verification. Thank you.

The queries in posts #5 and #6 refer to your explanation of the process in post #4. Post #2 referred to your answer to part 2. I don't see any responses that challenged your answer to part in post #1.

Anyway, it was not clear to me whether your explanation in post #4 was just oversimplified or a genuine misconception. The process of answering part 2, with the balance of potentials/forces, should help you to word it better.

 

[Hak]

The queries in posts #5 and #6 refer to your explanation of the process in post #4. Post #2 referred to your answer to part 2. I don't see any responses that challenged your answer to part in post #1.

Anyway, it was not clear to me whether your explanation in post #4 was just oversimplified or a genuine misconception. The process of answering part 2, with the balance of potentials/forces, should help you to word it better.

Thank you for the clarification. My reply to post #4 is not an oversimplification, it should be a genuine misunderstanding. What is excessively wrong with it? Is it all to be thrown away? Could you clarify for me what is right and what is wrong with this post, also in light of the talk of potentials and forces you introduced me to? Thank you very much.

 

[haruspex]

Thank you for the clarification. My reply to post #4 is not an oversimplification, it should be a genuine misunderstanding. What is excessively wrong with it? Is it all to be thrown away? Could you clarify for me what is right and what is wrong with this post, also in light of the talk of potentials and forces you introduced me to? Thank you very much.

You wrote

the tides are mainly caused by the gravitational attraction of the Moon and the Sun on the Earth’s oceans. … 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

If it were that simple, the high tide would only be on the side facing the sun/moon.

Potentials and forces are just two ways of analysing it. You can use either, but it is easier to conceptualise the result using potentials; the water will always flow towards the lowest potential, so forms an equipotential surface. As discussed, these potentials/ forces have three sources (in the two body case): the two gravitational fields and the rotation.

The challenge is to sum up this up in words. For Earth+Sun we could say "The Earth rotates around the Sun because of Sun's gravity. The water on the side nearest the Sun is both more attracted by that gravity and less 'thrown out' by the rotation. Conversely, that on the far side is less attracted and more thrown out."

For Earth/Moon it is the same, but it becomes necessary to refer to rotation about the common mass centre. We do not normally think of the Earth as orbiting the Moon!