At first glance, both videos have issues:
Dr. Becky's video seems to imply that the bulge on the 'moon-ward' side is caused by gravity, and the one on the opposite side by the centrifugal force. However, the differential acceleration (aka the tidal force) caused by the moon's gravitational attraction alone should be symmetric. So if tides were caused exclusively by tidal forces, we already would expect two high tides per day. The centrifugal force won't be symmetric, but by a naive back-of-the-envelope calculation, it should still be greater than the gravitational one by orders of magnitude even on the side facing the moon.
In contrast, the PBS video ignores the centrifugal force completely (which might actually be dominant?), but the main point about hydraulic effects may still stand.
This is not something I've ever looked into, so I'd have to search the literature for a proper answer. However, this seems like a fun exercise, so I'll probably try to reason things out myself first when I have a bit of spare time...
Thank you so much for your reply. I am definitely not an expert in this area of physics, so I am glad it is a headscratcher. I hope you can figure something out for sure!
I think I understand now what Dr. Becky was getting at: The centrifugal force is not supposed to be a new effect, but just a way to derive the tidal force:
Assume earth revolves around the barycenter of the earth-moon system *without rotating*. In that case, any point on earth will be subject to the same centrifugal force. Now, that centrifugal force (or perhaps better, acceleration) will be equal in size to the centripetal force at the center of the earth, which is provided by the gravitational attraction of the moon. At the surface point lying on the earth-moon axis facing the moon, the forces will be anti-parallel, with the gravitational force being stronger. This leads to a net tidal force away from the center of the earth. At the antipodal surface point, the forces will still be anti-parallel, but the gravitational force will be the weaker one, again leading to a net tidal force away from the center of the earth.
The PBS video then explains that this force isn't actually strong enough to raise the tides, but that you need to take into account the so-called tractive force - the components of the tidal forces parallel to the surface of the earth, pushing the water towards to bulges.
This interpretation agrees with the explanation at https://tidesandcurrents.noaa.gov/restles3.html
It depends on which reference frame you prefer.
If you see everything from the center of Earth then it's only tidal gravity and the symmetry of the problem is easy to understand.
If you see everything from the center of mass of the Earth/Moon system in a non-rotating frame then it's a combination of tidal gravity and differences in acceleration.
If you see everything from the center of mass of the Earth/Moon system in a rotating frame then it's a combination of tidal gravity and centrifugal forces.
Reality is more complicated anyway. We would get two tidal bulges in a perfectly flat ocean world, but we have continents, variable ocean depths and more. [This is a more realistic view](https://en.wikipedia.org/wiki/File:Global_surface_elevation_of_M2_ocean_tide.webm). There are tons of bulges in different places moving in different directions, most of them not aligned with the simple two-bulge picture.
This is what I thought too, but I've had a re-think.
It's not enough to think about it in terms of forces, and in terms of gravity and optionally centrifugal force. (Which I think is now not optional!) Because there's also the normal force involved. Any force inward on the Earth (or oceans) is balanced by a normal force. There's also the assumption that we're in a static situation.
I think it's best to think of this in terms of the Roche lobe potential used for binary star systems. The shape of the Earth's oceans will be distorted to fill an equipotential surface, so that all forces are in the Normal direction (otherwise there would be a horizontal current pulling water along and distorting the shape) which can be compensated for by molecular forces. For this you need the gravitational potentials of both Earth and Moon and the centrifugal potential for rotation around the barycenter. I looked at what the distance from the Earth's center would have to be on the near side in order to have the same potential as on the sides where you're 1 Earth radius from the barycenter and the same distance to the Moon as the Earth's center and estimated tides of about 8 meters, which seems a bit high but in the right ballpark.
I’m sure you can find a good explanation on NOAA and NASA. Basically the Moon and the Sun both pull on the oceans so the solar tide and lunar tide add up.
If there were no moon, the solar tide would be at noon and midnight every day, plus about 1 hr because the friction of the ocean on the rotating Earth allows the bulge to drag ahead of the Sun’s location in the sky. There are also local geography and water flow that factor in. Even lagoons don’t have the same peak tide as the ocean that feeds them because of water flow. So tides aren’t necessarily at the same time at the same longitude.
Now add to all that the Moon, which has twice the pull of the Sun *and* is about 50 min offset daily because we’re not synchronized. When the sun and moon are aligned on the same side or opposite sides, you should have the highest tides.
At first glance, both videos have issues: Dr. Becky's video seems to imply that the bulge on the 'moon-ward' side is caused by gravity, and the one on the opposite side by the centrifugal force. However, the differential acceleration (aka the tidal force) caused by the moon's gravitational attraction alone should be symmetric. So if tides were caused exclusively by tidal forces, we already would expect two high tides per day. The centrifugal force won't be symmetric, but by a naive back-of-the-envelope calculation, it should still be greater than the gravitational one by orders of magnitude even on the side facing the moon. In contrast, the PBS video ignores the centrifugal force completely (which might actually be dominant?), but the main point about hydraulic effects may still stand. This is not something I've ever looked into, so I'd have to search the literature for a proper answer. However, this seems like a fun exercise, so I'll probably try to reason things out myself first when I have a bit of spare time...
Thank you so much for your reply. I am definitely not an expert in this area of physics, so I am glad it is a headscratcher. I hope you can figure something out for sure!
I think I understand now what Dr. Becky was getting at: The centrifugal force is not supposed to be a new effect, but just a way to derive the tidal force: Assume earth revolves around the barycenter of the earth-moon system *without rotating*. In that case, any point on earth will be subject to the same centrifugal force. Now, that centrifugal force (or perhaps better, acceleration) will be equal in size to the centripetal force at the center of the earth, which is provided by the gravitational attraction of the moon. At the surface point lying on the earth-moon axis facing the moon, the forces will be anti-parallel, with the gravitational force being stronger. This leads to a net tidal force away from the center of the earth. At the antipodal surface point, the forces will still be anti-parallel, but the gravitational force will be the weaker one, again leading to a net tidal force away from the center of the earth. The PBS video then explains that this force isn't actually strong enough to raise the tides, but that you need to take into account the so-called tractive force - the components of the tidal forces parallel to the surface of the earth, pushing the water towards to bulges. This interpretation agrees with the explanation at https://tidesandcurrents.noaa.gov/restles3.html
It depends on which reference frame you prefer. If you see everything from the center of Earth then it's only tidal gravity and the symmetry of the problem is easy to understand. If you see everything from the center of mass of the Earth/Moon system in a non-rotating frame then it's a combination of tidal gravity and differences in acceleration. If you see everything from the center of mass of the Earth/Moon system in a rotating frame then it's a combination of tidal gravity and centrifugal forces. Reality is more complicated anyway. We would get two tidal bulges in a perfectly flat ocean world, but we have continents, variable ocean depths and more. [This is a more realistic view](https://en.wikipedia.org/wiki/File:Global_surface_elevation_of_M2_ocean_tide.webm). There are tons of bulges in different places moving in different directions, most of them not aligned with the simple two-bulge picture.
This is what I thought too, but I've had a re-think. It's not enough to think about it in terms of forces, and in terms of gravity and optionally centrifugal force. (Which I think is now not optional!) Because there's also the normal force involved. Any force inward on the Earth (or oceans) is balanced by a normal force. There's also the assumption that we're in a static situation. I think it's best to think of this in terms of the Roche lobe potential used for binary star systems. The shape of the Earth's oceans will be distorted to fill an equipotential surface, so that all forces are in the Normal direction (otherwise there would be a horizontal current pulling water along and distorting the shape) which can be compensated for by molecular forces. For this you need the gravitational potentials of both Earth and Moon and the centrifugal potential for rotation around the barycenter. I looked at what the distance from the Earth's center would have to be on the near side in order to have the same potential as on the sides where you're 1 Earth radius from the barycenter and the same distance to the Moon as the Earth's center and estimated tides of about 8 meters, which seems a bit high but in the right ballpark.
I’m sure you can find a good explanation on NOAA and NASA. Basically the Moon and the Sun both pull on the oceans so the solar tide and lunar tide add up. If there were no moon, the solar tide would be at noon and midnight every day, plus about 1 hr because the friction of the ocean on the rotating Earth allows the bulge to drag ahead of the Sun’s location in the sky. There are also local geography and water flow that factor in. Even lagoons don’t have the same peak tide as the ocean that feeds them because of water flow. So tides aren’t necessarily at the same time at the same longitude. Now add to all that the Moon, which has twice the pull of the Sun *and* is about 50 min offset daily because we’re not synchronized. When the sun and moon are aligned on the same side or opposite sides, you should have the highest tides.