This is a very nice lecture. I will try and reconstruct it for my upcoming class. My only comment is that the centrifugal force needs to be clarified since the outward force per unit mass is R*Omega^2, where R is the distance of the object from the center of rotation and Omega is the angular velocity of the object relative to the center of rotation . The point is that the centrifugal force, because of gravitational-rotational equilibrium, can be expressed in terms of Newton's one over r squared law of gravitation. This step is missing and would be quite helpful to clarify for the students. The discussion on the dependence of the tide generating force on inverse cube of the distance and the force arrows is EXCELLENT otherwise!!!! Thank you, Nick, for constructing this material for the community. PS: I Caught your jab about spherical footballs for us Americans!!! Ha, Ha! !!!
Hi Mike, glad you’re enjoying the lectures, thanks for your generous comments ! Good point about the centrifugal force, you're right I didn’t actually give a separate formula for it. In fact I also glossed over the explanation of why it is the same at all points on the earth. This requires a bit of thought. If you look at support material (link on main banner) there are some diagrams that address that point (page 104). Also see Tribus Montibus (link in this video description) for some very careful explanations of tidal forces.
Hi Mick and Nick, if I may, I'd like to make a polite interaction here, which you both might find helpful? If you were going to reconstruct this excellent lecture for your upcoming class. Then here's a couple of things, which you might like to include... Firstly, not only does the real effect of centrifugal force need to be clarified, as the result of the revolving, orbital motion around the barycentre, due to inertia but, it might be nice to also include an actual description of how "gravity" really works, seeing as it can also be considered as an apparent force in an inertial frame of reference. As for the D cubed factor in the tidal force equation listed here. There was one symbol, which was inexplicably missing, and may lead to some confusion. And, that is the inclusion of a (+/-) symbol at the beginning. When you input the correct values into the equation you will then get a plus and minus value for the tidal force and a zero value at the centre of the earth, because it's actually in a freefall motion around the barycentre. I hope that helps and if you need any further clarification, then don't hesitate to contact me. Good luck with the future presentations to your classes.
It’s nice to see that there are still some teachers who are willing to take on the contentious subject of “fictitious” forces head on, and explain them so vividly. Thank you. Although, I’m sure you’ll find it’s called “football” and not “soccer” sir? Cheers.
Trying to write a compliment but somehow it gets blocked. Let me try line-for-line: Excellent presentation Nick. Too bad that this hasn't gotten more attention, while some tide explanations get thousands every day, further confusing the curious. They are typically saying that high tides occur because the moon and sun are 'dragging water upwards'. You are one of the very few who get it right in their videos. Good job! p.s. Nice 'football'-comment
Hi, thanks for your comment, I have very little influence on the TH-cam algorithm ! For the moment, my channel is aimed at students with some mathematical background rather than at the scientifically curious in general, so I'm not expecting too much in terms of popularity. When I get time to produce more content I will try to reach a wider audience. There is a lot of good stuff out there already. I just checked out your channel. I like your clear well paced explanations - especially your emphasis on real world numerical values, which can often get neglected in more conceptual or theoretical presentations. Thanks also for helping reply to my other viewer question. When I give that lecture again I will certainly add a bit more on the centrifugal force and your comment has helped me to get a clearer picture of that !
So wait it can lift up entire oceans but it does nothing to dust? Does it also cause sandstorms? Or mild tidal forcing in the dessert? Increases snow drift?
@@infinitevoid5141 that doesn't actually explain anything but I will move on a bit. The moon affects our water. Inversely the earth affects the moon with the same forces. As we "saw" moon dust is fine. With the strength of Earth's gravitational pull why is there not a cone on the moon pointing towards the center of the earth?
@@1goldinga this is part of the cause of tidal locking of the moon. moon is not made of powder, it is solid and have some amount of stiffness holding its shape, still tending to turn in to egg shape but microscopically.
From 3:37: Centrifugal forces (red arrowes) are shown constant, bud it should depend on distance from the barycenter (mrw^2). And on the right side of the earth cetrifugal force vector should show to the right, shouldn't (?)
@@NickMJHall Why can you neglect gravitational attraction to the centre of the earth? It is 10 million times bigger than the shown gravitational attraction to the Moon.
@@vassmiklos Sorry my previous reply was misleading so I removed it for the general good. I think the best way to visualise this is to think of the earth as a cloud of particles that are all separately interacting with the moon. If you look at it that way the Centrifugal force is pretty much the same for all of them. Of course the earth's gravity is generally balanced by the pressure gradient force. I can't think of a better way of describing it right now - if you come up with something let us know !
Hello Miklós. I see your point and applaud your curiosity. If you analyze the whole thing mathematically you will invariably find the following (so I'm only telling you what your curiosity would eventually have led you to find). You are absolutely correct. If you account for motion around a center of rotation (in this case the earth-moon barycenter) any inertial or centrifugal force shall point away from it. So, indeed, if the earth in the picture was a cardboard disk, rotating around a pushpin pinned through that barycenter, it would experience a constant centrifugal force of fixed magnitude, pointing away from the pushpin. But if you plot the movement of all the earth's surface points (do yourself a favor; pretend the earth's rotation about its axis as zero, for now) you will see that all these points translate through inertial space over circles with the same radius. Their centrifugal forces point away from the centers of their respective circles, which are all equal in diameter. The 'pushpin' isn't the center of rotation for the surface points here. Hence, astronomers use a definition of centrifugal force, which is equal in magnitude and direction for any point of the planet. Doing the analysis for any individual point, from a standpoint of every individual movement, is much more complicated and will yield the same result in the end. You will find that you have to subtract the influence of earth's rotation (which you first added to make your model) when you subtract for coriolis force. A rapid earth rotation will create an 'upward' centrifugal force at different latitudes, but will be the same at any given latitude. Thus it doesn't contribute to the tides, as it isn't time dependent. Hope this wasn't confusing. Kind regards
This is a very nice lecture. I will try and reconstruct it for my upcoming class. My only comment is that the centrifugal force needs to be clarified since the outward force per unit mass is R*Omega^2, where R is the distance of the object from the center of rotation and Omega is the angular velocity of the object relative to the center of rotation . The point is that the centrifugal force, because of gravitational-rotational equilibrium, can be expressed in terms of Newton's one over r squared law of gravitation. This step is missing and would be quite helpful to clarify for the students. The discussion on the dependence of the tide generating force on inverse cube of the distance and the force arrows is EXCELLENT otherwise!!!! Thank you, Nick, for constructing this material for the community. PS: I Caught your jab about spherical footballs for us Americans!!! Ha, Ha! !!!
Hi Mike, glad you’re enjoying the lectures, thanks for your generous comments !
Good point about the centrifugal force, you're right I didn’t actually give a separate formula for it. In fact I also glossed over the explanation of why it is the same at all points on the earth. This requires a bit of thought. If you look at support material (link on main banner) there are some diagrams that address that point (page 104). Also see Tribus Montibus (link in this video description) for some very careful explanations of tidal forces.
Hi Mick and Nick, if I may, I'd like to make a polite interaction here, which you both might find helpful?
If you were going to reconstruct this excellent lecture for your upcoming class. Then here's a couple of things, which you might like to include...
Firstly, not only does the real effect of centrifugal force need to be clarified, as the result of the revolving, orbital motion around the barycentre, due to inertia but, it might be nice to also include an actual description of how "gravity" really works, seeing as it can also be considered as an apparent force in an inertial frame of reference.
As for the D cubed factor in the tidal force equation listed here. There was one symbol, which was inexplicably missing, and may lead to some confusion. And, that is the inclusion of a (+/-) symbol at the beginning. When you input the correct values into the equation you will then get a plus and minus value for the tidal force and a zero value at the centre of the earth, because it's actually in a freefall motion around the barycentre.
I hope that helps and if you need any further clarification, then don't hesitate to contact me. Good luck with the future presentations to your classes.
It’s nice to see that there are still some teachers who are willing to take on the contentious subject of “fictitious” forces head on, and explain them so vividly. Thank you. Although, I’m sure you’ll find it’s called “football” and not “soccer” sir? Cheers.
Thank you for wonderful video.
Trying to write a compliment but somehow it gets blocked. Let me try line-for-line:
Excellent presentation Nick. Too bad that this hasn't gotten more attention, while some tide explanations get thousands every day, further confusing the curious. They are typically saying that high tides occur because the moon and sun are 'dragging water upwards'. You are one of the very few who get it right in their videos. Good job!
p.s. Nice 'football'-comment
Hi, thanks for your comment, I have very little influence on the TH-cam algorithm ! For the moment, my channel is aimed at students with some mathematical background rather than at the scientifically curious in general, so I'm not expecting too much in terms of popularity. When I get time to produce more content I will try to reach a wider audience. There is a lot of good stuff out there already. I just checked out your channel. I like your clear well paced explanations - especially your emphasis on real world numerical values, which can often get neglected in more conceptual or theoretical presentations. Thanks also for helping reply to my other viewer question. When I give that lecture again I will certainly add a bit more on the centrifugal force and your comment has helped me to get a clearer picture of that !
@Prime Thanatos Thanks for clearing that up for us Einstein
So wait it can lift up entire oceans but it does nothing to dust? Does it also cause sandstorms? Or mild tidal forcing in the dessert? Increases snow drift?
Fg=GMm/r^2 mass of dust is smol so force is smol🥸
@@infinitevoid5141 that doesn't actually explain anything but I will move on a bit.
The moon affects our water. Inversely the earth affects the moon with the same forces. As we "saw" moon dust is fine. With the strength of Earth's gravitational pull why is there not a cone on the moon pointing towards the center of the earth?
@@1goldinga this is part of the cause of tidal locking of the moon. moon is not made of powder, it is solid and have some amount of stiffness holding its shape, still tending to turn in to egg shape but microscopically.
@@calvinlee1127 like does that even make sense to you after you said it. Does your knowledge also cover tidal nodes?
From 3:37:
Centrifugal forces (red arrowes) are shown constant, bud it should depend on distance from the barycenter (mrw^2). And on the right side of the earth cetrifugal force vector should show to the right, shouldn't (?)
@@NickMJHall Why can you neglect gravitational attraction to the centre of the earth? It is 10 million times bigger than the shown gravitational attraction to the Moon.
@@vassmiklos Sorry my previous reply was misleading so I removed it for the general good. I think the best way to visualise this is to think of the earth as a cloud of particles that are all separately interacting with the moon. If you look at it that way the Centrifugal force is pretty much the same for all of them. Of course the earth's gravity is generally balanced by the pressure gradient force. I can't think of a better way of describing it right now - if you come up with something let us know !
Hello Miklós. I see your point and applaud your curiosity. If you analyze the whole thing mathematically you will invariably find the following (so I'm only telling you what your curiosity would eventually have led you to find).
You are absolutely correct. If you account for motion around a center of rotation (in this case the earth-moon barycenter) any inertial or centrifugal force shall point away from it. So, indeed, if the earth in the picture was a cardboard disk, rotating around a pushpin pinned through that barycenter, it would experience a constant centrifugal force of fixed magnitude, pointing away from the pushpin. But if you plot the movement of all the earth's surface points (do yourself a favor; pretend the earth's rotation about its axis as zero, for now) you will see that all these points translate through inertial space over circles with the same radius. Their centrifugal forces point away from the centers of their respective circles, which are all equal in diameter. The 'pushpin' isn't the center of rotation for the surface points here.
Hence, astronomers use a definition of centrifugal force, which is equal in magnitude and direction for any point of the planet. Doing the analysis for any individual point, from a standpoint of every individual movement, is much more complicated and will yield the same result in the end. You will find that you have to subtract the influence of earth's rotation (which you first added to make your model) when you subtract for coriolis force. A rapid earth rotation will create an 'upward' centrifugal force at different latitudes, but will be the same at any given latitude. Thus it doesn't contribute to the tides, as it isn't time dependent.
Hope this wasn't confusing. Kind regards