bedouin
Well-Known Member
Well we have a second moon - called the sun. It also causes tides but much weaker but enough to make the difference between springs and neaps
AT last a clear explanation of how we get 2 tides per day !
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Well we have a second moon - called the sun. It also causes tides but much weaker but enough to make the difference between springs and neaps
grumpy_o_g
Well-Known Member
The vector forces are the Moon's gravitational force vs centrifugal force due to the Earth's rotation. Perhaps one way to think about it is that the force exerted by the Moon on the far side of the Earth (I can't get Pink Floyd out of my head while I'm writing this
) is less so the water is thrown higher by the rotating earth. It's totally counter-intuitive to me and, whilst I understand the vectors involved, I just accept that that's how it is.Try this video - I now have my head completely round it. If you're in a hurry jump in at 2 minutes and look closely at the diagram at around 3 minutes. It's the 30 secs from 2 min 30 sec to 3 mins that finally made it more than just maths for me.
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Lon nan Gruagach
Well-Known Member
I think I see a bit of confusion here.... That is the Earth's spin (making night and day) and the Earth orbiting the Moon.
The Earth spinning round is what makes the tides move, but it is (on one side) the Moon's gravity, and (on the other side) the centrifugal force (bleagh), that makes the bulges.
To simplify...
Imagine that the Earth does not spin (relative to the Moon), in much the same way that the Moon does not spin (relative to the Earth [the same side always faces us]). Viewed from a long way away, the bulges would (almost) look the same. The side facing the Moon dragged up by its gravity, the side away from the Moon flung out due to the mutual orbits.
Now, spin the Earth. This will smear the bulges a bit, but not that much, and for someone on Earth you will have 2 (almost) high tides per day.
The Earth spinning round is what makes the tides move, but it is (on one side) the Moon's gravity, and (on the other side) the centrifugal force (bleagh), that makes the bulges.
To simplify...
Imagine that the Earth does not spin (relative to the Moon), in much the same way that the Moon does not spin (relative to the Earth [the same side always faces us]). Viewed from a long way away, the bulges would (almost) look the same. The side facing the Moon dragged up by its gravity, the side away from the Moon flung out due to the mutual orbits.
Now, spin the Earth. This will smear the bulges a bit, but not that much, and for someone on Earth you will have 2 (almost) high tides per day.
paulburn
Well-Known Member
thankyou - this makes sense in an intuitive way.
only the Nottingham Uni video didn't reference centrifugal effects - only gravity - which didn't to make sense to me to give the 'far side' bulge !
only the Nottingham Uni video didn't reference centrifugal effects - only gravity - which didn't to make sense to me to give the 'far side' bulge !
guernseyman
Well-Known Member
The standard explanation relies on the gradient in the moon's gravitational pull: the moon is pulling the earth more strongly than the sea on the far side, so that sea is left behind as it were. I used to accept that, but I am now more convinced by considerations of centrifugal force.
KINGFISHER 9
Well-Known Member
I blame Brexit.
bedouin
Well-Known Member
As has been said several times - there is no such thing as centrifugal force (in an inertial frame) - the only force involved is gravity. The Nottingham Uni video describes all forms of the tidal force that does not require the bodies to be in orbit around each other - and I think that makes it easier to understand.
Think first about the differential impact of gravity on two bodies (or collections of bodies) passing each other and see that the nearer objects are deflected more than those in the middle, which in turn are deflected more than those that are further away - hence the "stretching" effect that is used to describe the tides.
guernseyman
Well-Known Member
I suggest that we have a choice: either to view the tides in an inertial frame not rotating with the the earth, or to view the tides from the rotating earth, which is not an inertial frame. In the latter case centrifugal force (or whatever alternative designation is preferred) cannot be denied. In the former it does not exist.
bedouin
Well-Known Member
I agree - two different ways of looking at the same system of forces in different frames of reference. But the Nott Uni video did not assume the two systems were in orbit and so took the approach of choosing an inertial frame.
pmagowan
Well-Known Member
I think most people prefer to use their own frame of reference.
lw395
Well-Known Member
You can't view the system without allowing for the moon orbiting the earth.
If the moon was stationary, earth moon and sea would all be pulled together by gravity.
So the simplest viable model has to be the earth and moon orbiting a point offset from the centre of the earth due to the mass of the moon....
This point is obviously towards the moon.
So the water away from the moon is orbiting this 'barycentre' at a (slightly) larger radius.
It's also slightly further away from the centre, so gravity is slightly less.
The force to make something orbit on a given radius is proportional to radius squared.
So on the side away from the moon, the required force to keep the water in the same orbit is more and the gravity is less, therefore the water is pulled to the centre less than at other points around the equator.
(that is what a centrifuge does, so if want to call that centrifugal force, when it's really a deficit in centripetal force, that's fine by me!)
On the side toward the moon, the point on the earth-moon axis is closest to the barycentre, so there is a component of gravity pulling the water towards the axis.
If the moon was stationary, earth moon and sea would all be pulled together by gravity.
So the simplest viable model has to be the earth and moon orbiting a point offset from the centre of the earth due to the mass of the moon....
This point is obviously towards the moon.
So the water away from the moon is orbiting this 'barycentre' at a (slightly) larger radius.
It's also slightly further away from the centre, so gravity is slightly less.
The force to make something orbit on a given radius is proportional to radius squared.
So on the side away from the moon, the required force to keep the water in the same orbit is more and the gravity is less, therefore the water is pulled to the centre less than at other points around the equator.
(that is what a centrifuge does, so if want to call that centrifugal force, when it's really a deficit in centripetal force, that's fine by me!)
On the side toward the moon, the point on the earth-moon axis is closest to the barycentre, so there is a component of gravity pulling the water towards the axis.
JumbleDuck
Well-Known Member
You can say it as often as you like, but it doesn't make it correct. In a rotating frame of reference centrifugal forces are real and have real effects.
bedouin
Well-Known Member
A rotating frame of reference is not an inertial frame.
JumbleDuck
Well-Known Member
They are equally useful and equally valid, though, and the forces in them are equally real.
Even in a non-rotating axis system centrifugal force can be a useful notion as a reaction force, just as centripetal force is in a rotating system. The blades of a propeller do pull away from the hub, however you look at it.