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Does the moon affect the human body?


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So is it just coincidence that on average a woman cycle is the same length as the lunar cycle?

 

I always wondered why this is, obviously not all animal's cycles are that length and I'd guess our cycles are frequent and year round as we're not tied in to the seasons where it comes to rearing young as other animals are. But it does seem strange it happens to be around about lunar monthly. Might have to look it up later!

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The force due to the moon is proportional to the mass of the object (ie if the sea is "X" times heavier than a person, then the force acting on the sea is X times that on the person).

 

However, the effect of a force on an object is directly proportional to the mass of the object (force = mass x acceleration). So the effect of the respective forces on their objects would be the same.

 

ie

 

Person weighs 100 kg, and force on person = F

 

Acceleration of person due to that force = F/100

 

Sea weighs X x 100, force on sea = X x F

 

Acceleration of sea due to that force = (X x F)/(X x 100) = F/100

 

 

So, regardless of the mass of the object, its acceleration would be the same, surely? If not, why not?

 

Oh gods, this is why I'm failing physics!

 

You're quite right the masses cancel, really should have known that. Go on then give me a lesson, I would appreciate that. Also, if possible could you stick to algebra instead of putting values in? For example what you've just demonstrated would be easier (for me at least) to understand if you think of it in terms of simply substituting F=Gmm/r^2 into a=F/m, giving a=Gm/r^2, which I really should have known anyway:blush:

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Oh gods, this is why I'm failing physics!

 

You're quite right the masses cancel, really should have known that. Go on then give me a lesson, I would appreciate that. Also, if possible could you stick to algebra instead of putting values in? For example what you've just demonstrated would be easier (for me at least) to understand if you think of it in terms of simply substituting F=Gmm/r^2 into a=F/m, giving a=Gm/r^2, which I really should have known anyway:blush:

 

Having said all that, it doesn't mean that the moon has any "real" impact on people. I don't know if it does or not, in which case I'd go for real statistical evidence. If that says there's no effect, then I'd go for that in preference to people's "feelings".

 

---------- Post added 20-06-2013 at 20:25 ----------

 

Oh gods, this is why I'm failing physics!

 

You're quite right the masses cancel, really should have known that. Go on then give me a lesson, I would appreciate that. Also, if possible could you stick to algebra instead of putting values in? For example what you've just demonstrated would be easier (for me at least) to understand if you think of it in terms of simply substituting F=Gmm/r^2 into a=F/m, giving a=Gm/r^2, which I really should have known anyway:blush:

 

I can't give any lessons. I just remembered Force = Mass x Acceleration, and didn't think it squared with F = G x M1 x M2/(r^2) in the way you suggested. TBH, I'd forgotten the Gravitational Force one. It's over 40 years since I was at school, and haven't used it since.

 

I've no idea whether the phases of the moon have any "real" effect on people, but I'm much more inclined to believe objective statistical evidence over people's feelings on the subject. Also, whether there is no moon, a quarter moon, half moon or full moon, it is still there. All that changes is how much of it that we see as lit by the sun. The moon is there, so any gravitational effects it has are still the same, with the possible exception of some effect due to the light it reflects towards us... ...assuming that the extra light associated with a full moon has some effect.

 

As I see it, the majority of the sea is not affected by the gravitational pull. It stays where it is. However, a bit of it, near the surface, is affected. Some, in the direct line of the moon, is lifted, and some other, further away, ie at 90 deg, drops to compensate, hence the tides. The effect of the moon on the sea is very obvious, because, in practise, the whole force of gravity due to the moon (which is very big as both the moon and the sea have large masses) only acts on a small part of the (mass of) the sea, ie the surface layer which is small in comparison with the sea as a whole. In the case of people, it doesn't make us grow by a couple of inches because it is acting on all of us, not just a small part as was the case with the sea.

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The gravitational effect of the Moon has nothing to do with phases.

The effect of the Moon could be bigger when we cannot see it (a perigee new Moon).

So it's difficult to see how animals respond to the 'gravity' of a full Moon.

Animals do respond to changing light levels from the Moon.

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I can't give any lessons. I just remembered Force = Mass x Acceleration, and didn't think it squared with F = G x M1 x M2/(r^2) in the way you suggested. TBH, I'd forgotten the Gravitational Force one. It's over 40 years since I was at school, and haven't used it since.

Don't worry it definitely does, I was being very dense the other day. The masses cancel out, the same reason why everything falls with the same acceleration near the surface of the earth. I made the exact same substitution you did, I just did it in a slightly different way that made it clearer for me.

 

I've no idea whether the phases of the moon have any "real" effect on people, but I'm much more inclined to believe objective statistical evidence over people's feelings on the subject. Also, whether there is no moon, a quarter moon, half moon or full moon, it is still there. All that changes is how much of it that we see as lit by the sun. The moon is there, so any gravitational effects it has are still the same, with the possible exception of some effect due to the light it reflects towards us... ...assuming that the extra light associated with a full moon has some effect.
Not on the gravitational field strength certainly.

 

As I see it, the majority of the sea is not affected by the gravitational pull. It stays where it is. However, a bit of it, near the surface, is affected. Some, in the direct line of the moon, is lifted, and some other, further away, ie at 90 deg, drops to compensate, hence the tides. The effect of the moon on the sea is very obvious, because, in practise, the whole force of gravity due to the moon (which is very big as both the moon and the sea have large masses) only acts on a small part of the (mass of) the sea, ie the surface layer which is small in comparison with the sea as a whole. In the case of people, it doesn't make us grow by a couple of inches because it is acting on all of us, not just a small part as was the case with the sea.
But the gravity acts on the whole sea, almost equally.

 

A quick google tells me the moon is 385,000 km away and the radius of the earth is only 6000 km, so the sea at 90 degrees from the direction of the moon really can't be affected all that much less, as you've only increased r by about 1-2%.

 

Also, how do you explain that there is also another high tide on the other side of the earth, the side further away from the moon?

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Don't let all this scientific mumbo-jumbo detract from the OP .....

 

 

The effect of mooning totally depends upon the relative dimensions of the human body.

 

The effect is also proportional (positive and negative) to how hot the body is.

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Don't worry it definitely does, I was being very dense the other day. The masses cancel out, the same reason why everything falls with the same acceleration near the surface of the earth. I made the exact same substitution you did, I just did it in a slightly different way that made it clearer for me.

 

Not on the gravitational field strength certainly.

 

But the gravity acts on the whole sea, almost equally.

 

A quick google tells me the moon is 385,000 km away and the radius of the earth is only 6000 km, so the sea at 90 degrees from the direction of the moon really can't be affected all that much less, as you've only increased r by about 1-2%.

 

Also, how do you explain that there is also another high tide on the other side of the earth, the side further away from the moon?

 

Disclaimer: This information is worth exactly what you are paying for it. ie, it could well be a load of rubbish, but as I understand it...

 

The ocean directly beneath the moon is attracted towards the moon, and moves towards it and reaches a point of equilibrium where the gravitational pull of the moon on the sea matches the gravitational pull of the core of the solid bit of the earth on the same bit of the sea. The bit of sea at 90deg to the moon has 2 forces (the earth and the moon) at 90 deg to each other, so the resultant force on that bit of sea is mostly towards the earth, but with a pull towards the moon. The pull towards the moon isn't opposed by the earth, as it is at 90 deg, so the sea moves towards to moon - well, a bit of it moves a short distance and in turn pushes the next bit etc. Most of the sea (ie that below the low tide mark) will basically stay where it is, as it's not free to move. It is just the top that moves between low and high tides. The moon acts on all of the earth, but that is mostly solid and so wouldn't deform (except for a slight flexing). However, as this top surface of the sea is free to move, all the effect is concentrated there.

 

As you move around towards the moon, and the angles change, then the result of the two forces changes, giving us the bulge shape, and, I presume, the "sine wave" shape of tidal patterns that you see at any coastal point.

 

The moon pulls the sea away from the earth, creating a high tide directly below it. Similarly it pulls the earth away from the sea on the far side of the earth. Basically, the whole earth is stretched, but its mainly the seas that deform because they are not solid.

 

Maybe?

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