Tide problem

NWilner

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The sun and moon jointly contribute to the tides. By observation the moon is about 6 times as influential as the sun.

The strength of the sun's gravity on earth can be calculated easily. The centripital acceleration of the earth in orbit is V^2/R, where V is 1 circumference per year (2Pi AU/yr) and R is an A.U. 93M miles. (You can assume all the acceleration is from the sun's gravity because the earth is so small in relation.)

The strength of the moon's gravity on earth can also be calculated. The centripital acceleration of the moon in orbit is V^2/r, where V is one moon circumference (2Pi R) per lunar month and R is about 240k miles (units are not important if consistent). Assume most of the acceleration comes from the earth's, gravity, not the moon's. Get the value for the earth. Then multiply by about 5/32, which is the fraction moon grav/ earth grav.

When you do this the sun's gravity on earth is much (about 16x) larger than the moon's gravity on earth. So why is the moon at least 6 times as powerful in affecting the tides?

Astronomers and astrophysicists and their families are not allowed to answer.
 
The thing is, earth tides are caused by the difference in gravity. The difference in the moon's gravitational influence on the near side, the center and the far side of earth pulls the earth out of shape. Because water is much easier to move around the ocean becomes more out of shape than the rest of the earth, one bulge on the near side because the pull of gravity here is greater than on the main bulk of the earth, and one bulge on the opposite side because gravity here is weaker than on the main bulk of the earth.
Because of the inverse square relation between gravity and distance the difference in the moon's pull is greater than the difference in the sun's.

Rough numbers:

Earth diameter = 1.27 * 10^7 m

Mean distance sun-earth = 1.5*10^11 m

Mean distance earth-moon = 3.84*10^8 m

Difference in gravitational influence on near and far side of earth:

Sun: (1.27 * 10^7 + 1.5*10^11)^2 / (1.5*10^11)^2 = 1.0001693
Or 0.017 %

Moon: (1.27 * 10^7 + 3.84*10^8)^2/(3.84*10^8)^2 = 1.0672
Or 6.72 %

Just did the maths to prove this to myself. :)
 
Easy. The two regular and cyclical events of the moon rotating around the earth (creating force pulling water towards the moon), and the earth spinning (creating centrifugal force pushing water away from the earth) have formed a resonant frequency with the oceans of the earth. It's purely coincidental that these two forces have created a harmonic frequency with each other, much like it's coincidental that the moon and sun appear to be roughly the same size when viewed from earth. In layman's terms, the oceans are "sloshing" around at a specific frequency we know as tides.
 
NWilner,

The expression for the gravitational attraction that exists between two objects is extremely well known, and you should be familiar with it. It is given by

F = G*m_1*m_2/r^2, ......... (1)

where G is the constant of gravitation ~ 6.67 * 10^-11 N m^2 kg^-2, m_1 and m_2 are the masses of objects 1 and 2 respectively, and r is the distance between their centers.

Now,

Earth-sun distance ~ 1.5 x 10^11 m,
Earth-moon distance ~ 3.84 x 10^8 m,
sun's mass ~ 1.989 x 10^30 kg,
moon's mass ~ 7.348 x 10^22 kg, and for completeness
Earth's mass ~ 5.974 x 10^24 kg.

Inserting these values into Eq. (1) will show that the gravitational attraction between the Earth and the sun is in fact nearly 170x greater than that between the Earth and the moon. However, as pointed out by bjornart, the tides are due to the difference in gravitational attraction exerted on different parts of the Earth. By taking the derivative of Eq. (1) w.r.t. to r you can see that this difference is proportional to the inverse cube of r. So in the present case, the ratio of the inverse cubes multiplied by the ratio of the masses is a little more than two (assuming my arithmetic is correct), and so the moon's influence is a little more than twice that of the sun.

For interest, when the Earth, moon and sun are nearly collinear the tides are, as you can guess, highest, and are called spring tides. When the sun and moon appear to be in perpendicular directions as viewed from the Earth, the tides are lowest, and are called neap tides.

Lucy.
 
Being that the hour is so late I will bother with the mathematical details of your question.

However, you may interested to know that the tides induced by the sun are termed 'Neap Tides'.
 
LucyR: What am I, invisible?
Click to expand...
Don't ya just hate that? This kind of thing happens a lot on this forum. I've lost track of the number of times I've wanted to smack someone upside the head and say, "Didn't you notice that I already posted that same info yesterday??" :D

What's really comical is when three people start three separate threads about the same exact news item, and then the next day, a fourth person starts yet another thread on the same thing.
 
There are lots of, um, interesting tidbits of info in this thread.

First, the Sun's contribution to the tidal force on the Earth is roughly 1/2 the Moon's (or 1/3 the total). As was pointed out, tides are the derivative of the gravitational force, so they drop as 1/r^3, but go linearly with mass. The Sun is 375 times the distance of the Moon, and 375^3 = 53 million. The Sun is about 27 million times the Moon's mass. 27/53 is about 1/2. Taadaa!

Second, a "neap tide" is when the Sun and Moon are at right angles to each other in the sky. The Sun pulls the tides one way, and the Moon pulls them the other. This results in slightly lower high tides, and slightly higher low tides. When the Sun and Moon line up in the sky (new and full Moon), the forces add, and you get higher high tides and lower low tides. When the Moon is nearest the Earth, and/or the Earth nearest the Sun (in January) you get proxigean tides, which are slightly higher because the distances are minimized.

As for why we have tides in the first place, well, I have something to say about that, too. :-)
 
I can't see that you've really added anything. Presumably you agreed with my calculation.
 
The Bad Astronomer said:
There are lots of, um, interesting tidbits of info in this thread.


What are you referring to?

Second, a "neap tide" is when the Sun and Moon are at right angles to each other in the sky. The Sun pulls the tides one way, and the Moon pulls them the other. This results in slightly lower high tides, and slightly higher low tides. When the Sun and Moon line up in the sky (new and full Moon), the forces add, and you get higher high tides and lower low tides.


Did I not just say this?
Click to expand...
 
Lucy, it's because you're so short. Here, stand on this box. Oh, and wear this orange wig. There, much better. I'm sure people will notice you now.
 
Ah, casual malice directed at me by one of my favourite posters!

Poor, unhappy Lucy! :(
 
Smiley face? :)

Box of chocolates?
box_chocolates.jpg


Flowers?
flowers.jpg


A10 built out of Lego?
lego_a10s.jpg
 
Pixy! Thank you so much! That's so sweet!

I was going to say something nasty about Shane Warne, but I won't now.

Also, did you build the A-10? It's really good.
 
LucyR said:
Also, did you build the A-10? It's really good.
Click to expand...
Alas, no; I'm a Lego nut but that's not one of mine. I did a Google search for an A10 picture to use as a peace offering, and that popped up. Obvious choice ;)
 
There I go, learning stuff again. I wondered what was wrong with my numbers. :)

The extra high tides at the new and full moon are called spring tides by the way.
 
LucyR, as it happens, I did miss what you said. That sometimes happens, even to veteran posters, and in general doesn't deserve a snide response. I made a mistake, that's all.
 
Astronomer,

It was your site that introduced me to the JREF in the first place.:cool:

Being a long-time fan I'm certainly prepared to forgive your uncharacteristic lapse.:)
 
NWilner,

The expression for the gravitational attraction that exists between two objects is extremely well known, and you should be familiar with it. It is given by

F = G*m_1*m_2/r^2, ......... (1)

where G is the constant of gravitation ~ 6.67 * 10^-11 N m^2 kg^-2, m_1 and m_2 are the masses of objects 1 and 2 respectively, and r is the distance between their centers.

Now,

Earth-sun distance ~ 1.5 x 10^11 m,
Earth-moon distance ~ 3.84 x 10^8 m,
sun's mass ~ 1.989 x 10^30 kg,
moon's mass ~ 7.348 x 10^22 kg, and for completeness
Earth's mass ~ 5.974 x 10^24 kg.

Inserting these values into Eq. (1) will show that the gravitational attraction between the Earth and the sun is in fact nearly 170x greater than that between the Earth and the moon. However, as pointed out by bjornart, the tides are due to the difference in gravitational attraction exerted on different parts of the Earth. By taking the derivative of Eq. (1) w.r.t. to r you can see that this difference is proportional to the inverse cube of r. So in the present case, the ratio of the inverse cubes multiplied by the ratio of the masses is a little more than two (assuming my arithmetic is correct), and so the moon's influence is a little more than twice that of the sun.

For interest, when the Earth, moon and sun are nearly collinear the tides are, as you can guess, highest, and are called spring tides. When the sun and moon appear to be in perpendicular directions as viewed from the Earth, the tides are lowest, and are called neap tides.
 
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