A TIDE IN MY AFFAIRS


Warning: This post will contain some mathematical formulae and terms, which may scare or otherwise turn off some.  I hope such formulae do not detract from the beauty of what will be seen, because indeed, mathematics is beautiful.  It answers questions.  Is that not beauty?  In a week, pictures of the result will be shown.

I’m going over to Newport, Oregon next week to see the King Tides, something I had once never heard of.  I am almost a true Oregonian, but when I led a trip to the coast the last week, I forgot to look up the tides. That’s inexcusable.  Always know the tides when you are at the ocean.

Tides matter.  A lot.  In nature, many species thrive at border zones between one ecosystem and another.  They allow for organisms to live in varying degrees of wetness, rather than always wet or always dry.  They allow for tidal pools to become cut off from the ocean, where periodically they get refilled or organisms shuffled.  Without tides, the Earth would be a very different, far less diverse place.

What are tides, anyway?  They are common throughout the universe.  If one object tugs on another, it can deform the latter due to gravitational attraction, which may cause buckling or movement of the surface of the attracted object.  Jupiter’s moon Io gets tugged by massive Jupiter, causing volcanic eruptions on its surface.  The first was spotted by a woman, Linda Morabito, who saw a plume on Io, which had been once thought once to be dead, then had volcanism predicted.  Io is the most volcanically active place known in the solar system.

Both the Sun and Moon tug on the Earth.  While the Moon is much smaller, a mass 1/27,000,000 that of the Sun (mass is the amount of “stuff” something has; weight is the effect of gravity.  Diet removes mass; being in zero gravity does not, but it makes you weightless), the Moon exerts a majority (55%) of the tidal activity on the Earth.

For a long time, that 55% bothered me, because gravitation is proportional to the product of the masses but inversely proportional to the square of the distance, the distance between the two centers, or d, and the numbers didn’t work.

F=G m1 m2/d^2.

where G is the gravitational constant, m1 the mass of one body, m2 the mass to the second, and  d^2=d*d, the distance between them multiplied by itself.  The Moon is smaller, less massive, but it is much closer than the Sun.  Still, if one compares the large mass of the Sun with its admittedly larger distance from us (400 times further from the Moon, and the distance varies, which is important), the Sun ought have an effect 170 times greater than the Moon upon us.  It doesn’t, and that bothered me.  I show this below.  Gravity is the reason we circle the Sun and not the Moon; the Moon circles both of us.  I did not consider tidal forces, those which work differentially on a body, more on the near side than the far side.  These Ah-hah moments are one of the joys of life, when one understands a concept that has been murky for years.

The Moon tugs on the Earth, the oceans are pulled towards the Moon. Tides are maximal in general when the Moon is either overhead or at the opposite side, although that can vary considerably due to other factors and local conditions, which give rise to enormous tides at the Bay of Fundy or tidal bores on Turnagain Arm in Alaska.  The tide is greater (spring tides, nothing to do with the season) when the Moon is lined up with the Sun and the Earth, occurring about every 15 days, and lesser (neap tides) when the Moon is not aligned.  The square of the distance means that anything decreasing distance increases the tide, so when the Moon is close to us, which happens every 27.5 days, even not well aligned with the Earth and Sun, the tides are significantly affected. The Earth is 3 million miles closer to the Sun in early January compared to early July, and this increases tides as well, because while the Sun’s force is slightly less than the Moon’s, its distance from us is the least for the year. That’s why we’re going to Newport.

In Newport, king tides occur at full Moon in January, near perihelion.  The full Moon is opposite the Sun, meaning that it is in the northern part of the celestial sphere, over the northern hemisphere, and therefore is closer to the coastal cities there.

I also didn’t know why the Moon had a greater pull, given the gravity equation.  The numbers didn’t work. I thought—incorrectly— it was all gravity.

The tidal force looks at slight changes in the distance between the two bodies; the force is proportional to the cube of the distance between the bodies, d^3, or d*d*d, and a simplified proof is shown below.  Cubes are volumes, and the three factors are length, width, and depth.  When we compare the gravitational equation using the cube of the distance and twice the mass product, the Sun is responsible for about 45% of the tidal force; the Moon the rest.

Additionally, the lowest tide is not in January, as one would think, but is in the late spring early summer and at New Moon.  Why?  In May, the Earth is further from the Sun, so the Sun’s pull is less.  But at New Moon, which aligns with the Sun, the Moon is over the northern hemisphere. There are issues with the lunar nodes and the tilt of the Earth’s axis at different times of the year.  Tides are more complex than I thought, not due to simple gravitational pull but to a differential force that must be accounted for. When I go to Newport, I will be watching a 3 meter high tide and the -0.5 meter low tide, both a full meter higher than normal.

 

 

F(S-E)=Gm (S)*m(E)/d(S-E)^2. The Sun-Earth gravitational force is proportional to the product of the masses and inversely proportional to the distance between their centers. The same holds for the Moon-Earth.  It also holds between you and your computer, too.

F(M-E)=Gm (M)*m(E)/d(M-E)^2

Let’s take the ratio of the Sun-Moon forces which is dividing the top by the bottom.  Stay with me, because G and m(E) will disappear when we divide, because they are part of both.

Ratio=m(S)/d(S-E)^2 divided by m(M)/d(M-E)^2

When we divide, we invert the divisor, which is the value that is “going into” something.

If we divide 1 by 1/3, we invert the 1/3 and have 1 *3/1 or 3.  One-third goes into 1 three times.

If we do this math, we invert the denominators and have

Ratio=m(S)*d(M-E)^2 divided by m(E)*d(S-E)^2

We know these ratios.  The mass of the Sun is 27,000,000 that of the Moon.  The distance to the Moon is about 1/389 the distance to the Sun.  Let’s call it 1/400.  By the way, in the sky, the Moon is about the same angular size as the Sun, which is why we can just have total solar eclipses. The Sun is about 400 times the diameter of the Moon and is about 400 times further away, so they have about the same size when viewed from the Earth, one of the greatest cosmic coincidences there is.

The ratio of forces is about 27000000/400^2, or 169.  But the Sun is actually less powerful as the Moon in producing tides.  Tidal forces are differential and work differently on one side of the body versus the other.  Tidal forces are not the same as gravitational forces. They work as the inverse cube, not as the inverse square.  A cube here is d*d*d or d^3.  We measure volume when we know three factors—length, height and depth.

The ratio can be done by subtracting the force of the two objects from the front by the force from  the back.  Or, and this is why calculus was invented, we can take the derivative of the gravitational force with respect to the distance, because only the distance is changing, not the masses, and derivatives of constants are zero, making life a lot easier.  Here, we deal with the change of distance.

The derivative of Gm1m2/d^2 with respect to d is -2Gm1m2/d^3.  The bottom line, literally, is a cube, and the differential force for tides is a function of the cube of the distance, not the square.  If we look at the above ratio, we get 27,000,000/400^3 and it is 0.42.  If we use the average figure of 389 times further away, we get 0.46.  Tides are much more complex, but the idea of the inverse cube ratio is why the Moon exerts a greater tidal force on us than the Sun.

A second proof for tidal forces being proportional to the inverse cube of the distance is abbreviated, but goes something like this:

Force of Sun (Fs)= G(SE)/d^2, where G is the gravitational constant and SE is the Sun Earth distance.  We could make it the lunar distance if we wanted to.

The distance is slightly different on the other side of the Earth, so we will call that p.

F(SE-near or s1)-F(SE far or s2)=G (SE)/d^2-G(SE)/(d+p)^2

=G(SE){(1/(d+p)^2)-(1/(d^2)}, d is much greater than p or d>>p.  We have factored out G(SE), which is common to both.

Look at the parentheses, and using common denominator subtraction,

(d^2+2dp+p^2-d^2)/(d+p)^2d^2

=2dp+p^2/(d^4+2d^3p+d^2p^2)

=2dp/d^4,  skipping some steps, since as d gets very large, the denominator approaches d^4,

=2p/d^3

From earthsky.org, which is nowhere near scale but shows where tides come from.

Screenshot 2017-01-06 09.52.28.png

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