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Treebeard's Stumper Answer
6 December 2002

Time and Tide

It was easy to miss the new moon on Wednesday, but we all noticed the beautiful full moon two weeks ago during the Leonid meteor shower. Surfers know that there were extreme tides both times, like the -1.5 ft low tide for Hike Club at Surf Beach yesterday. We all know the moon causes tides, but why do full and new moons both cause extreme tides? Why don't they work against each other? This is related to another classic stumper. The earth turns once every day on its axis under the moon, so why are there two high and low tides every day instead of just one?

It was like being at the end of the world skimboarding on the tidal flats in the fog at Surf Beach last week under the new moon. There's a 8.3 foot difference between that -1.5 ft low tide and the 6.8 ft high tide just a few hours earlier. It's impressive to look up at your extended arm and think: sea level was up there!

The photo on the right was an even lower -1.7 ft tide at Devereux Point on the night of the winter solstice on December 22, 1999. See my photo of the rising full moon that evening at my More Lunacy (21 Jan 2000) stumper.

The moon is on opposite sides of the earth relative to the sun during new and full moons, so how can they both cause extreme tides?


The distant but massive sun tugs the earth more than the moon, but the moon has a greater effect on tides because it's closer and so pulls the near side of the earth much more than the far side. It's the difference that makes tides. There's high tide under the moon, but there's another high tide across the earth where water is left behind in an equal tidal bulge. That's why we have extreme tides during both full and new moons when the earth, moon, and sun are lined up in syzygy. It's also why we have two tides every day as the world turns, and not just one.

Notes:

I didn't understand tides when I wrote this stumper, or I could have made it even more of a stumper. Most of us would say without thinking that the moon tugs the earth more than the sun, and so causes greater tides. In fact the sun's gravitational pull on the earth is almost 200 times greater than the moon's. Of course, we orbit the sun! But the moon still has the greater tidal effect because the differential pull on opposite sides of the earth is greater than the differential pull of the massive but distant sun. It's the difference in gravity across an object that makes tides, not the sheer force of it.

It helps to think of an extreme case, like falling feet-first into a super-massive black hole. Your feet are closer to the source, and so are attracted more than your belly and much more than your head. The result is that you get stretched out on both ends. (Of course you would have other problems too.) Larry Niven's classic sci-fi story "Neutron Star" (collected in Three Books of Known Space (1996)) tells what it might feel like. Differential tidal forces around massive planets like Jupiter and Saturn can break apart moons that are too close into a ring of fragments or keep a moons from forming. It's enough to literally crack the surface of a moon apart, and cause volcanoes, and change its rotation. We can see the results in our solar system.

This is more technical than my usual stumper. I wanted to work through the math and dig into some physics to really understand this. I won't try to explain every step, but it helped me to go through the math. I had help from Mikolaj Sawicki's Myths About Gravity and Tides.

The gravitational attraction between two objects depends their mass, but it it drops off with the square of the distance by Newton's classic formula:

          Mm
    f = G --
          d2

The relevant facts (from The Nine Planets and my ancient Halliday & Resnick) are:

                                               m^3
    G  = gravitational constant = 6.67x10^-11 ------
         (big G)                              kg s^2

                                                    m
    g  = gravitational acceleration on earth = 9.8 ---
         (small g)                                 s^2

    Ms = mass of the Sun = 1.989x10^30 kg
    Mm = mass of the moon = 7.35x10^22 kg
    Me = mass of the earth = 5.972x10^24 kg
    Ds = earth to sun distance = 1.496x10^11 m
    Dm = earth to moon distance = 3.84x10^8 m
    Re = radius of earth = 6,378 km = 6.38x10^6 m

The gravitational tug (acceleration) on a 1 kg unit mass on earth from the sun and moon are therefore:

                  Ms                  m^3       1.989x10^30 kg                 m 
    sun  = as = G --- = (6.67x10^-11 ------) x ----------------- = 5.93x10^-3 ---
                  Ds2                kg s^2    (1.496x10^11 m)^2              s^2

                  Mm                  m^3       7.35x10^22 kg                  m 
    moon = am = G --- = (6.67x10^-11 ------) x ---------------   = 3.32x10^-5 ---
                  Dm2                kg s^2    (3.84x10^8 m)^2                s^2

The ratio between these pulls is:

          sun    as   5.93x10^-3 m/s^2
    ratio ---- = -- = ---------------- = 178.4 times stronger
          moon   am   3.32x10^-5 m/s^2

The pull of earth's gravity is (small) g,

            m
    g = 9.8 --
            s2
So we can compare the pull of the sun and the moon to that of the earth:
                            m
                5.93x10^-3 ---
                           s^2
    sun  = as = -------------- = 6.05x10^-4 g
                        m  
                   9.8 ---
                       s^2

                           m
               3.32x10^-5 ---
                          s^2
    moon = am = ------------- = 3.39x10^-6 g
                       m  
                  9.8 ---
                      s^2

Both sun and moon gravitation effects are miniscule with ordinary objects here on earth. We don't weigh much less at noon or under a full moon. Don't bother waiting to weigh yourself, the position of the sun and moon don't matter!

That's gravity, so why are there tides at all? Tides depends on the differential pull across an object from head to foot or near to far. Above I really calculated gravity to the center of the earth and ignored the fact that the earth has a diameter.

The differential tidal pull (ta) on a 1 kg unit mass between the center of a planet (a) and the far side at radius r, caused by a body with mass M at distance d, can be calculated generally:

         GM
     a = ---
         d^2

         GM      GM
    ta = --- - -------
         d^2   (d+r)^2

         GM (d+r)^2 - GM (d^2)
       = ---------------------
            (d^2) (d+r)^2

         GM d^2 + 2 GM rd + GM r^2 - GM d^2
       = ----------------------------------
                   (d^2) (d+r)^2

          2 GMr (d+r)      2 GMr          2 GMr 
       = ------------- = ----------- = ------------
         (d^2) (d+r)^2   (d^2) (d+r)   d^3 + (d^2)r

At first I was stumped how to simplify this, but then I realized that the radius r is much smaller than the distance d, and so the (d^2)r term in the denominator is much smaller than the d^3 term, so we can ignore it as a minor nuisance. The fraction then reduces to:

         2 GMr     2r
    ta = ----- = a --
          d^3      d

(That's also the differential of GMrd-2 = -2GMrd-3. Don't worry about the minus sign unless you're in a college physics class!)

A similar calculation gives the same difference between the center and the near side of the planet. The earth as a whole orbits the sun in a free fall orbit determined by the sun's gravitational pull on the earth's center as. But the far side of the earth lags behind the center with acceleration as - ts and the near side has the extra acceleration as + ts. That's why there are tidal bulges on both sides of the earth, both under and opposite the sun and moon. Fortunately the earth is fairly rigid. If the differential force was too much, it would pull the earth apart into a ring of fragments. Think about Saturn and Jupiter and the asteroid belt. Even so, I've read that tidal forces can stretch the solid earth as much as 40 cm, so you actually are a bit taller (measured from the center of the earth, not your feet) when the moon is overhead!

Gravitational force varies with the inverse square of distance, but the differential tidal force varies with the inverse cube of distance. That's why the closer moon has more impact on tides than the more massive but distant sun, despite its much weaker gravitational pull. The earth has a radius Re of about 8,000/2 miles = 6.38x10^6 m, so we can now calculate the differential tidal pull from the sun and moon on opposite sides of the earth compared to the center of the earth:

         2 GMr      2Re                m    2 x 6.38x10^6 m               m  
    ts = ----- = as ---- = 5.93x10^-3 --- x --------------- = 5.05x10^-7 ---
          d^3        Ds               s^2    1.496x10^11 m               s^2

         2 GMr      2Re                m    2 x 6.38x10^6 m               m  
    tm = ----- = am ---- = 3.32x10^-5 --- x --------------- = 1.10x10^-6 ---
          d^3        Dm               s^2     3.84x10^8 m                s^2

The ratio of these forces is interesting:

                            m
                5.05x10^-7 ---
    sun    ts              s^2
    ---- = -- = -------------- = 0.459
    moon   tm               m 
                1.10x10^-6 ---
                           s^2  
The gravitational pull of the sun on the is 178 times greater than the gravitational pull of the moon, but the solar tidal pull is only 46% of the lunar tidal pull: ts = 0.46 tm. That's why the moon dominates the tides, but it also explains why the allignment of the sun and moon at new moon and full moon still makes a noticeable difference.

This typical illustration (from here) also explains another stumper. Imagine you're looking down on the earth from above the north pole. Tidal forces aren't strong enough to actually lift the ocean 8 feet above normal, but there is a tangental force that pulls water from the low tide sides towards the high tide bulges. Tides are really tidal currents that pull water around the earth rather than lift it up and down. Friction and local topography influence these sideways currents, which is why our tides always seem to lag behind the moon by 8+ hours, as shown by this data from the TBone Tide Calculator for Gaviota Beach on December 3-4, when we were skimboarding in the fog under the new moon:

    2002-12-03  23:36 PST              New Moon
    2002-12-04  08:58 PST   6.80 feet  High Tide
    2002-12-04  16:10 PST  -1.42 feet  Low Tide

That's a lot of (rusty) physics. It did me good to work through this the hard way, and it was gratifying to get the expected results. Despite all the math, this analysis is rudimentary. Orbits are elliptical, so the distance to the sun and moon vary over the year. The earth is closest to the sun (at perihelion) in early January. If the moon is closest to the earth (at perigee) at about the same time, then the tides will be even greater. Extreme Proxigean Spring Tides result when all these factors peak at the same time. Did that happen on the solstice night of December 22, 1999? (See my photo above and my More Lunacy (21 Jan 2000) stumper.)

That's plenty for a stumper answer, but I became interested in tide prediction while researching this. We take newspaper tide tables for granted, but how are they made? After doing my recent stumper on Sound Tracks (8 Nov 02), I had a moment of deja vu as I read at NOAA that tides are calculated as the algebraic sum of simple integer harmonics of only a few sinewave variables of the form t = h cos(at+p):

Local data for these "harmonic constants" is available, but I'm not sure how to use the raw data. The info might be hidden on the NOAA CO-OPS Publications page. It's fascinating to think of tides as a physical chord of harmonics. Maybe I will program my own Harmonic Analysis tide predictor in good old BASIC to understand the algorithm and raw data for my next DMS science fair project.

Here are some Web links for your own research on tides.

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Copyright © 2002 by Marc Kummel / mkummel@rain.org