A Little Help?… ( Calling All Geeks!)

Was watching a doco on TV tonight and it mentioned the topic of the Moon and the affect it has on earth’s tides.  (Ocean and tectonic).

It’s long been my understanding that both the Sun and the Moon have their own effects on Earth’s tides, whereby, due to their individual masses, they pull the more ‘mobile’ oceans (as compared to the crust) towards themselves causing a ‘bulge’ of raised water levels at the point over the earth’s surface closest to their respective centre’s of mass. This bulge ‘moves’ west to east ( yes?) owing to the faster diurnal rotation of the earth than the respective planetary /lunar orbit. ( not too important re: direction now). I have always heard the one with the greatest tidal effect is the Moon.

Being the careful guy i like to think i am when it comes to just accepting information on ‘faith’ ( in this case in the reporters) alone and knowing Newton’s Formula for gravitational force between two respective bodies ( F = g M1M2/r^2) i decided to do a quick mental calculation of the respective tidal force from both the Moon and the Sun.

Problem!  in my head i figured something unexpected. Mental maths came up with the astounding conclusion that the Sun exerts a greater pull on earth than the moon does by a factor of about 5000!

That can’t be right i thought!

So i checked my assumptions, put the corrected ones to paper and guess what?? i was WRONG!!

Bigger problem… the ‘correct’ answer came out to the Sun having only 168 times as much pull on our oceans as does the moon! I had over-estimated solar mass by a factor of 3 and misplaced the moon radius squared by  a single decimal place to make a ‘small’ factor of 30 error!

Could someone kindly explain where the error ( if any) is in the following reasoning?

Force due to Moon (m)  on Earth (e)                             Force due to Sun (S) on Earth (e)

M = mass of ; D = distance between

Me*Mm/D m^2                                                                  Me*Ms/DS^2

Mm ~ Me/81; Dm =d                                                        MS = 330000 Me; DS = 397 d

Substituting

Me^2/81*d^2                                                              Me^2 *330000/(397d)^2

Gives:

(Me/d)^2 /81                                                  (Me/d)^2 * 330000/157500 (approx)

or a ratio of 1 : 2.1/81  = 1 : 168 give or take??

So tides due to Sun should be 168 times stronger than those due to the moon even though the moon is much much closer to us. Huh?

Help!

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6 Comments

  1. I personally love physics geography and science so this would be Perfect for me to figure out however I just can’t! 😦
    I loveee your blog 😀

    Reply
  2. Thank you Zoey, you are very kind 🙂

    I am 99% sure i am ‘missing’ something.. but i cannot think what it might be. Maths does not lie!

    Reply
  3. Bob, you’re looking at this wrong. The calculation you should make is not the force the Sun exerts on the Earth, but the difference between the force on one side of the Earth and the force on the other side, which will give you the net force for tides. Given the Moon is much closer, the difference should be greater for the Moon, even though the overall pull is smaller. If you redo your calculations like that I’m confident it will make sense.

    Reply
    • That is, Force(tides Sun) = F (Sun-Earth, far side) – F (Sun-Earth, close side)

      Reply
      • Thanks for the contribution David. I do recall reading something along those lines previously but i have less confidence than you that it makes sense! 🙂

        Having done the math i can see that the net difference between the force on one (far side of the planet) to the near side for the Moon is roughly 385 times greater than the equivalent for the Sun.
        Considering their relative strengths gives a (reasonable – ish) force of about 2 times greater diff for the Moon than the Sun which is more in line with ‘tidal theory’.

        As to the making sense however, i’m still missing something ‘important’.

        What you have helped me see is the maximum difference between gravitational effects on opposite sides of the planet (Close side/far side). My reading suggests the tidal forces are vector forces (i get the idea) but surely for a single body (such as the Moon) all gravitational forces are directed towards it’s centre of gravity (as is the Earth’s) so on the far side from the Moon the Earth and Moon are pulling the water in the same (vertically downward) direction, while on the closer side these forces are in opposite directions and it is the ‘upward’ force of the Moon’s gravity that causes the water to bulge upwardly toward it – no? So should not the water on the far side be pulled even more strongly toward the centre of the Earth (and Moon) causing a ‘dip’ (low tide) and the opposite on the opposite side?

        All diagrams/explanation of tides say this is not so and there is another bulge on the far side?

        I’m wondering ( trying to figure out what the ‘weaker’ forces away from the points equidistant from the moon’s CoG compared to the stronger forces closer to the Moon) what if any difference it might make to my understandings if the world’s oceans were frozen solid instead of fluid?

        Of course, the theory may not matter all that much if my observation in the second post ‘Time and Tide…’ is accurate, as that shows the bulge does not exist where, or when, it ‘should’ be. That is not a matter of poor understanding of the theory but demonstrated discrepancy between theory and experimental (practical) observation.
        Any thought there?

  1. Time and Tides Wait for No Moon! | The World Is My Soapbox

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