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Why moon causes tides

  1. Jul 19, 2004 #1
    I know the moon causes tides, but I am unsure why. I did a quick calculation and found that the sun has a greater gravitational effect on the Earth than the moon does. My question is, how come the sun does not affect tides despite the fact that the gravitational force is stronger than the moon?
  2. jcsd
  3. Jul 20, 2004 #2
    It does. This is why the tides are strongest when the moon and sun are aligned.
  4. Jul 20, 2004 #3
    The tidal effect is proportional to the inverse cube of the distance to an object. The sun is about 30 million times the mass of the moon and about 400 times the distance. That means that the effect of the sun is 30000000/64000000 or about half of the effect of the moon
  5. Jul 20, 2004 #4
    Yes, choron is correct.

    To make it simple.

    Tides are caused by the difference of gravitation pull from one point to the next. So to calculate it you would figure the gravitation pull on the far side of the earth due to the moon and then calculate the pull on the near side of the earth due to the moon. Then you can do the same for the sun, and you will clearly see the difference.

    F= G m1m2/r2

    It is just a plug and chug from there. After that you can clearly see high tides low tides ect... (although it is rather easy to visualize).

  6. Jul 22, 2004 #5
    Oh so basically the moon causes a greater difference in gravitational pull than the sun does.
  7. Jul 22, 2004 #6


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    Agreed, the moon is closer. Do the distance squared thing and you will find it has much more tidal effects than the sun.
  8. Jul 22, 2004 #7
    Last edited by a moderator: May 1, 2017
  9. Jul 22, 2004 #8
    I did the distance *squared* thing, and the sun's pull came out much larger.

    But chronon showed the calculation for distance *cubed*, and then the moon's pull is about half that of the sun... So which one is it? And if the latter, where is the extra distance multiplication coming from??? :confused:
  10. Jul 22, 2004 #9
    It's cubed.

    If the position of the Moon, or Sun relative to the Earth centre is given by [tex]\vec{s}[/tex] and the position of an ocean point is given by [tex]\vec{r}[/tex] then the tidal acceleration is given by:

    [tex]\vec{a} = GM \left( \frac{\vec{s} - \vec{r}}{| \vec{s} - \vec{r} |^3} - \frac{\vec{s}}{| \vec{s} |} \right) [/tex]

    expanding of the first term between brackets, filling in, and neglecting small terms gives something that is proportional to ][tex]\frac{GM \vec{r}}{s^3}[/tex]

    more details later..maybe
  11. Jul 22, 2004 #10

    Doc Al

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    What matters in causing the tides is not the gravitational pull (which goes as distance squared) but the change in gravitational pull across the planet. It's that change that goes as distance cubed.

    While it's true that the sun's pull is much stronger, the moon, being closer, exerts a greater tidal effect since its pull (think "force per unit mass") varies more from one side of the earth to the other. Again, it's the variation in gravitational pull that causing the "stretching" effect that is the tides.
  12. Jul 22, 2004 #11


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    What's the Sun's gravitational pull at the point on the Earth closest to the Sun? ditto, furthest from the Sun? What's the difference?

    Now do the same calculations, but use 'Moon' instead of 'Sun'.

    For the nitpickers, assume mid-solar (or mid-lunar) eclipse and a point on the Earth's equator.
  13. Jul 22, 2004 #12


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    Read this one again:
    The diameter of the earth is 13,000km. The distance from the earth to the moon is 385,000km, and the distance from the earth to the sun is 150,000,000km.

    So the earth's diameter is .0087% of the distance to the sun, but 3.4% of the distance to the moon. That's why the moon has a bigger tidal influence.
  14. Jul 22, 2004 #13
    Thanks everyone.
  15. Jul 22, 2004 #14
    on the same note.... the Sun accelerates the Moon more than the Earth does :) really!
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