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Solar vs Lunar Tides

  1. Sep 30, 2011 #1
    I have a question about the effects of solar and lunar tides. I know that the effect of lunar tides is twice that of solar. However when I calculated the force of gravity of the Sun on the EArth vs the Moon on Earth using F=GmM/r^2 the force due to the Sun was much greater. Can anyone help me explain this? Thank you!
  2. jcsd
  3. Sep 30, 2011 #2


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    If you call the radius of the earth ε, then you can see that the gravitation of the sun actually applies to points on earth over a distance which varies from r-ε for the nearest point, to r+ε for the furthest point. It's this difference in gravity due to the difference in distance that generates the tides.

    Now if you recompute the gravity difference between near and far points, you will see why the lunar tide is so large (hint: it has to do with the ratio ε/r).
  4. Oct 1, 2011 #3
    Last edited by a moderator: Apr 26, 2017
  5. Oct 1, 2011 #4
    The tide raising force is indeed inversely proportional to the cube of the distance between the bodies.

    Looking more closely at the link I agree it was an inappropriate reference as it is about something else entirely, so my apologies.

    However I am suprised that no one has come in to describe the mechanics of how the inverse cube relationship occurs so I will do that when I have time- It really is quite simple in principle. As oliversmsun says it is because the force due to gravitational attraction is not directly the tide raising force (TRF). The TRF stems from the difference between the forces of gravitational attraction and the forces due to the earth-moon or earth-sun mechanical systems revolving about their respective centres of mass.

    Edit, thank you Gannet you posted whilst I was writing the above.
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