Solar vs Lunar Tides: Explaining the Difference

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    Lunar Solar Tides
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SUMMARY

The discussion centers on the differences between solar and lunar tides, emphasizing that lunar tides exert approximately twice the effect of solar tides despite the Sun's greater gravitational force on Earth. The gravitational force is calculated using Newton's law of universal gravitation (F=GmM/r^2), but the tidal effect arises from the differential gravitational pull experienced across the Earth's surface. This difference is influenced by the varying distances from the Earth to the Moon and the Sun, leading to the inverse cube relationship in tidal forces.

PREREQUISITES
  • Understanding of Newton's law of universal gravitation (F=GmM/r^2)
  • Basic knowledge of tidal mechanics and forces
  • Familiarity with the concept of differential gravitational pull
  • Awareness of the Earth-Moon and Earth-Sun systems
NEXT STEPS
  • Research the mechanics of tidal forces and their calculations
  • Explore the inverse cube relationship in gravitational forces
  • Study the effects of celestial mechanics on tidal patterns
  • Investigate the role of distance in gravitational interactions
USEFUL FOR

Students of physics, oceanographers, and anyone interested in understanding the dynamics of tidal forces and their implications on Earth.

The Head
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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!
 
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The Head said:
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!

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).
 
Found this
http://oceanworld.tamu.edu/resources/ocng_textbook/chapter17/chapter17_04.htm"

Check out equations (17.12), (17.13), & (17.14)
 
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You're kidding me. You linked to a website that promotes a crackpot theory? Please don't tell me you believe that nonsense.

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 into 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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