Physics 430: Lecture 20 Non-Inertial Frames


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physics430 lecture20

9.2 The Tides

  • November 16, 2010
  • What causes the tides? You may know that it has something to do with the Sun and Moon, and in fact, both have a contribution. To simplify matters, however, we can consider only the effect of the Moon and ignore the Sun.
  • You may think it plausible that the Moon attracts the water on the side nearest the Moon, so that the situation is like this:
  • However, it is observed that the tides occur twice a day, not once, and in fact the true situation is like this:
  • You can think of the ocean bulge (here grossly exaggerated) as being fixed while the Earth rotates under it. How can we understand the bulge in the direction AWAY from the Moon?

The Tides-2

  • November 16, 2010
  • Basically, what causes the tides is the gradient of the Moon’s gravity across the finite size of Earth. Let’s make this quantitative.
  • The force on any mass element m near the Earth’s surface (say a drop of water) is its weight mg + any buoyant (or other) force Fng + the gravitational pull of the Moon
  • But the acceleration of the Earth (considered as a point mass at the center of the Earth) is:
  • so that
  • Or, if we combine the two terms involving we identify the tidal force

Magnitude of the Tides

  • November 16, 2010
  • The figure below may make this more clear. It shows the forces on various parts of the Earth due to the Moon, and then again relative to the center of the Earth.
  • To find the magnitude of the height difference between high and low tides, notice that the surface of the ocean has to be an equipotential surface—a surface of constant potential energy.
  • You can see the truth of this last statement by considering the three forces acting on a drop of water (gravity, pressure, and tidal [Moon]). No liquid can exert shearing forces, so the pressure force must be normal to the surface of the ocean. Another way to say this is that any non-normal pressure force will push sideways on the drop until it vanishes. Since the total force is zero (the drop is in equilibrium) it follows that mg + Ftid must be normal to the surface.

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