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On the origin of an global ocean phenomenon

The general definition of the term tide according to Morrison and Owen [3] is that:

Specifically on Earth the term tide is also used to describe the rise and fall of sea levels caused mainly by the combined effects of the gravitational forces exerted by the Moon and the Sun. This is a bit misleading since focussing on the sea level is ignoring the fact that tides also affect the solid earth crust. Moreover the observable rise and fall of the sea level is influenced strongly by shoreline topography, ocean currents and the distribution of the continents on earth. For now We will investigate the tides based on the definition given by Morrison and Owen and ignore the effects of topography and ocean currents that lead to more complex tidal cycles on Earth. These effects will be explaind later in the section Tidal Cycles on Earth.

Skip the explanation, just show me the tidal simulation applet!

This web page is based partially on another article about the "Computation of Tides" by Paolo Sirtoli [1] as well as the article "Tidal Misconceptions" by Donald E. Simanek [4]. I will extend the ideas brought forward by those two articles with up to date visualization. We will start be first looking into the gravitational pull that the Moon is excerting on the Earth.

The theoretical basis for the computation of the tidal effects caused by Moon and Sun is the same. For simplicity we will first examine the effect of the Moon. We will also ignore Earths rotation in our models since the rotation of Earth has no influence on the existence of the tidal bulges. So lets start by looking at the Earth, the Moon and the law of gravity:

At first we will exclusively look at the acceleration caused by the gravity of the Moon on objects lying at the surface of Earth. For the moment we also assume both Earth and Moon are pinned down in space and not orbiting one another. By substituting the force in Equation 1 with Newtons second law of motion (\(\vec F = m \cdot \vec a\)). We can quickly rearrange this equation to calculate the acceleration of a body lying on the earths surface due to the gravitational pull of the moon (Equation 2):

$$\vec a(\vec r) = \gamma \cdot m_m \cdot { { \vec r_m - \vec r } \over |\vec r_m - \vec r|^3}$$ Where:- \(\vec a\) is the acceleration of the body laying at the earths surface
- \(\vec r\) and \(\vec r_m\) are the position position vectors of the body and the moon
- \(m_m\) is the mass of the moon in \(kg\)

By doing this for a variety of points at the Earths surface as well for its center of mass we get the following result:

The sizes of Earth and Moon in the image above are to scale. You may use the button to toggle between a visualization that is correct with respect to distance and one that shows the moon unrealistically close to Earth. In latter setup you can grab the moon with the mouse and move it around to see how the acceleration vectors change. For the sake of completeness i have also added a marker to denote the common center of mass of the system Earth-Moon. There are a few things you will notice:

- The acceleration is greater on the side of Earth facing the Moon. This effect is smaller if the Moon is farther away.
- The accelerations vectors are pointing to the Moons center of mass
- If the Moon is far away the acceleration vectors are almost parallel

In the visualization above we focus exclusively on the gravitational forces caused by the moon. Earths own gravity is ignored. If you do the math you will realize that the acceleration caused by the moon is by orders of a magnitude lower that the acceleration caused by earths own gravity (3.3e-5 m/s² vs. 9.8 m/s²). It is also slightly lower on the side of earth opposing the moon.

This concludes our look into the gravitational pull Moon is excerting onto Earth. In the next chapter we will look into the motion of the Earth and Moon as they circle one another and what role this motion plays in the computation of tidal accelerations. Go there now...

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