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On the origin of a global ocean phenomenon

So far we have computed the gravitational pull of the Moon on the surface of Earth. We learned that unsuprisingly the acceleration vectors caused by the gravity of the Moon are in fact pointing towards the Moon. Lets ignore for a moment, that we used a bit of unscientifc magic and fixed the positions of Moon and Earth arbitrarily in space. Water following in the direction of the acceleration vectors could form a single tidal bulge on the side of Earth facing the Moon. This alone however is not enough to explain the tides. To explain why lets look a bit into the system Earth-Moon:

Earth needs 24 hours to complete a rotation around its axis. The Moon takes 27.32 days to complete a single orbit around Earth. In a single day the moon travels an angular distance of:

$$360°/27.32 d = 13.18°$$Consequently a place on Earth will have the same position with respect to the moon after 24 hours and 52 minutes. Observing the time between two high or low tide events at most ocean sites on earth yields 12 hours and 25 minutes (notable exceptions will be discussed later). That is pretty much exactly half of the previous number. The logical conclusion is that there must be two tidal bulges on opposing sides of the planet. Our current model does not account for that because we have ignored that Earth and Moon are gravitationally bound to one other. They do not sit still in space but orbit one another around common center of mass (also called the barycenter) as illustrated in Animation 2.

For this investigation we have selected a reference frame located at the center of mass of the Earth-Moon System. The system itself does not rotate or accelerate and remains fixed with respect to the rest of the universe. (We are ignoring the Sun and Earths orbit around it for the moment.) In physics such a system is referred to as an inertial frame of reference.

The Earths center describes a nearly circular orbit of radius \(d\) around the common center of mass with the angular velocity \(\omega\). The average Distance of Moon and Earth is 385000 km. The distance of the common center of mass from the center of Earth can be computed with the following equation:

$$ r_{cm} = 385000 km \cdot {m_m \over {m_e + m_m}} = 4678 km $$ where:- \(m_e\) - Mass of the Earth (5.9721986e24 kg)
- \(m_m\) - Mass of the Moon (7.3459e22 kg)

The first thing you might notice is that the common center of mass of the system Moon-Earth is still within Earth at a depth of 1700 km below Earths surface. If we now trace a point on the earths surface we will discover that each point is following a circular path with the same radius as the distance between Earths center and the common center of mass. Each one of these paths has a different center. In our reference frame each of the points on and within our (non rotating) earth is experiencing the same centrifugal acceleration of:

$$ a_z = \omega^2 \cdot r_{cm} $$caused by Earths motion around the common center of mass of the system Earth-Moon. Lets briefly compute the centrifugal acceleration based on the length of the lunar month (27.32 days) we obtain an angular velocity of:

$$ \omega = 2 \cdot \pi \cdot { 1 \over { 27.32 d \cdot 86400 s} } = 2.662 \cdot 10 ^{-6} s^{-1} $$Which translated into a centrifugal acceleration of:

$$ a_z = (2.662 \cdot 10 ^{-6})^2 \cdot 4678000 = \underline { \underline { 3.31 \cdot 10^{-5} {m \over s^{2} } }} $$By subtracting this acceleration from the acceleration caused by the gravitational pull of the Moon we obtain the tidal acceleration. But before we do that hold on a second.

First we will have a closer look at the center of the Earth. The centrifugal force computed above is the same in any point on and within Earth and is always pointing in the same direction. This means it is also acting on the center of Earth itself (see orange arrow in the animation above). If we have a centrifugal force acting on the Earths center of mass why is Earth not moving away from the Moon? What is keeping it in its place? The gravity of the moon does! If we compute the acceleration caused by the gravitational pull of the moon acting on the Earths center we obtain:

$$ a_g = \gamma \cdot { m_m \over d } = 6.675\cdot10^{-11} \cdot { {7.349 \cdot 10^2 } \over 385000000} = \underline { \underline {3.31\cdot10^{-5} {m \over s^{2} }} } $$ $$ a_g = a_z $$The centrifugal force is kept in balance by the gravitational force of the Moon which is acting as the centripetal force. Instead of computing the centrifugal acceleration it is more convenient to compute the gravitational force excerted by the Moon to the center of Earth and subtract this force from the acceleration vectors obtained by Equation 2 in order to obtain the tidal acceleration.

The same principles explained here also apply to the Suns gravitational effect on Earth. The tidal effect caused by the Sun is a bit smaller than the effect caused by the Moon. If Sun, Moon and Earth are aligned along a straight line the tidal acceleration is strongest since the effects of Moon and Sun add up.

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