User:Anorderofmagnitude/sandbox

The following is a calculation for the height water on the surface of an earth without any land features. It requires an understanding of calculus and a first semester of calculus based physics. Disclaimer: this is my original derivation so it may be totally wrong. But it is a start. An assumption that is made is that the moon is very far away compared to the size of the earth. I also make an approximation using the binomial expansion, during the derivation that introduces a very small error but greatly simplifies the math.

= Diagram and Definitions =

Lets start with an initial drawing of the problem.



$$\vec{r}_{em}$$ is from the center of the earth to the center of the moon.

We need a detailed view of the earth because that is where all the action is.



$$\vec{r}_e$$ has the magnitude of the radius of the earth and is normal to its surface pointing away from the center

$$\vec{c}$$ is from the center of the earth to the the barycenter of the earth moon system.

$$\vec{D}$$ is the distance to a particular spot earth we are interested in to the moon.

and using the law of cosines D is given as

$$D=D=({r_{em}^2-2r_{em}r_e\cos{\theta}+r_e^2})^{\frac{1}{2}}$$

Lets go to a diagram of of a small wedge of the earth from its center to its surface. This diagram is drawn severely out of scale.



$$\vec{a}_T$$ is the tangential acceleration on the small piece of water.

$$\vec{a}_N$$ is the normal acceleration on that piece of water.

$$d\theta$$ is an infinitesimal angle coming from the center of the earth.

$$r_e$$ is the radius of the earth.

$$dh$$ is an infinitesimal height increase of the water across $$r_e d\theta$$.

$$d\theta$$ is drawn rather large. This drawing serves two purposes. It shows that the slope on the surface of the water is tilted because of a net tangential acceleration (or force if you like).

The first insight to the problem is that the net accelerations or force acting on the surface of the water is normal to the surface of the water. Why? Well if the acceleration (or forces) acting on the water were anything but normal it would have a component that would act tangentially to the surface of the water and therefore it would flow in the direction of the acceleration (or force). So for the tide to be in equilibrium the net acceleration (or force) is normal to the surface.

For the rest of the derivation I am going to refer to acceleration acting on the surface of the water rather than the force, because F=ma. So the accelerations are proportional to the force through the constant of mass and the mass of the a small piece of water is always the same.

From the diagram and the geometry of the surface of the water and the geometry of the accelerations we can write:

$$\frac{a_N}{a_T}=\frac{r_e d\theta}{dh}$$

or

$$dh=\frac{a_T r_e d\theta}{a_N}$$

We take the earth moon system to be stationary in a rotating frame of reference. The frame of reference is rotating relative to the inertial frame at the angular velocity equal to the rotation of the moon earth orbiting each other. In other words a frame of reference where the earth and moon are stationary in their orbits around each other. Commonly symbolized by $$\Omega$$. Now we define our reference frame so that $$\Omega$$ is rotating in the opposite direction that the earth and moon are, $$\omega$$. $$\omega$$, speed that the earth and moon are rotating about each other is derived elsewhere and given by

$$\omega^2 = \frac{G(M_e+M_m)}{r_{em}^3} $$

$$G $$ is the universal gravitational constant.

$$M_e$$ is the mass of the earth.

$$M_m$$ is the mass of the moon.

We press on with more physics. To find what the normal and tangential accelerations on the water of the surface of the earth are we introduce the general equation relating the inertial and fictitious accelerations a body experiences in a rotating frame of reference. We use this system because it simplifies the derivation.

$$\vec{a_b} = \vec{a}_{int} - 2\vec\Omega\times\vec{v}_b-\vec\Omega\times(\vec\Omega\times\vec{x}_b) - \frac{d\vec\Omega}{dt}\times\vec{x}_b$$

where:

$$\vec{a_b}$$ is the accelerations a body experiences in the rotating frame of reference.

$$\vec{a}_{int}$$ is the inertial accelerations a body experiences.

$$\vec{v}_b$$ is the velocity of the body in the frame.

$$\vec{x}_b$$ is the position of the body in the frame of reference.

Lets break down the right side of the equation piece by piece then we will add them up.

= Inertial Terms  $$\vec{a}_{int}$$ =

The only two inertial accelerations are the ones due to the earth and moon's gravity

$$\vec{a}_{int} = \vec{a}_{ge} + \vec{a}_{gm}$$

$$\vec{a}_{ge}$$ is the acceleration due to the earth's gravity.

$$\vec{a}_{gm}$$ is the acceleration due to the moon's gravity.

The acceleration due to gravity is approximately g=9.81 m/s^2 and points toward the center of the earth.

$$\vec{a}_{ge} = -g \hat{r}_e$$

The vector that points toward the center of the earth can be written as a sum of components.

$$\vec{r}_e=r_e(\cos{\theta}\hat{i} -\sin{\theta}\hat{j})$$

It's normal can be seen to be

$$\hat{r}_e=\cos{\theta}\hat{i} -\sin{\theta}\hat{j}$$

The acceleration due to the moon is from the universal law of gravitation.

$$\vec{a}_{gm} = \frac{GM_m}{|D|^ 3}\vec{D}$$

Again $$\vec{D}$$ is the distance to the moon from a particular point from the diagram is given as

$$\vec{D}=r_{em}\hat{i}-\vec{r}_e$$

$$=r_{em}\hat{i}-r_e(\cos{\theta}\hat{i}$$$$ -\sin{\theta}\hat{j})$$

Now we go after the square of the magnitude of the distance

$$D^2=(r_{em}-r_e\cos{\theta})^2 +(r_{em}\sin{\theta})^2$$

$$=r_{em}^2-2r_{em}r_e\cos{\theta}+(r_e\cos{\theta})^2 +(r_{em}\sin{\theta})^2$$

$$D=({r_{em}^2-2r_{em}r_e\cos{\theta}+r_e^2})^{\frac{1}{2}}$$

= Angular Velocity of Rotating Reference Frame  $$\Omega$$ =

Now about $$\Omega$$ terms

$$\omega = \frac{d\theta}{dt} $$

$$\omega$$ is both the magnitude of the angular velocity of the frame of reference and since the earth is fixed with respect to the background stars it is also the angular velocity of the earth in the frame of reference but in the opposite direction..

So the angular velocity of the frame of reference is $$\vec\Omega=-\omega\hat{k}$$

and this will help us compute the rest of the terms on the right side.

= Euler Term $$\frac{d\vec\Omega}{dt}\times\vec{x}_b$$ =

The reference frame is not accelerating. So

$$\frac{d\vec\Omega}{dt}=0$$

And that knocks out the 4th term of the accelerations on the body.

= Coriolis Term  $$2\vec\Omega\times\vec{v}_b$$ =

The velocity of a point on the surface of the earth $$\hat{v}_b$$ is

$$\vec{v}_b=\frac{d}{dt}\vec{x}_b$$

From the diagram

$$\vec{x}_b+\vec{c}=\vec{r}_e$$

So

$$\vec{x}_b=\vec{r}_e-\vec{c}$$

Where the vector from the center of the earth to the barycenter is

$$\vec{c}=c\hat{i}$$

c is the distance from the center of the earth to the barycenter and happens to be

$$c=\frac{r_{em}M_m}{M_e+M_m}$$

Substitution into above

$$\vec{v}_b=\frac{d}{dt}(\vec{r}_e-\vec{c})$$

$$\vec{r}_e=r_e(\cos{\theta}\hat{i} -\sin{\theta}\hat{j})$$

When we differentiate, $$\vec{c}$$ is a constant so its derivative is zero. So we get $$\vec{v}_b$$

$$\vec{v}_b=r_e\omega(-\sin{\theta}\hat{i} -\cos{\theta}\hat{j})$$

The Coriolis acceleration can now be calculated

$$- 2\vec\Omega\times\vec{v}_b=-2\omega\hat{k} \times r_e\omega(-\sin{\theta}\hat{i} -\cos{\theta}\hat{j})$$

$$=-2r_e\omega^2(-\sin{\theta}\hat{j} +\cos{\theta}\hat{i})$$

= Centrifugal Acceleration $$\vec\Omega\times(\vec\Omega\times\vec{x}_b)$$ =

The centrifugal term is

$$-\vec\Omega\times(\vec\Omega\times\vec{x}_b) = -\vec\Omega\times(\vec\Omega\times(\vec{r}_e-\vec{c}))$$

First, substitute

$$= -\omega\hat{k}\times(\omega\hat{k}\times(r_e(\cos{\theta}\hat{i} -\sin{\theta}\hat{j })-c\hat{i}))$$

and taking cross products gives

$$=-\omega^2 r_e(-\cos{\theta}\hat{i} + \sin{\theta}\hat{j})-\omega^2c\hat{i}$$

= Combining Fictious Forces=

$$- 2\vec\Omega\times\vec{v}_b-\vec\Omega\times(\vec\Omega\times\vec{x}_b) - \frac{d\vec\Omega}{dt}\times\vec{x}_b$$

plugging in what we calculated above.

$$=-2r_e\omega^2(-\sin{\theta}\hat{j} +\cos{\theta}\hat{i}) -\omega^2 r_e(-\cos{\theta}\hat{i} + \sin{\theta}\hat{j})-\omega^2c\hat{i} - 0$$

with some algebra becomes

$$=\omega^2 ((\frac{-M_m r_{em}}{M_e+M_m}-r_e\cos{\theta})\hat{i} + r_e\sin{\theta}\hat{j}) - 0$$

= Calculating Tangential Acceleration of Inertial Forces=

We define a unit tangential vector at the surface of the earth.

$$\hat{u}_T\equiv\sin{\theta}\hat{i} + \cos{\theta}\hat{j}$$

We can use $$a_T$$ as the the magnitude of the acceleration in the clock wise direction. When it is negative we understand that the tidal force is in the counter clockwise direction

First we will look at the contributions of the inertial accelerations. Those being the gravity of earth and moon.

$$ \vec{a}_{int}\cdot\hat{u}_T = \vec{a}_{ge}\cdot\hat{u}_T + \vec{a}_{gm}\cdot\hat{u}_T$$

Not surprisingly the tangential acceleration due to gravity of earth is zero.

$$\vec{a}_{ge}\cdot\hat{u}_T=-g(\cos{\theta}\hat{i} -\sin{\theta}\hat{j})\cdot(\sin{\theta}\hat{i} + \cos{\theta}\hat{j})=0$$

However this is not true for the moons gravity.

$$\vec{a}_{gm}\cdot\hat{u}_T=\frac{GM_m}{|D|^ 3} (r_{em}\hat{i}-\vec{r}_e) \cdot (\sin{\theta}\hat{i} + \cos{\theta}\hat{j})$$

$$\vec{a}_{gm}\cdot\hat{u}_T=\frac{GM_m}{|D|^ 3} (r_{em}\hat{i}-(r_e(\cos{\theta}\hat{i} -\sin{\theta}\hat{j}))) \cdot (\sin{\theta}\hat{i} + \cos{\theta}\hat{j})$$

The component normal to the surface of the earth, $$\vec{r}_e$$, disappears and we are left with

$$\vec{a}_{gm}\cdot\hat{u}_T=\frac{GM_m}{|D|^ 3} r_{em}\sin{\theta} $$

= Calculating Tangential Acceleration of Fictitious Forces=

Now we can look at the contributions to the tangential acceleration due to the fictitious forces.

$$(-2\vec\Omega\times\vec{v}_b-\vec\Omega\times(\vec\Omega\times\vec{x}_b))\cdot\hat{u}_T$$

$$=\omega^2 (\frac{-M_mr_{em}}{M_e+M_m}-r_e\cos{\theta})\hat{i} + r_e\sin{\theta}\hat{j}) \cdot (\sin{\theta}\hat{i} + \cos{\theta}\hat{j})$$

$$=\omega^2 (\frac{-M_mr_{em}}{M_e+M_m}\sin{\theta}-r_e\cos{\theta}\sin{\theta}+ r_e\sin{\theta} \cos{\theta})$$

$$=-\omega^2 r_{em}\frac{M_m}{M_e+M_m}\sin{\theta}$$

$$\omega$$ can be put in terms of actual values of the system

$$\omega^2 = \frac{G(M_e+M_m)}{r_{em}^3} $$

So the tangential acceleration due to the fictitious forces reduces to.

$$(-2\vec\Omega\times\vec{v}_b-\vec\Omega\times(\vec\Omega\times\vec{x}_b))\cdot\hat{u}_T$$

$$= - \frac{GM_m\sin{\theta}}{r_{em}^2} $$

= Total Tangential Acceleration =

Combining the inertial and fictitious tangential forces.

$$a_T =\frac{GM_m}{|D|^ 3} r_{em}\sin{\theta} - \frac{GM_m\sin{\theta}}{r_{em}^2} $$

factoring out

$$ =GM_m\sin{\theta} (\frac{r_{em}}{|D|^ 3} - \frac{1}{r_{em}^2}) $$

substituting $$D$$

$$ =GM_m\sin{\theta} (\frac{r_{em}}{({r_{em}^2-2r_{em}r_e\cos{\theta}+r_e^2})^{\frac{3}{2}}} - \frac{1}{r_{em}^2}) $$

factoring

$$=GM_m\sin{\theta} (\frac{r_{em}}{r_{em}^3({1-2\frac{r_e}{r_{em}}\cos{\theta}+\frac{r_e}{r_{em}}^2})^{\frac{3}{2}}} - \frac{1}{r_{em}^2})$$

more factoring

$$=\frac{GM_m\sin{\theta}}{r_{em}^2} ( \frac{1}{(1 - 2\frac{r_e}{r_{em}} \cos{\theta} +(\frac{r_e}{r_{em}})^2)^{\frac{3}{2}}} - 1)$$

$$=\frac{GM_m\sin{\theta}}{r_{em}^2} ( (1 - 2\frac{r_e}{r_{em}} \cos{\theta} +(\frac{r_e}{r_{em}})^2)^{-\frac{3}{2}} - 1)$$

Now a binomial expansion. For some $$y$$ and $$\alpha$$ real.

$$ (1 + y)^\alpha \approx 1 + \alpha y$$

$$a_T = \frac{GM_m\sin{\theta}}{r_{em}^2}(1+(\frac{-3}{2})(- 2\frac{r_e}{r_{em}} \cos{\theta} +(\frac{r_e}{r_{em}})^2)-1)$$

for $$(\frac{r_e}{r_{em}})^2 \approx 2.7x10^{-4} \ll 1$$

$$a_T = \frac{GM_m\sin{\theta}}{r_{em}^2}(3\frac{r_e}{r_{em}} \cos{\theta} -\frac{3}{2}(\frac{r_e}{r_{em}})^2)$$

Finally we can write

$$a_T = \frac{3GM_mr_e\sin{\theta}}{r_{em}^3}(\cos{\theta} -\frac{1}{2}\frac{r_e}{r_{em}})$$

This is the tangential acceleration to calculate the height of the tides.

= Location of Minimum and Maximum Heights =

$$a_T = \frac{3GM_mr_e\sin{\theta}}{r_{em}^3}(\cos{\theta} -\frac{1}{2}\frac{r_e}{r_{em}})$$

has four zeros. Two at $$\sin{\theta}=0$$ This gives us our maximums at $$0$$ and $$\pi$$. Also known as the point closest to the moon and the point furthest from the moon.

Two are at

$$0=\cos{\theta} -\frac{1}{2}\frac{r_e}{r_{em}}$$

Which is at $$\theta$$=+/- 89.52 degrees. The low tide is not when the moon is on the horizon but when it is approximately 1/2 a degree below.

= Calculation of the Tide Height =

From the top

$$dh=\frac{a_T r_e d\theta}{a_N}$$

Now for some hand waving. The normal force on the surface of the earth is dominated by g and the force due to the moon is very small. Where g is the acceleration due to gravity.

$$dh=\frac{a_T r_e d\theta}{g}$$

We are ready to integrate. From 0 to H on the left side. Where H is the height of the water above the lowest point that we set to be zero. And $${\pi/2}$$ to 0 on the right side.

$$\int\limits_0^H dh = \int\limits_{\pi/2}^0 \frac{1}{g} a_T r_e d\theta$$

The left side is trivial

$$\int\limits_0^H dh = H$$

Now the right side. We integrate from $$\pi/2$$ to 0. Integrating from $$\pi/2$$ rather than 89.5 degrees introduces a small error but one that is tolerable.

$$ \int\limits_{\pi/2}^0 \frac{1}{g} a_T r_e d\theta$$

$$= \int\limits_{\pi/2}^0 \frac{3GM_mr_e\sin{\theta}}{g r_{em}^3}(\cos{\theta} -\frac{1}{2}\frac{r_e}{r_{em}}) r_e d\theta$$

Now

$$ \int\limits_{\pi/2}^0 sin{\theta}\cos{\theta}d\theta = \frac{1}{2}$$

and

$$ \int\limits_{\pi/2}^0 sin{\theta}d\theta = 1 $$

finally, equating the two sides

$$ H = \frac{3GM_m{r_e}^2}{2gr_{em}^3}(1 -\frac{r_e}{r_{em}})$$

However

$$ g = \frac{GM_e}{r_e^2} $$

substituting this into our equation for H yields:

$$H = \frac{3M_m{r_e}^4}{2 M_e r_{em}^3}(1 -\frac{r_e}{r_{em}})$$

$$ G $$ has now canceled out. The tidal height is independent of the gravitational constant.

If we plug numbers into the above and calculate, we get.

$$ H = 52.9 cm$$

This is the change in height due to tidal forces of an earth covered with water.