Some Definitions

The position vector, velocity vector, and acceleration vector are show below.

The r, r dot, and r double dot, is just another way of expressing x, v, and a.

Another notation used is to denote the different components of a vector.

The i with the carrot on top denotes the x direction, where the j and k denote the y and z direction respectively. For example. If a vector is pointing 3 meters in the x direction, 4 meters in the y direction, and 0 meters in the z direction you would write,

We will just assume that all r's are vectors, so we won't include the vector arrow.

Another vector notation, which is pertinent to our proof includes the following:

The e sub r refers to the direction that radiates from the center of the Earth, the e sub R denotes the direction that radiates from the center of the moon to the small point mass m on the Earth's surface, and the e sub D refers to the direction emanating from the center of the moon to the center of the Earth.

Force on m due to Earth and moon

Let's break this equation down. The left side is the ma part. You have the point mass m, and the accleration of that point mass relative to the inertial frame r prime. The right side shows the gravitational force, using Newton's Universal Gravitational equation, from the Earth as denoted in the e sub r component, and from the moon as denoted in the e sub R component. Notice how in the e sub r direction, the distance in the gravitational equation is r, and in the e sub R direction, the distance is R. That is because the point mass is a distance r from the center of the Earth, and a distance R from the center of the moon.

They are both negative because that is the usual convention used with gravity. The point is, equation one is simply an F = ma equation.

Equation two is similar to equation one in that it is also an F = ma equation. Since we are considering the force of the moon on the Earth, we have the mass of the Earth times its acceleration relative to the inertial frame on the left side, and the gravitational force equation on the right side. Notice again how the distance in the gravitational force is D, which is the distance between the moon's center and the Earth's center.

Move to the Earth centered frame of reference

Now that we have established the two gravitational equations, we need to move from the inertial frame to the noninertial frame, namely the Earth centered frame of reference. To do this, we simply employ vector addition properties. Consider the picture above and notice how,

The vector r sub e prime plus r equals r sub m prime. This is illustrated in the picture below.

The three vectors in question are highlighted in yellow, and you can see how r sub e prime and r do add up to r sub m prime. Using simple algebra, we can solve equation three for r.

Now we can use a tool of calculus called the derivative. For those not familiar with the derivative, it is used to go from position to velocity, and velocity to acceleration. In other words, the derivative of position equals velocity, and the derivative of velocity equals acceleration. So, if we take the second derivative of equation four, meaning two derivatives, we go from position vectors to acceleration vectors.

Now we turn our attention back to equations one and two. In equation one, we can divide out the point mass m, since m is in all three terms.

That leaves us with,

Then we do the same thing to equation two and divide out the Earth's mass, M sub E,

which leaves us with,

Notice how equation six is solved for r sub m prime, and equation seven is solved for r sub E prime. We now take our results from equation six and seven, and plug them in place of the r sub m prime and r sub E prime in equation five.

Notice in the last two terms, there is the mass of the moon, M sub m. We can factor that with G, and get,

As equation nine explains, the far left term is the acceleration due to the Earth's gravity. The other two terms in brackets is the acceleration due to the moon. As denoted by e sub R and e sub D, this acceleration is a difference between the force of the moon on a point mass on the Earth's surface, the e sub R direction, and the force of the moon on the center of the Earth, the e sub D direction. From here on out, we will only focus on this difference term, which is equation ten.

F sub t is the tidal force. Notice the extra m added to the equation. That was necessary because equation nine only represents acceleration. Since F = ma, adding the point mass back into the equation converts it to a force equation.

Tidal Force: X - direction

Looking at equation ten, you can see when the point mass is located on the side of the Earth closest to the moon, R is less than D, which makes the tidal force negative. When the point mass is located on the side of the Earth opposite the moon, R is greater than D, which makes the tidal force positive. A negative graviational force means towards the object, in this case the moon. Therefore, the picture below illustrates the directional signs along the x - axis.

Since we are considering only the tidal force along the x-axis, we can drop the e sub R and e sub D directional notations, and have,

Considering the main picture, we can substitute D + r for R, since again we are only considering the x direction. This may seem incorrect, since if the point mass is located on the side closest to the moon, R would actually equal D - r. However, since the radius of the Earth, r, is so much smaller than D, it is really insignifant. In fact, D is approximately 60 times larger than r, so adding r or subtracting r does not change the value significantly.

The reason for this substitution is to ultimately simplify the equation. In doing so, we factor a D squared term out of the (D + r) quantity leaving us with,

Now we can factor the D squared term completely out leaving,

Now we can incorporate what is called a binomial expansion. the 1/(1 + r/D)^2 term can be expanded into an infinite series given as:

So, if we plug equation 15 in place of the (1 + r/D) term in equation 14, we have,

Here is where the simplifying comes in. Notice the first number one cancels with the last negative number one. Also, we can actually neglect all of the terms after 2r/D. This is because D is so much larger than r, when you start to square and cube those values, like you do in the infinite series after 2r/D, those values become so close to being zero, they are insignificant. Therefore, we can have,

Which reduces to,

Equation 18 is the tidal force due to the moon in the x direction! We will come back to it after we get the y-force to see what it means.

Tidal Force: Y - direction

The Y-direction is somewhat easier to obtain. Consider the picture below.

Going back to equation ten,

We need to look at the e sub R and e sub D directional components, and find their y-components. Looking at the picture above, you can see that since the distance between the Earth and moon is so large, R approximately equals D. Therefore, we can write equation ten as,

Now it looks like the two terms in the brackets cancel, but we first have to see how they break into their respective x and y components. Since the e sub D direction is totally in the x-direction, we can write e sub D as,

The e sub R direction, as shown in the picture, has both an x direction, since it extends to the Earth, and a y direction, which is the part that extends from the center of the Earth to either of its poles. Using the trig functions cosine and sine to obtain the x and y components, we get,

Again, due to the immense distance between the Earth and the moon, the angle between R and D, as shown in the picture, is very small. Therefore, we can make some more approximations. The cosine of a very small angle is nearly equal to one, and the sine is nearly equal to the angle itself. Furthermore, the angle theta is approximately equal to r/D. Making those substitutions gives us,

Distributing the 1/D squared term yields,

Now we plug the results from equation 23 and 20 into equation 19.

Notice how the i directional components are the same and opposite in sign, which means they cancel, which leaves us with just the j direction, which remember is the y direction. Thus, the equation reduces to,

Analysing the Results

We now have equations for the gravitational force of the moon on the Earth for the x and y direction.

Before we see what these results mean, it is necessary to do one more substitution. Instead of using just r in both equations, we need to designate where the point mass is located in terms of the angle, which depends on where one is located on the Earth's surface. Therefore, we use cosine for the x directional force, and sine for the y directional force.

Now let's see how these forces work at different points on the Earth's surface.

When the angle is zero degrees, the point mass is located on the side of the Earth directly opposite the moon. The Fy force is zero because the sine of zero is zero, and the Fx force is a positive force because the cosine of zero is positive one. This means, the force is actually directed away from the moon!

When the angle is 90 degrees, the point mass is located at the north pole. The Fx force is now zero because the cos of 90 is zero, and the Fy force is a negative force, because the sin of 90 is a positive one, but the Fy force starts out as negative.

When the angle is 180 degrees, the point mass is located on the side facing the moon. The Fy force is now zero again since the sin of 180 is zero. The Fx force is now negative because the cos of 180 degrees is a negative one.

Finall, the when the angle is 270 degrees, the point mass is located at the south pole. The Fx equation is zero again since the cos of 270 is zero, and the Fy equation is a positive because the sin of 270 is negative, which cancels the original negative in the equation, leaving it a positive value.

As a result of these equations, the tidal forces acting on the Earth are illustrated below.

The cool thing is, the Earth actually undergoes a stretching and squeezing from the Moon's gravity. It is stretched along the x-direction, and squeezed along the y-direction. This is what causes the tides twice a day.

The reason for this stretching and squeezing effect has to do with the difference in strength of the Moon's gravitational force at the four places we analyzed on the Earth above. There is a big difference in strength between the point closest to the moon, and the point on the Earth opposite the moon. In other words, the moon is pulling the Earth more at the point closest to it, and pulling less at the point farthest from it. The reason is because the Earth is so large, that difference in distance to the moon is very noticeable. Compare this to the tidal force on your body from the Earth. Because we are so small compared to the Earth, the difference in gravitational force or pull from our feet to our head is so miniscule, it is virtually the same. Therefore, we don't feel a stretching and squeezing effect.

The following diagram attempts to illustrate this point.

Since gravitational force has an inverse square law for distance from the object, the graph of the gravitational force curves as shown above. Since the Earth is so large, if you place the Earth on the graph corresponding to how far it is from the moon, you notice how the curve of the force graph is much higher at the point on the Earth closest to the moon than it is on the point farthest from the moon. (Graph not to scale of course.)

The Sun's Tidal Effect

Does the sun produce tidal forces on the Earth? Is it larger than the moon's tidal force? The answer to both questions is a yes and no. All objects will produce a tidal force from its gravity, but again there has to be a considerable difference in gravity between two points for the effect to be felt. As a result, the sun's tidal effect is a lot smaller than the moon's because the Earth is so much farther from the sun.

Notice how the Earth in the above diagram is lying along the curve of the Sun's force that is more shallow. As a result, the difference in gravitational force between the two points on the Earth closest and farthest away from the sun virtually feel the same force. Therefore, the tidal effect is very small.