Students in physics courses have (or should have) drummed into their heads the idea that one evaluates forces only in inertial frames of reference. An inertial frame of reference is one that is moving with a constant velocity - that is to say, it's not accelerating. What's an example of a non-inertial frame? Look at a car going around a curve. Relative to the car, you feel like you're being pressed up against the outside part of the curve. Circular motion is a kind of acceleration, since the velocity vector is changing. If you put yourself in the frame of reference of the object undergoing circular motion, strange things happen - fictitious forces arise. In the example of the car going around the corner, you aren't really experiencing a force, what is happening is your body's inertia is trying to keep you moving in a straight line with a constant velocity, but the car underneath you is accelerating as it rounds the curve, which presses against you to make you change direction along with the car. This 'centrifugal force' is an artifact of you evaluating the motion in an accelerating frame of reference.
Physicists, by-and-large, stick to evaluating forces only in inertial frames of reference. This avoids the ambiguity of fictitious forces arising.
In describing the static theory of tides, more often than not, one hears of a description involving the tidal bulge on the earth's side facing the moon as due to the higher strength of the moon's gravity, and the away side bulge being due to a 'centrifugal force'. I *have* seen on many websites variants of these terms, like 'centripetal force' or 'centripetal acceleration'. I think that, in part, people can imagine that the gravitational pull of the moon is stronger on the side of the earth facing the moon, giving rise to a near side bulge. But, they have difficulties understanding that the diverging field lines give a symmetric away-side bulge, and hence the idea that a 'centrifugal force' is necessary to describe the that bulge.
In point of fact, one can make a derivation of the tides in a rotating frame of reference where fictitious forces appear. For a derivation using an accelerating (rotation) frame of reference see, for example, Introduction to Classical Mechanics by David Morin (Cambridge International Student Edition, 2007) p 471-477, but look at note 3 at the end of the derivation.
Derivations in the rotating frame and inertial (static) frame should yield the same answers, and they do, in fact. For me, however, I prefer to evaluate the forces in an inertial frame, since I feel it gives more insight into the origin of the tides. In either case, one can talk about the "meaning" of the tides in making the transformation from one frame to the other.
So, how to describe tides in an inertial frame? Actually this is quite straight forward and can serve as an instructive exercise for a student even in introductory physics. The following is a simple derivation, followed by a piece of mathematica code that analyzes the shape and strength of the tidal forces.
First, let's set up a coordinate system. We'll take the origin as the center of the earth. The $x$ axis points from the center of the earth to the center of the moon. The $y$ axis is perpendicular. The angle $\theta$ takes its zero along the $x$ axis and I'll take the positive sense of rotation of $\theta$ to be counterclockwise (with apologies to those who have gotten used to my clock-wise representation of azimuth from the northing axis).
Figure 1 Coordinate system for evaluating tidal forces.
In this derivation, I'll take the positions of the moon and earth to be fixed. Although they are both orbiting a common center, called a barycenter, it's not necessary to worry about the orbital motion to find the tidal forces. Just imagine that this snapshot of the earth-moon system is fixed and for our purposes it's static.
Now, gravitational field lines diverge from the object, and this creates an inverse square law force:
$\vec{F_G} = G {\Large {M m \hat{r} \over r^2}}$
This is illustrated in the figure below. The symbol $\hat{r}$ is a directional vector in a line linking the two masses, $M$ and $m$, meaning that the gravitational force is directed along that line. The inverse square law means that the force drops off like one over the square of the distance between the two objects. In this case, we're just considering point-like objects.
Gravitational force between two point-like objects.
In reality, objects are not point-like in nature, but are extended in space. This brings up two important theorems, which I will not prove, but only state the results. You can dig these up in typical freshman texts on physics.
1.) In dealing with an extended object, one adds up all the forces acting at each point on the object. The net acceleration of the object is then considered as acting at the center-of-mass of the object.
2.) In gravity, with the inverse square law, the sum of all the forces on a spherical mass from another mass can be taken as if all the mass were concentrated at the center-of-mass of the sphere. This is called the "shell theorem".
These two theorems greatly simplify the calculation. To get the motion of an extended spherical mass in a gravitational field that goes like $1/r^2$, one takes the gravitational attraction as if all the mass were concentrated at the center of the sphere. This is the same thing as saying that the sphere would seem to be immersed in a uniform gravitational field.
All the above means that any residual forces acting across the surface of the sphere must add up to zero. This is illustrated in the figure below, which is the taking-off point for my calculation of the static tides. The diverging field lines associated with the moon can be factorized into a set of parallel lines that represent the average force - the force acting acting as if the earth were a point-like object located at the center of the earth, PLUS a set of lines of field that sum to zero. If one adds the parallel lines plus the residual field, one ends up with the diverging field lines in the top of the figure.
Diverging field lines represented as the sum of uniform field lines plus tidal lines that average out to zero.
The above plan is the setup, so now we have to get down to the calculation. Call $r_o$ the radius of the earth, and $R_o$ the distance from the center of the moon to the center of the earth. From the geometry, shown below, and the law of cosines, the distance from the center of the moon to a point on the surface of the earth is
$R_s=R_s (\theta) = \sqrt{r_o^2+R_o^2-2r_o R_o \cos \theta}$.
Geometry of the problem.
Next, we find the construction of the forces. Since we're interested in the effect of the gravitational field of the moon on a small piece of the earth on the surface, we factor out $M$, the mass of the earth, in the force law and just concentrate on the gravitational field at the surface. This will have units of Newtons/kg. To get the force on a mass on the surface of the earth, we only need to multiply the field by the mass. Hence, we're evaluating the field, $\vec{F}$
$\vec{F}=\Large{{Gm \hat{r}} \over{ R_s^2}}$
Where $G= 6.42 \times10^{-11}$ $m^3 \space kg^{-1} \space sec^{-2}$ in MKS units, $\hat{r}$ just denotes the direction of the field, and $R_s$ is the distance from a point on the surface of the earth to the center of the moon. The figure below shows what we need to evaluate. $\vec{F}_{avg}$ is the average field directed purely along the $x$ axis, by construction. $\vec{F}_{surf}$ is the field at a specific point on the earth's surface and is a function of $\theta$.
The average field and the field at the surface of the earth.
We seek the tidal field. This is simply the difference between the field at the surface and the average field.
$\vec{F} = \vec{F}_{surf}-\vec{F}_{avg}$
A lot of times people like to solve problems using vector notation throughout and only take components at the last minute, the thought being that the physics is independent of the coordinate system, which is true. However, I want to move as quickly as possible to some expressions that I can put into Mathematica to generate plots, so I'm less concerned with elegance and more concerned with writing code. We can boil down the expressions to something more elegant and take approximations later on.
So, the average field is purely along the $x$ axis:
$F_{x,avg}=\Large{Gm \over R_o^2}$
$F_{y,avg} =0$
Where the subscript denotes the component. Getting the components of $\vec{F}_{surf}$ is a bit tricky, but you can see how it works out in the figure below. Draw a line parallel to the $x$ axis at the surface point you want to evaluate. Denote the angle between that line and the line from the surface point to the center of the moon as $\beta$. The actual field at the surface is directed from that point to the center of the moon. You can see in the construction that there's a right triangle that is formed with a base of $R_o - r_o \cos \theta$, a height of $r_o \sin \theta$, and the hypotenuse of $R_s(\theta)$ as defined above.
Finding the angle $\beta$ to construct the components of the field on the surface of the earth.
From the above figure, we can find $\cos \beta$ and $\sin \beta $:
$\cos \beta = \Large{R_o-r_o \cos \theta \over R_s(\theta)}$
and
$\sin \beta = \Large{r_o \sin \theta \over R_s(\theta)}$
This allows us to find the $x$ and $y$ components of the field at the surface as:
$F_{x,surf} = Gm \Large { \{ {R_o - r_o \cos \theta \over R_s^3 ( \theta)} \} }$
and
$F_{y,surf} = Gm \Large{ \{ {-r_o \sin \theta \over R_s^3(\theta)} \}} $
The minus sign for the $y$ component is because it is pointing in the negative $y$ direction. Since the common factor $Gm$ appears in all the field expressions, I'm going to call this $\kappa$ to simplify things. Now we go back to the construction and subtract off the average field. The $x$ component of the residual (tidal) field is:
$F_x = \kappa \Large { \{ {R_o-r_o \cos \theta \over R_s^3(\theta)} - {1 \over R_o^2} \} }$
Since the average field is pointing along the $x$ axis, the $y$ component of the field at the surface is the residual, tidal field:
$F_y = \kappa \Large{ \{ {-r_o \sin \theta \over R_s^3(\theta)} \}} $
So, that's about it. We have our answer. Note that the tidal field goes like $1/R_s^3$, which is typical for tidal fields. We can now just punch these equations into Mathematica. Here's the Mathematica code that I wrote to evaluate the components of the fields and makes a nice vector field plot, so you can see that we do indeed get the tidal forces as advertised above. Note that I use the same notation in the code snippet as I do above, so you can keep track of things.
Begin code snippet here
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(*Calculates tidal forces in static limit *)
(*Data input, distances are in meters, Forces in Newtons, gravitational field in Newtons/kg *)
Ro = 3.85*10^8(*earth-moon distance meters*)
ro = 6.378*10^6(*earth radius meters*)
m = 6.3*10^22(*moon mass kg*)
M = 5.97*10^24 (*earth mass kg, not that we need it *)
G = 6.42*10^-11(*Newton's constant*)
kappa = G m (* Constant appearing in front of expressions. *)
(* Theta is only variable in problem, is angle from earth-moon axis, starting at zero at closest approach *)
(* Find distance at surface of evaluation*)
Rs = Sqrt[ro^2 + Ro^2 - 2*Ro*ro*Cos[theta]];
(* Find x-component of residual grav. field, note 1/R^3 dependence *)
Fx = kappa*(((Ro - ro*Cos[theta])/Rs^3) - (1/Ro^2));
(* Find y-component of residual grav field *)
Fy = kappa*((-1.*ro*Sin[theta])/Rs^3);
(* Find x and y values as a function of Rs and theta*)
xds = ro*Cos[theta];
yds = ro*Sin[theta];
(* Plot x values, y values on surface of earth *)
Plot[xds, {theta, 0, 2 Pi}]
Plot[yds, {theta, 0, 2 Pi}]
(* Plot force components *)
Plot[Fx, {theta, 0, 2 Pi}]
Plot[Fy, {theta, 0, 2 Pi}]
(* Create table for vector plot *)
tide = Table[{{xds, yds}, {Fx, Fy}}, {theta, 0, 2 Pi, 0.1}];
(* Make vector plot*)
ListVectorPlot[tide, DataRange -> {{-10^7, 10^7}, {-10^7, 10^7}}]
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End code snippet
Below is the output of the vector field plot from Mathematica, where I'd drawn in by hand the approximate size of the earth as the blue disk.
Output of Mathematica code snippet with earth drawn in by hand.
Here is a link to the next post on the origin of the tides.
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