# Testing Copernicus

### Harys Dalvi

### October 2021

In school, we are taught that the earth goes around the sun. This is correct. I'm not trying to say heliocentric theory is wrong. (A physics major disagreeing with heliocentrism is not a good look at all.)

I'm not trying to say heliocentric theory is wrong, *but*
(this sounds bad so far) I am going to look at one way in
physics that geocentric theory could be correct too. (Could be,
but it's really not. I'm not saying it is.)

Now that we have firmly established that I believe the earth and all the planets go around the sun, not the other way around, I think I can safely start my critique of heliocentrism.

There will be aspects of this that have to do with the rotation of the earth, the Coriolis effect, and the theory of relativity. I will neglect these ideas, as they are topics in themselves, and aren't really necessary to look at heliocentrism compared to geocentrism.

## Frame of Reference

From the sun's perspective it is clear that the earth goes around the sun. If people could live on the sun, they would look up at the fiery sky and observe the planets rising and setting, just like we observe the sun rising and setting here on Earth.

But why is the earth's perspective any less valid? After all, physics
dictates that **there is no privileged frame of reference**
and **the laws of physics are the same in all inertial reference
frames**.

It turns out that while the perspective on the sun is an inertial reference frame, the perspective on Earth is not. This is the reason why heliocentrism is accepted while geocentrism is not.

For our purposes today, an **inertial reference frame** is one that
follows Newton's first law of motion, or one in which real forces are the only
way in which things can accelerate. If you don't see how a reference frame might
break Newton's first law, consider this thought experiment.

You and I are in space. First, my perspective: I believe that I am stationary,
and I see that you are stationary as well. Then, you use a jetpack to produce
a force on you that accelerates you at 1 m/s^{2}. Does this follow
Newton's first law of motion?

As for me, I am at rest, and I remain at rest because I am not acted upon by an outside force. As for you, you start at rest, but you are then acted upon by an outside force to accelerate you. This all follows Newton's first law of motion, meaning my frame of reference is inertial.

But now let's think about your perspective. You start at rest, and you see me
at rest. But then, in your perspective, *I* start to accelerate towards
*you* at 1 m/s^{2} when you start your jetpack. It's like
when you're in a car, it seems like you are sitting still while everything
else moves past. Does this follow Newton's first law of motion?

As for me, I accelerate from rest, but there is no force on me. This doesn't make sense according to Newton's first law. As for you, you remain at rest, even though the jetpack exerts a force on you. This doesn't make sense either. How do we deal with this?

Since this is a non-inertial reference frame, we can solve the problem by creating
a **fictitious force**. In this case, we will need a fictitious force to
accelerate both you and me by 1 m/s^{2} in the opposite direction of
the force on you due to the jetpack. Now everything makes sense: this fictitious force
cancels out with the force of the jetpack on you, so you remain at rest.
This fictitious force is the only force acting on me, so it causes me to
accelerate at 1 m/s^{2}.

Even though we now have a reference frame that follows the laws of physics, we needed to invent a fictitious force to get there. This fictitious force is fictitious, as the name implies. If we need a fictitious force, the frame is non-inertial.

Now, why do we need a fictitious force to produce a geocentric theory?

## Comparing Heliocentric and Geocentric Models

### Heliocentrism

I'll start with a heliocentric model. Let's make things very simple so we can focus on the differences between heliocentrism and geocentrism rather than on the details of orbital mechanics. We'll only consider the earth and the sun, and we'll assume the earth's orbit is circular.

Then the earth is at a constant distance \(R\) from the sun. The acceleration of the earth is given by \(GM/R^2\), where \(M\) is the mass of the sun and \(G\) is the gravitational constant. Since the mass of the sun is much greater than the mass of the earth, we can take the acceleration of the sun to be 0. Now we will ask the same question: is this an inertial frame of reference? Does this follow Newton's first law?

Looking at the earth, it feels a force \(GMm/R^2\) from the sun, where \(m\) is the mass of the earth. Based on this, the acceleration of the earth is \(GM/R^2\). We will say that this is just the acceleration we need in order to keep a stable orbit at this speed. (We can do this with basic algebra-based physics.)

## Proof: keeping a stable orbit

Let's say the earth orbits the sun at a speed \(v\), and continues this speed in its circular orbit. In order to do so, the earth needs the right amount of centripetal force. This centripetal force must be provided by gravity, the only force on the earth. Therefore, we can set the force due to gravity and the required centripetal force equal to each other. $$\frac{GMm}{R^2} = \frac{mv^2}{R}$$ $$v = \sqrt{\frac{GM}{R}}$$ So as long as the earth keeps going at \(v=\sqrt{GM/R}\), it can keep its circular orbit stable.Since \(GM/R^2\) is both the acceleration calculated for earth based on the forces it experiences, and the acceleration it must be observed to experience in order to keep a circular orbit, Newton's first law is followed: the earth is an object in motion that is being acted upon by an outside force and responding predictably to that force. This is analogous to the space example: the acceleration we would calculate for you because of your jetpack, is also the same acceleration I observed in my frame of reference. Therefore, this is an inertial frame of reference.

The acceleration of the sun is calculated as \(Gm/R^2\). We can assume this is very small, near 0,
since \(G\) is very small while \(R\) is very big. Since our frame of reference is centered on the sun,
this should be 0, so this makes sense. (In order for it to add up perfectly, we would need a frame of
reference centered on the *barycenter* of the sun and earth, but this is close enough to
the sun that we can say a heliocentric model works.)

### Geocentrism

Now let's shift to a geocentric model. Now, since we are centering our frame of reference on the earth, the acceleration of the earth should be 0. Now the sun is going around the earth at the same speed \(v=\sqrt{GM/R}\) that the earth went around the sun in the heliocentric model, and at the same distance \(R\). This means that the centripetal acceleration needed for the sun is equal to \(GM/R^2\), which is what was needed for the earth before.

Is this an inertial frame of reference? Let's look at the sun. Imagine you are an observer on earth with access to the position of the sun, its speed, its mass, its distance, the mass of earth, all these relevant variables. What would you see if you watch the sun?

You would see it rise and set as it orbits around the earth. Based on its observed speed \(v=\sqrt{GM/R}\) and distance \(R\), you can calculate its acceleration to be \(GM/R^2\). Does this acceleration make sense? The only force on the sun is gravity from the earth, with magnitude \(GMm/R^2\), meaning you would predict the acceleration of the sun to be \(Gm/R^2\). This is much less than the \(GM/R^2\) you observe.

As for the earth, it experiences the same force \(GMm/R^2\), predicting an acceleration \(GM/R^2\). But since you are on earth, it seems to you that the acceleration is 0. This is off too. Therefore, your frame of reference is not inertial.

We can resolve this by creating a fictitious force. Remember that we just found the predicted acceleration of the sun to be \(Gm/R^2\). Let's approximate this to be 0. (We are doing this because we will neglect any difference between the barycenter of the solar system and the center of the sun.) The acceleration for the sun that we observe in a geocentric model is \(GM/R^2\), so we will apply this acceleration to the sun in the direction towards the earth.

The earth feels a pull to the sun, but we find that our new fictitious force opposes this. An acceleration vector of \(GM/R^2\) from the sun towards the earth, if applied to the earth, becomes an acceleration vector of \(GM/R^2\) in the direction opposite the sun. It's easy to see how this will cancel the force of gravity on the earth due to the sun.

So by creating a fictitious force that produces an acceleration \(GM/R^2\) in the direction from the sun to the earth, we have now built a working geocentric model.

Of course, this fictitious force isn't real. That's why we say geocentrism isn't real. Since we need a fictitious force to account for geocentrism, it makes more sense to use a heliocentric model.

### Barycentrism?

I said that the acceleration of the sun due to the earth's gravity is \(Gm/R^2\), and I approximated this to be 0. What if I hadn't done this approximation?

Then we would find that a strictly heliocentric model actually needs a slight correction with a fictitious force, similar to what we found with a geocentric model. The acceleration of the sun should be 0 in a heliocentric model, but this doesn't match up with the fact that there should be a (slight) acceleration of the sun due to gravity. We need a fictitious force to fix this.

So if geocentrism needs a fictitious force, and heliocentrism needs a fictitious force,
which model is *really* correct? In order to really have no fictitious force,
we need to set a frame of reference centered at the center of mass of the solar system
(called the **barycenter**) rather than the center of the sun.
Since the sun's mass is such a huge piece of the solar system's mass, it turns out that
the actual barycenter of our solar system is *inside* the sun. But the mass of the other stuff is
there too, so the barycenter isn't exactly at the middle of the sun.

In some systems, such as binary star systems, the barycenter might not be inside any of the bodies. For example, if two stars of equal mass orbit each other, their barycenter is right between them, not inside either one. In this case, we would set up a coordinate system at this barycenter in order to avoid fictitious forces. This makes a lot of sense for a system of binary stars, and might help see why it applies even to our solar system, since it can be unintuitive at first to center a model around a point slightly off balance from the center of the sun. We might think at first that we have to pick a particular body to center our system around, like the earth or the sun or Jupiter, but the binary star example shows that the barycenter is really what we need.

## Conclusion

Based on the idea that all frames of reference are valid in physics, I tried making a geocentric model of the solar system. But then we found that I needed a fictitious force for the model to work, so this isn't a very good model. We found the heliocentric model to be better.

But taking a closer look at the heliocentric model, we realized it depends on approximation. If we remove this approximation, we end up with a barycentric model, centered at the center of mass of the solar system. Since the sun has such a big piece of the mass of our solar system, we usually approximate this to say that heliocentrism is the true model. But in reality, barycentrism works a bit better, although it is only slightly different from heliocentrism.

Using the idea of barycentrism, you can tell your friends that you don't believe in heliocentrism. Then, when they call you a conspirator and ask if you're a flat-earther too, you can explain why a barycentric model is technically superior to both heliocentrism and geocentrism for the same reason that heliocentrism is super to geocentrism: it cuts out fictitious forces. I don't know if this is actually the kind of stuff you tell your friends, but if it is, I hope you enjoy.