Deriving the Gravitational Potential Energy
Let's file this post under "mistakes that Rhett made that he doesn't want to make again." If I don't write it down, I won't learn from my mistakes.
How about I start this with the following problem.
Suppose I have a rock that is very far away from a moon and I release it. How fast will the rock be moving when it is a distance r from the center of the moon?
Of course, I will include a picture - it's what I do.
So, where do you start with a problem like this? Since this is essentially a one dimensional problem, let me put the origin at the center of the moon and let the rock start somewhere on the x-axis. This means that I can write the x-component of the gravitational force as:
What next? I know the common idea is to say "hey, let's use the momentum principle". Nice try, but it won't be easy. Why? Because this gravitational force is not constant. Technically, you could set up something to find the momentum when it gets to the final location, but it won't be fun. Instead, when there is a problem defined in terms of displacement, you should always consider the work-energy principle.
The work-energy principle says:
For the rock, the force on it is the gravitational force and it changes in kinetic energy. So, let me calculate the work done by this force. I already have the x-component of this force, so the integral becomes:
Here is where I was making a mistake. I want to point out what I was doing wrong and why it was wrong. Before, to calculate the work, I wrote the gravitational force and the displacement as the following vectors:
Just to be clear, I am using the vector notation that my favorite calculus based textbook uses (Matter and Interactions) where the three numbers presented are the x,y,z components. And moving along, I can find the dot product of these two vectors as:
All of this is PERFECTLY OK. Note the this work during this short displacement is positive, as it should be. Right? And here is where I went wrong. I then wrote:
Which is CLEARLY WRONG. Sorry for "yelling" with capitol letters, but really I am yelling at myself for making this mistake. If you don't yell at Rhett, he just makes the same mistakes over and over. Why is this wrong? Well, if I integrate this, I am going to end up with a negative work. That isn't good, is it?
My error: Maybe you already see this. My mistake was that I tried to account for the direction of the integration twice. I let the small change in displacement be -dx AND I integrated from infinity to r. Doing this would be like having your cake AND EATING your cake. Clearly, you can't have your cake and eat it too. Once you eat it, it isn't cake, is it? So, that is what I did wrong.
Now back to the problem. I calculated the work on the rock by gravity. What does this work do? It changes the kinetic energy. Since the rock started from rest, I can write this as:
Now for a quick check. The closer the rock gets to the center of the moon, the smaller r would be and the faster the rock would be moving. Check. Also, what if the mass of the moon is larger. This would also produce a faster moving rock. Double Check.
Gravitational Potential Energy ——————————
Let me look at this problem again. The integral of the gravitational force over some path doesn't actually depend on the path. Try it for a simple two simple cases like this:
For the red path, the rock goes past the final point and back. Do these two integrals, you need to break it into two pieces, and you will see you get the same value as above since you end at a distance r from the center of the moon. For the green path, the rock takes a little curved detour and then back. During this circular path, the gravitational force is perpendicular to the displacement. This means the dot product (and thus the work along this path) is zero. Both the green and red paths give the same work because they start and end at the same place. WARNING: not all forces have works that do this. You have been warned.
Let me write the work-energy principle as:
So, instead of having the work done by gravity, I have this change in gravitational potential energy term. If I let the gravitational potential energy at an infinite distance be zero Joules, then:
Yes, the gravitational potential energy in this way would always be negative. Don't worry. Be Happy. Eveything's going to be ok. Why? Because who really cares about the potential energy anyway? All we really care about is the CHANGE in the gravitational potential energy. For this rock, falling towards the moon, the potential energy gets more and more negative (with smaller r) so the change in potential will be negative. This means that the change in kinetic energy will be positive.
Here is where you have to be careful. You can either have work done by the gravitational force, OR you can have a change in gravitational potential energy. You can't do both. That would be like having your cake and eating it too.
The System ———-
While I am talking about gravitational potential energy, let me emphasize the system. If you want to include a gravitational potential energy term, you need to have both the rock AND the moon in your system. Why? Well, what if you only had the rock? You would then do everything essentially the same as above and you would be happy. But what if the moon was another rock? In that case your value for the final speed would be wrong. It would be wrong because you failed to include the increased speed of the other rock. Both rocks move and speed up.
If you don't have both objects in your system, then you would be double counting the work done by gravity. There is a gravitational force on both objects, but really this is just one force. Recall that forces are an interaction between two objects. It isn't two different forces.
#Gravity
#physics
#potential energy