Gravitational Potential Energy for Systems of Masses - Online Tutor, Practice Problems & Exam Prep

In a system with multiple masses, the total gravitational potential energy is the sum of the potential energies between each pair of masses. For three masses, this includes energies \( u_{12} \), \( u_{13} \), and \( u_{23} \). The formula for gravitational potential energy is given by \( U = -\frac{G m_1 m_2}{r} \), where \( G \) is the universal gravitational constant. For an equilateral triangle configuration, distances are equal, simplifying calculations. By substituting known values, students can find the total energy, reinforcing concepts of scalar quantities and gravitational interactions.

Gravitational Potential Energy for Systems of Masses

Hey, guys. So in the last couple of videos, we saw that any 2 masses apart from each other have some gravitational potential energy. Well, in this video, you are going to see what happens when you have 3 or 4 masses in a system. Alright. Let's go ahead and check it out. So the idea is that if I take a look at this triangle that I got over here, I have 3 masses. Right? And I know that between masses 1 and 2, there's some distance in between them, so I could figure out what the gravitational potential energy is. And I'm gonna call that u₁₂ because it's between masses 1 and 2. But if you take a look, this isn't just the only pair of masses that I have in this diagram. Because I also have a pair between mass 1 and mass 3, and I have another pair between mass 2 and mass 3 as well. So the idea is that all of these pairs of masses actually have energies between them. So I have u₂₃ over here, and I've got u₁₃. So if I were to be asked on a problem, what's the total amount of gravitational potential energy of this whole system at masses, all I have to do is simply add the energies for each individual pair. And the reason I can do this is that energies, unlike forces which are vectors, are scalars. So we don't have to deal with any sines or cosines or vector decomposition or trig or anything like that. We can just simply add these numbers together for each of these pairs because scalars are just numbers. So in general, if I have a system of masses, all I have to do is just add the energies, u₁₂, u₁₃, u₂₃, and so on and so forth, depending on how many numbers of masses I have. Usually, you'll see just about 3 or 4 in a given problem. Alright. So let's just say I magically knew what all of these masses were and the distances between them. I'm gonna just make up some numbers here. Let's say I knew that the potential energy between these masses was negative 2 joules, and I have this was negative 3 joules and negative 4 joules. And if I were to be asked what the total gravitational potential energy is, the sum of all of the gravitational potential energies is just gonna be adding those negative numbers up. So I'm gonna have negative 2 plus negative 3, plus negative 4. And so, if you add all of these things up together, you're gonna get negative 9 joules as your total gravitational potential energy. Alright. So it's very very straightforward. Let's go ahead and just jump into a real example. So we're being asked to calculate what the gravitational potential energy is in this equilateral triangle system of masses that we have. So you've got these 10-kilogram and 20-kilogram masses that are all positioned in an equilateral triangle. Now, if we wanna figure out what the total gravitational potential energy is, so in other words, sigma u_g, that sigma just means the sum, all you have to do is just add all of these things up together. Now, what I like to do in these kinds of problems is I like to label my masses. So I'd like to label them just so I don't lose track. I've got this as m₁, m₂, and this is gonna be m₃. Okay? So I've got u₁₂, u₁₃, and then u₂₃. So I'm just gonna gonna go ahead and write in those individually out. So I've got the total gravitational potential energy is negative g, the product of mass 1, mass 2, and then I've got gm₁m₃, and now I've got negative gm₂m₃. And you've got to remember that these negative signs are here because you're adding together these potential energies which are negative. Okay. So there's always negative signs right there. Alright. Now, the only thing we have to do is just figure out what the distances between each of these things are. Now, we're told this is actually an equilateral triangle. Now, what that means is that all of these distances, 60 centimeters, are the same between all of these points right here, which is actually really nice because it ends up simplifying things for us. So that means that all of these distances right here, I don't have to go find them individually, they're all just equal to 60 centimeters, which is nice for me. So all I have to do is just plug in the masses and just the gs, and then just divide by 60 centimeters, and that's it. So that means the total gravitational potential energy is negative g. Now, I'm not gonna keep on plugging this \(6.67 \times 10^{-11}\). You guys can do that and just make sure that you get the right answer. So I'm gonna have negative g, and now I've got to just plug in the masses. I've got 10 and 10, and then divided by 60 centimeters which is 0.6. Now, minus g. Now, I've got 10 and 20 divided by 0.6, minus g 10 20 divided by 0.6. Okay. Now you can just go ahead and add all of these things up together. One thing you could do is you could just say that this 20 is exactly twice of what this number is and all these other numbers are the same. So you can kinda use shortcuts like that to make the math a little bit easier. Anyway, you should get the total gravitational potential is negative \(5.55 \times 10^{-8}\) in joules. Alright. And that's our final answer. Let me know you got if you guys have any questions with this, and let's go ahead and get some more practice.

What is the total gravitational potential energy of this system of masses?

• A. $-1.52\times 10^{-9}$ J
• B. $-3.49\times 10^{-8}$ J
• C. $-7.78\times 10^{-8}$  J
• D. $-1.75\times 10^{-7}$ J

Energy Conservation in Three-Mass System

Hey, guys. Let's work out this example together. So, in this first part, we're supposed to figure out what the total gravitational potential energy of the system is. To do that, I need to have all of my masses labeled, and I need all the distances. So, fortunately, I have that. The total potential energy is going to be the sum of the potential energies between all of these masses right here. So, I've got U_12+U_23+U_13. I can go ahead and start writing out all of those expressions. U_12 is going to be -GM_1M_2∕r_12. I've got GM_2M_3∕r_23-GM_1M_3∕r_13. Let's just make sure that I have all these variables before I start plugging stuff in. I might need to go off and figure something out. I've got the masses of all of these things in these diagrams, and I also have the distances between them, 50, 50, and 60 centimeters. So, that means I have everything I need to start plugging into these formulas. Let's go ahead and do that. The total gravitational potential energy is going to be -G I'll keep this as a letter since the number is right here; I don't want to have to write it out over and over again. Then I've got 25 and 10 divided by the distance between them, 0.5 because we are 50 centimeters. I've got -G1025∕0.5. Notice that these numbers are the same. 25×10 is the same thing as 10×25. Both of these things, basically, you just need to plug them into your calculator once and then multiply by 2. We've got -G and the product of 25x25 divided by the distance between M_1 and M_3, which is 60 centimeters, not 50. So, we've got 0.6. If you go ahead and start working this stuff out, again, just plug this in once and multiply it by 2. You're going to get this as -6.68×10-8, and then this guy over here is -6.95×10-8. We have to subtract him because here, everything is subtracted. The total gravitational potential energy is equal to -1.36×10-7, and that's in joules. Great. So that's the first part.

In part B, M_2 is going to be released from rest. As it's released from rest, it's going to be pulled directly towards the center because of symmetry. The mass is exactly the same, and the distances are symmetrical. So, it's going to be pulled in this direction, and we're supposed to figure out what the velocity is going to be when it reaches this point and passes through the center. So that's our target variable, V_final. In this part, first, we're going to start from our V_final. If we're talking about velocities and we're talking about gravitation, we have to use energy conservation. So, let's write out the equation. We've got kinetic initial plus potential initial plus any work done by non-conservative forces equals final kinetic and final potential. One thing we have to remember though, is that because we're working with multiple masses, we have to use energy conservation for the entire system, all three masses. We have to see if anything has any kinetic energy or potential energy. We must look at the entire system as a whole. We're told that all of these objects are fixed or are released from rest. So that means nothing is moving at the beginning, which means all the kinetic energy is equal to 0. There's nothing moving. There is some gravitational potential, and that's what we figured out in part A. Let me go ahead and scoot this down a bit actually. Just so that it doesn't get, and let's see. Once we release these objects, the only force that acts on them is gravity, so there's no work done by any non-conservative forces. Now, the final kinetic energy is going to come from the fact that anything is moving in the system after its final state. We know that mass 2, when it gets towards the middle, is going to be moving off in this direction. So, there is going to be some kinetic energy, but these two objects are fixed, which means they don't. The only thing that contributes some kinetic energy is going to be 12M_2V_final2, and that's where our target variable is. So that's our V_final. And then we have the final potential energies, in which there's going to be some distances involved. There's always potential energy. Great. So, let's go ahead and start writing out what those formulas are. The initial gravitational potential is going to be just the U12 initial, U23 initial, and U13 initial. Then the final gravitational potential energy is going to be U12 final, U23 final, and U13 final. Now, the reason I had finals and initials is that this object moves, this M2 moves towards the center. As that happens, these distances, the 50 centimeters, will change. They'll go from 50, and finally over here, it'll be 30, 30 over here. So, because the distances are changing, our gravitational potential energies are going to be changing. So, what we can do is we can write out all of these expressions right here. We've got again, this is just going to be, I'm just going to go ahead and copy this guy right over here, and then move that down over here. We know that this is the expression for all of the gravitational potentials squared, then -G², and then -GM_1M_2∕r_12final, -GM_2M_3∕r_23final, -GM_1M_3∕r__13final. I know this looks like a huge expression, but we can actually use some of this. We can look at this diagram and see if we can eliminate some of these terms from this equation. We know what these distances are, all of the distances in the initial state. We have a number for these guys. These are going to simplify down to -6.68×10-8. Now let's look at these two terms, the mass 13 and mass 13 final. The mass between 13 and the distances between them don't change as M2 gets closer. We can see that because here, they're separated by 60 centimeters. And then when this thing gets finally close to the center, these things are fixed in place. So they still remain at 60 centimeters. In other words, these distances are the same. So both of those numbers are going to be the same. So we can basically cancel those out and keep going with our equation. We've got 12M_2V_final2, and this is going to be -G. Now we've got M_1M_2�D70.3, and it's going to be the exact same thingG1025∕0.3. So if you work this out, these 2 together, again, multiply just, you know, calculate 1 and then multiply by 2. We're going to get -6.68�D7;10-8 equals 12M_2, and that's going to be -1.12�D7;10-7. So we're going to move this guy over and add it. And this term on the left becomes 04.52×10-8, and it's a positive number equals 1210V_final2. So now, all we have to do is basically multiply by 1 half and divide by 10, which is like dividing by 5. When you move all this stuff over, this is just going to be 9.04×10-9, and that's going to be equal to V_final2. So the last thing you have to do is just take the square root, and we get that V_final=9.51×10-5m/s. That's equal to V_final. Let me know if you have any questions. It's very important that when you do energy conservation, you have to apply it to the entire system and then see what you can cancel out. Anyway, let me know if you have any questions.