Gravitational Potential Energy - Online Tutor, Practice Problems & Exam Prep

Gravitational Potential Energy

Hey, guys. So now we've seen how masses exert gravitational forces on each other, we're going to see how to solve problems involving gravitational potential energy. So, we're going to involve questions involving changing energies and velocities, but we're going to see that we need a new equation for that potential energy to solve these problems. Okay? So that's the idea. So let's think back to a second when we were first introduced to energy. We had one object that was above the ground. All we needed was just m, g, and h. We just ignored that the Earth had mass. But now when we're dealing with gravitation, we're always going to have problems that involve multiple masses and distances. So now the new equation that we're going to use from now on for gravitational potential energy is U=−gmMmr. Okay? So, we have to be very careful when we use this equation because there are two things: 1, there's a negative sign here, it kind of looks like Newton's law of gravity with this negative sign, and there's also only one power of r in the denominator. Now, just remember that little r represents the center of mass distance, big R + H. So if you're talking about a planet, you have to go all the way from center to center. Okay? Now, in fact, what actually happens is that this equation becomes this equation here on the left, kind of like the simplified version for very small changes in height. So, we could use this equation here because we always assume that the heights were very, very small. So for very small Δh, this equation actually becomes this one. Alright. So, we used U=mgh as basically an alternative for when we solved kinematics problems. So let's just say, for example, that I had a question involving dropping this space rock from some height, and I wanted to figure out what the velocity was when it hit the ground, v final. Now I had two approaches. I could use kinematics, right, so I could figure out what g is. I just needed to know the Δy or the height. I have the v₀, and I could use kinematics to solve these things. The reason I could use kinematics is because we just assumed that this g was 9.8, and we just assumed that it was constant. So we could use our kinematics equation, things like that, because the acceleration is constant. We also could have used the conservation of energy. Well, the difference is that in gravitation, the main difference is that we cannot use kinematics to solve these problems. The reason for that is that as these two these two masses, as they get farther and farther apart as there are changes, the gravitational forces and the accelerations between them will also change. So the reason you cannot use kinematics is because as r changes, g is not constant, it's changing. But the good thing is that we can still use gravitational potential energy or sorry, we can still use energy conservation to solve these problems. The only thing that changes is that instead of just using this mgh for UiandUf, we're just going to use this new equation, U=−gmMmr. That's the whole thing. So, we're just going to replace that thing over there. Okay? So let's go ahead and take a look at an example. We've got an asteroid that is falling to Earth from some distance above the center, and we're asked what's going to be the impact speed. So in other words, we've got this space rock over here. We've got the Earth. Alright. So, we've got these two things. Now I've got the initial distance that's going to be over here. Now what happens is that the asteroid falls towards the Earth and it's going to hit right here at the center. So what I'm supposed to find out is I'm supposed to find out what is the v final. So I know that when it hits the surface, there's still going to be some r over here. So let's go ahead and set up my energy conservation because I have an initial and I have a final right here. I've got initial velocity and a final velocity. I definitely need to use, energy conservation. I can't use kinematics because the g changes across this distance. So I've got initial, actually, I can just go ahead and I got it right here. Boom. So I've got my energy conservation. First things first, is anything moving in the initial and is there any kinetic energy? Well, I've got the space rock is dropped, so it's at rest, and then the Earth, we can just assume is not moving. So there's no kinetic energy over there. Now, there is some potential energy because we have 2 masses. Right? We've got the mass of the asteroid, the mass of the Earth separated by some distance. So I've got −gMm1rinitial. So I had to use that new gravitational potential energy. Now is there any work done by nonconservative forces? Well, we're always talking about gravity, and gravity is a conservative force, so there's no work done by nonconservative forces. Now, finally, I have the kinetic energy final. Well, that just is going to be one half of the Earth velocity squared plus 1 half of the asteroid velocity squared. Now we can just assume that the Earth is so massive that it's not actually moving. So in other words, this, term is going to cancel out. It won't always cancel out in some cases, but in this case it does. And this is actually where our target variable comes from. This is where we're going to get that velocity. And, now, finally, we have the potential energy. Let's take a look here. Is there some potential energy when it hits the surface? Well, actually, there's still some center of mass distance in between the asteroid and the Earth. So that means that there is actually still some gravitational potential energy even at the surface. Remember that this r final is going to be big R+H. But when you're at the surface, what happens is your h just turns to 0, but your little r just becomes big R. So that means that there's still some gravitational potential energy at the surface, it's just this the little r turns into big R, so that's what we're going to put in over here. Alright. That's basically it. So now we just have to go ahead and rearrange. I'm just going to move this term all the way over to the left and it's going to become positive because of the negative sign, And then I'm going to have gmM1R−gmM1rinitial=m2vfinal2. So what I'm going to do is I'm going to move this one half over to the other side and I'm also going to pull out this GMM as a greatest common factor, see how it appears on both sides here. So I've got GMM, then I've got one over r Oh, sorry. There's a 2 out here. One over r minus one over r initial equals vfinal2, and actually, I missed an m. So there's an m v final squared. So this is my target variable here. Let's just go ahead and plug in everything, but there's one last thing. If you look through, I've got g, I've got the mass, and I've got the r's. And because all of these things are constants over here, the only thing I'm missing is the mass of the asteroid. But just like what happens in normal energy calculations, this, this m will actually cancel out because it appears on both sides of the equation. There we go. Okay. So now all we have to do is just take the square roots. So we've got the final velocity is going to be the square roots and I'm going to have 26.67 x 10^-11. Mass of the Earth is 5.97 x 10^24. And now, I've got, the distances. So I've got 1 over 6.37 x 10^6 minus 1 over 6 x 10^7. So, you have to be very careful when you plug that all into your calculators. You should get a final velocity, 1.06 x 10^4 meters per second, and that's the answer. Alright. So, that's how you solve these gravitational potential energy problems. Let me know if you guys have any questions.

How much energy is required to move a 1000-kg object from Earth's surface to a height twice Earth's radius?

• A. $3.13\times 10^{10}$ J
• B. $4.17\times 10^{10}$ J
• C. $7.83\times 10^{23}$ J
• D. $8.46\times 10^{23}$ J

Collision of Two Small Planets

Hey, guys. So let's check out this problem. We have 2 identical masses, 2 identical planets. They're initially at rest, but they're eventually going to drift together due to gravity and collide. And we're supposed to be finding what the speed of each of these planets are. So because I'm talking about speed and gravitation, I know I'm going to need to use some energy. I can't use kinematics because the forces are constantly changing; gravity constantly changes. So if I'm using energy considerations, I need an initial and a final position. So let's go ahead and draw a diagram. I've got these two masses that are some distance apart, and I'm told what the masses of each of those things are. So m equals \( 7 \times 10^{22} \). And I'm told that their center of mass distance here is equal to, which is our initial, is equal to \( 5 \times 10^{10} \), and that's in meters. So eventually, these things are going to crash into each other. They initially start from rest, so we know the initial velocity is equal to 0. But eventually, when these two things collide with each other, their surfaces are going to touch like that. And we know that the distance between them is going to be this final distance, \( r_f \). And they're both going to have some final velocity, \( v_f \). And that's actually what we're looking for. What is the speed of each of these planets as they collide? So that is our initial kinetic and potential, plus any work that's done is equal to the final kinetic and potential.

So let's see. These things are initially starting from rest, so we know that the initial velocities of both of them are 0. What that means is that the kinetic energy initial is going to be 0. We also know that we're talking about gravity. Gravity is pulling these things together, and it's a conservative force. So that means that there is no non-conservative force acting on this. So let's go ahead and write out each of these equations now. We know that the initial gravitational potential energy is \( g \times m^2 \) divided by the initial radius \( r_i \). What about the kinetic energy final? Well, normally, if we were talking about one object, we would just do \( \frac{1}{2} m v^2 \). But we're actually talking about this energy conservation law for the entire system. Now remember that this whole entire equation for energy conservation applies to all objects that are in the system. So it's not just \( \frac{1}{2} m v^2 \), we also have \( \frac{1}{2} m v^2 \) for the other object. But because the masses are the same for both of them, I don't have to write \( m_1, m_2 \). It's just \( m \). And then I've got the gravitational potential. So \( g \times m^2 \) again divided by the final distance.

So, really, what my target variable is in this equation are these \( v_{\text{final}} \). So I can go ahead and clean up this equation a little bit. So I'm going to write this as \( g \times m \) squared because both of those masses are the same. So I've got \( gm^2 \) divided by our initial equals \( \frac{1}{2} m v^2 + \frac{1}{2} mv^2 \). So that's going to be \( m_{\text{vfinal}}^2 \). And then I've got minus \( g \times m^2 \) divided by \( r_{\text{final}} \). But when it reaches this spot here and these two things are colliding, what is that final distance between them, the centers of masses? Well, think about it. If this is just the radius of 1 and the radius of another, their surfaces have to be touching. So that means that this final distance here is actually 2 times the radius. That's the point at which they finally collide. So 2 times big \( R \), which we actually know.

So now basically I've got to isolate this \( v_{\text{final}}^2 \) by moving this term over to the other side. Also, what I can do is I can actually cancel one \( m \) factor that appears in each one of these things. So now that I've finally cleaned this up, I've got \( g \times m \) divided by, well, this thing is going to be when it goes over, so it's going to be \( 2r \), and then minus this term over here. So it's going to be \( gm \), whoops, so I've got \( gm \) over \( r_{\text{initial}} = v_{\text{final}}^2 \). So now finally all I can do is pull out this \( gm \) as a common factor. So I can have \( gm \) and then I've got 1 times, or one over, then I've got \( 2 \times 2 \times 10^6 \), and that's in meters, minus 1 over, and then I've got \( 5 \times 10^{10} \). And then that's I'm going to have to square root that. So I've got to square root that whole entire thing, and that's going to equal \( v_{\text{final}} \). So that's going to be let's see. I've got square roots, \( 6.67 \times 10^{-11} \). Then I've got the mass of the planets, which is \( 7 \times 10^{22} \). And then I've got, let's see, 1 over \( 4 \times 10^6 \) minus 1 over \( 5 \times 10^{10} \). And then you plug all of that stuff in just very carefully. You should get a final velocity is equal to \( 1.08 \times 10^3 \) meters per second. That is the velocity of both of these planets as they finally collide into each other. So let me know what you think about this solution. Let me know if you have any questions.

You launch a rocket with an initial speed of $5\times 10^3$ m/s from Earth's surface. At what height above the Earth will it have ¼ of its initial launch speed? Assume the rocket's engines shut off after launch.

• A. $9.57\times 10^5$ m
• B. $1.47\times 10^6$ m
• C. $7.84\times 10^6$ m
• D. $2.02\times 10^{22}$ m