Energy of Circular Orbits - Video Tutorials & Practice Problems
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Energy of Circular Orbits
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Hey guys. So we've talked about energies and we've also talked about satellites in orbit. So we're gonna put those two things together in this video and talk about the energy of orbits because a lot of times press questions will ask you a satellite that is changing from one orbit out to another. So let's go ahead and see how that works. So we've basically got that a satellite is gonna go from some orbital distance. So I'm just gonna call that R and a corresponding velocity that keeps it in orbit V one. And at some later time, it's gonna have a different R so R two and it's gonna have a different corresponding velocity that keeps that in orbit. We know the relationship between that distance and the speed is related by the V Sat equation. So what happens is we're changing from one orbit in which we have R one and V one and we're gonna change it to another one in which it has R two and V two. So in other words, we're going from an initial sort of state or position to a final. And when we do that in gravitation, we can't use kinematics to solve initial versus final, we have to use energy conservation. So we've got our energy conservation equation over here. It's gonna be exactly how we've always used it before. But in terms of these actual uh terms that we use the kinetic energy and the potential energy, we can actually see from this little table here that if a satellite is in orbit. So in other words, if it has some velocity over here, then that means that it has some kinetic energy that's just given by one half MV squared. But if that satellite is some distance away from the mass, then that means it also has some gravitational potential energy. So that means that these are the terms that we're gonna use over here for the kinetic and potential energies. Now, if we're going from one orbit out to another orbit, then it means that we have to do some work. So in other words, to change orbits, there must be some work that's done. And that's usually what these kinds of questions will ask you. Now, there's two basic kinds of questions that you'll see. So let's just go ahead and start working them out. One is when you're changing an orbital distance from one to another. So in other words, a question will give you an initial orbits and you're gonna go to a final distance and you're gonna have to figure out using energy conservation how that works. So let's go ahead and check out this example, we got the work that's needed for a 200 kg spacecraft. So we've got how much work is needed. So let's just go ahead and um so we've got a spacecraft going from some orbit out to a higher one. So I'm just gonna draw a quick little sketch. I've got the earth like this, I've got some orbits and then I've got some later time, some other orbit. So in other words, I'm going from some initial orbit out to some final orbit are final. All right. So I know I'm gonna have to use energy conservation. But rather than just me writing that new equation all over again, I'm just gonna start from uh the kinetic and potential energies. So I've got that the one half M, the initial squared minus GMM over our initial plus any work done equals one half MV, final squared minus GMM over our final. OK. So if I take a look here, I've got the target variable is the work that's needed. But if you go and look through these variables, I actually don't know what the velocities are, but I do know what the initial positions are. So I've got a situation where I've got, let's see, I've got initial, the initial that I don't know. V final, I don't know. And I'm looking for this work. I have three unknowns in this equation. So I can't solve it. I have to use some else. What we can do about this is we can use the relationship between V sat and the R. So we know that's equal to square root of GM over R. So what we can do in this equation is we can replace these vs with this expression here to get everything in just terms of RS which we know. So let's go ahead and do that real quick. So I've got one half and then M and I've just got to substitute these things in. So I've got GM over our initial because this was the initial, I'm going to square that minus GMM over our initial plus any work that's done equals one half M. Now, I've got square root GM oops over our final squared minus GMM over our final. OK. So now if we actually go ahead and work this out and you combine these two terms together, what this ends up being is, this ends up being negative GMM over two, our initial and you can go ahead and work out out for yourselves, just make sure that kind of uh makes sense and then this actually becomes negative GMM over to our final. So if you took a look at what's happened here, we've actually gotten rid of the velocities and we've put it in terms of the RS which we actually know we have the variable, we actually have the numbers for that. So if you're trying to figure out what the work is all you have to do is just plug this into your equation. I'm not actually going to do it for you. You guys can go ahead and, and see for yourselves. Now, let's look at the other kind of problem that you'll see. You'll often see problems in which you have a changing velocity instead of orbital distance. In other words, we're going some from some initial velocity to some final velocity, we have to use energy conservation to solve it. So we've got how much work is needed. So there we go, we're looking for the work again for 200 kg spacecraft to travel from a circular orbit in which it's traveling at this speed and then it's gonna be traveling at this speed. So let's just go ahead and write our energy conservation again. We can't draw a diagram because we don't know what the distances involved are. So I've got one half MV, initial squared minus GMM over our initial plus the work done, it's equal to have to be final GMM over our final. OK. So we've still got, we're looking for the work. But now what happens is that before, like we actually knew what the initial and final distances are here. Those things are actually unknown. So these are our unknowns and we actually know what the initial and final velocities are. So how do we get those two unknowns down to something that we actually do know? So how do we get rid of these unknown variables. Again, we use the relationship between vat and R. So what happens is we actually go ahead and solve this equation just for R. So we're gonna have to square root both sides and then replace the two and sort of just swap them. What we're gonna get is that R is equal to GM over V squared. So now what we can do is take this expression and plug it in for those unknown variables and then we'll get rid of those unknowns. So let's see how that works. We're gonna get one half M the initial squared minus GMM over. Now you've got to plug in GM over the initial squared plus any work done and then we have the final expression. So minus GMM over and then again, this is gonna be GMM this is gonna be GM over V final squared. So we've got to make sure that we're keeping track of our V finals and V initials. OK. So now in a very, very similar way to how we did this. If you sort of merge these two expressions together, you're gonna cancel some f some stuff out like the G MS, the G MS will cancel and you'll flip some stuff. What we're gonna get is we're actually going to get uh negative one half M, the initial squared plus the work is equal to negative one half MV, final squared. So now we've gotten that we are still looking for the work. But now we've gotten in terms of the variables that we know. So we've gotten rid of those two distances that were unknown. We've got in terms of the loss, the velocities that we do know. So let's take a look at what happened in both of these cases, we went from a kinetic energy and a potential energy in both of these situations. And we combined them down to a singular term that involved the variables that we knew. In this case, we involved uh in terms of the R initial and our final. And in this case, we put it in terms of the V initials and V finals. So in other words, this term right here is just the total amount of energy because we basically put together the kinetic and the potential energy. Remember that the total amount of energy for anything is just K plus U. So that means that the energy for a circular orbit has two different forms. It has negative GMM over two R or it has negative one half MV squared. So when you're dealing with problems that involve changing distances, you're gonna use this equation in order to solve it. If you're dealing with a problem that involves changing velocities, you're gonna use this equation to solve it. All right. The last thing I wanna mention is that to increase or enlarge an orbit, you have to do some work that is positive. So in other words, you have to put in work into the system so that this thing achieves a higher orbit. And so what happens is that the R is going to increase, but the velocity is going to decrease now, basically the same exact thing. But opposite. For when you're shrinking or decreasing an orbit size, the work has to be negative in order for that orbit which has a lot of energy to shrink down to something that is smaller. So in other words, the R is going to be decreasing and the velocity is going to be increasing. All right guys, that's it for this one. Let me know if you guys have any questions.
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Problem
Problem
The 12,000-kg Lunar Command Module is in a circular orbit above the Moon's surface. If it spends ¼ of its fuel energy ( J) bringing it to a circular orbit just above the surface, how high was its original orbit?
A
m
B
m
C
m
D
m
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Problem
Problem
a) How much work do you have to do on a 100-kg payload to move it from Earth's surface to a height of 1000 km?
b) How much additional work must you do to put this payload into orbit at this altitude?
A
(a) WAB=3.37×108 J;
(b) J
B
(a) WAB=8.48×108 J;
(b) J
C
(a) WAB=3.37×108 J;
(b) J
D
(a) WAB=8.48×108 J;
(b) WBC=8.1×109 J
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