Energy of Circular Orbits - Online Tutor, Practice Problems & Exam Prep

Understanding the energy of orbits involves the conservation of energy when a satellite changes its orbital distance or velocity. The kinetic energy (KE) is given by $\frac{1}{2}m{v}^2$ , while gravitational potential energy (PE) is $-\frac{GmM}{r}$ . To change orbits, positive work is required to increase altitude, resulting in decreased velocity, while negative work is needed to decrease altitude, increasing velocity. The relationship between orbital radius and velocity is crucial for solving these problems.

Energy of Circular Orbits

Hey, guys. So we've talked about energies, and we also talked about satellites in orbit. So we're going to 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 about a satellite that is changing from one orbit to another. So let's go ahead and see how that works. We basically got that a satellite is going to go from some orbital distance, so I'm just going to call that r, and a corresponding velocity that keeps it in orbit v1. And at some later time, it's going to have a different r, so r2, and it's going to have a different corresponding velocity that keeps that in orbit. We know the relationship between that distance and the speed is related by the VSAT equation. So what happens is we're changing from one orbit in which we have r1 and v1, and we're going to change it to another one in which it has r2 and v2. 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 going to be exactly how we've always used it before. But in terms of these actual 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 12mv2. But if that satellite is some distance away from the mass, then that means that it also has some gravitational potential energy. So that means that these are the terms that we're going to use over here for the kinetic and potential energies. Now, if we're going from one orbit 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 are 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 orbit, and you're going to go to a final distance, and you're going to 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-kilogram spacecraft. So we've got how much work is needed. So let's just go ahead and, so we've got a spacecraft going from some orbit to a higher one. So I'm just going to draw a quick little sketch. We've got the Earth like this. I've got some orbits, and then I've got at some later time, some other orbit. So, in other words, I'm going from some initial orbit to some final orbit, rfinal. Alright. So I know when I have to use energy conservation, but rather than just me writing that new equation all over again, I'm just going to start from the kinetic and potential energies. So I've got 1 2 m v initial 2 − G M m / r initial + any work done = 1 2 m v final 2 − G M m / r final . Okay? 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 vinitial that I don't know, vfinal I don't know, and I'm looking for this work.

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 ($-1.74\times 10^9$ J) bringing it to a circular orbit just above the surface, how high was its original orbit?

• A. $1.9\times 10^5$ m
• B. $1.6\times 10^6$ m
• C. $1.8\times 10^6$ m
• D. $3.7\times 10^6$ m

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) $W_{AB}=3.37\times10^8$ J;(b) $W_{BC}=1.99\times 10^9$ J
• B. (a) $W_{AB}=8.48\times10^8$ J;(b) $W_{BC}=2.7\times 10^9$ J
• C. (a) $W_{AB}=3.37\times10^8$ J;(b) $W_{BC}=5.97\times 10^9$ J
• D. (a) $W_{AB}=8.48\times10^8$ J;(b) $W_{BC}=8.1\times10^9$ J