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

In elliptical orbits, kinetic energy (KE) and potential energy (PE) vary, unlike in circular orbits where they remain constant. At periapsis, where the distance is minimal, KE is maximal, and PE is minimal. Conversely, at apoapsis, KE is minimal, and PE is maximal. The relationship between velocities at two points can be expressed as v1 r1 = v2 r2. Transitioning between circular and elliptical orbits involves energy conservation, requiring work to change the orbit's shape.

Speed and Energy of Elliptical Orbits

Hey, guys. So now that we've talked about energy conservation for circular orbits, we're going to be talking about elliptical orbits in this video. There are a couple of really important things that you need to know, so let's go ahead and check it out. The first is that, unlike for circular orbits, the kinetic and the potential energies are not going to be constant. They're going to be changing. Let's go ahead and see why. Now, we said for circular orbits, all we needed was the distance little r, and we need the velocity at this point, and we can relate that to the total amount of energy. The kinetic energy was related to the velocity, and the potential energy was related to the distance. Well, the idea is that for elliptical orbits, what happens is this distance of little r is actually constantly changing throughout an orbit. So what happens is the energies are going to change.

Let's take a look at a spacecraft that's at the very, very farthest point in its orbit. Now, just remember that this farthest point over here is called the apoapsis distance and that's the farthest. Right? So, it has some velocity at this point, and gravity is going to pull it in throughout its orbit. And as it does that, it's going to speed up and accelerate. So then it gets to this point over here, which is the periapsis distance that is the closest that it gets, and it has some really really high velocity because gravity has pulled it in. So it's at maximum speed. Then as it goes out, gravity slows it down, so that means the velocity when it gets out here is going to be minimum. So, if it's at the closest distance of periapsis, which is over here, then that means that the speed is maximum. And if the speed is maximum, then that means that kinetic energy is maximum over here. Now, because this energy is still conserved. Right? Because we're still talking about gravity, if the kinetic energy goes up, the potential energy has to go down in order to keep it balanced. So that means that U is minimum here.

Well, then it gets to the other side of the orbit, so it gets to its apoapsis distance over here. And we know that the velocity is minimum over here. So if the speed is minimum, that means the kinetic energy is minimum. And, again, because the total amount of energy is conserved, as K goes down, U has to go up. So the potential energy is maximum here at apoapsis. Alright. So that's one really important point. So, basically, to wrap it up, the total amount of energy in circular orbits is constant and K and U are all constant. But for an elliptical orbit, the total amount of energy is constant, but between K and U, it is constantly changing. So what happens is one goes up, the other one goes down and vice versa.

Now the second most the second really important thing or the most important thing you need to know is how to compare the velocities at 2 different points in an elliptical orbit. So the idea is over here at any point, you can actually compare the velocity to between these two points over here. So if all we need are the distances and the speed. So if we have r so this is v₁ and r₁, and we have v₂ and r₂, we can basically draw a relationship between these two velocities and distances by using this really important equation, v1 r1 = v2 r2. So what happens is in problems, you're usually going to be given 3 out of these 4 variables, and you can use this equation to solve for the 4th one. Obviously, this is just an example, but you can solve for any one of those unknown variables. Now, this equation actually comes from a different conservation law that we haven't talked about in a while, which is conservation of angular momentum around any object is equal to mvr. So, if we're talking about an elliptical orbit, we can use conservation of angular momentum to compare any two points in that orbit. But because we're talking about the same mass, this little m will cancel, and you will just be left with this relationship over here. Alright? So that's basically it. Let's go ahead and see how we use it in an example.

So we've got a planet that is observed orbiting a distant star. The closest distance over here is going to be, remember, that's periapsis, and its velocity is given over here. So that's the velocity of periapsis. So now we're supposed to find out what is the velocity at its farthest distance, and this is a different distance over here, which is equal to r_a. That's apoapsis. Right? Remember that's the farthest distance. So we've got that we're comparing 2 different speeds and velocities, and we're looking for, in this case, we're looking for v₂. So what this means is that these two points so we have v₁ is equal to the periapsis distance, and this r₁ over here is equal to r_p, and this is equal to v_a times r_a. So in this case, it happens to be the apoapsis and the periapsis, but it doesn't necessarily have to always be those 2. So we actually have what 3 of these variables are, and we can go ahead and solve for that third one, and it's as simple as that. So if we've got, we're just going to move this over to the other side, and we've got vprpra = va. So in other words, we have, the velocity, which is 59 kilometers per second, and 0.31 for the distance divided by 0.47. Now, some of you might be wondering why I've chosen to put in those units instead of having to convert them. And the reason is that when we're plugging in for v and r, they actually can be non SI as long as they're consistent. So as long as we use the same units for VP and RP, then or in the same units for RP and RA, then we'll just get the correct units. So these things just have to be consistent with each other in order for us to plug them in. Okay? So if you work this out, you're going to get a velocity of 38.9. And because of the units, we're going to get kilometers per second. So just make sure that you're able to do this with your professor. If you see this on a homework problem and you're told to put it in meters per second, you're going to have to convert it. Alright? Anyway, so that's it for this one. Let me know if you guys have any questions.

Pluto travels in a fairly elliptical orbit around the Sun. At its closest distance of $4.43\times 10^9$ km, its orbital speed is 6.12 km/s. At its farthest, its orbital speed reduces to just 3.71 km/s. How far is it from the Sun at this point (in km)?

• A. $2.69\times 10^{12}$ km
• B. $7.31\times 10^{12}$ km
• C. $2.69\times 10^9$ km
• D. $7.31\times 10^9$ km

Comet Traveling Around the Sun

Alright, guys. Let's work this one out together. So we have a comet that's traveling around the sun, travels once every 2000 years, and we're told some information about its closest approach. And in part a, we're supposed to figure out how far what is its farthest distance from the sun. So you should recognize that as the apoapsis. Let's just go ahead and draw a quick little diagram. So we've got the sun right there, and I'm going to have this highly elliptical orbit like this. Now if we're trying to figure out the farthest distance that corresponds to that length right there, that's apoapsis, and that's what I'm really looking for. Now there are only 2 other pieces of information I know about this problem. The first is that the entire orbital period for the whole thing is equal to 2000 years, so I've got that. And I've got that the apoapsis distance, the closest that it gets to the sun, which is r_p, is just given by some number, 3 times 10 to the 8th kilometer. So I've got that. So let's look through my equations. Now you might be tempted to sort of use the equation which we compare 2 parts of an elliptical orbit. That's v₁r₁ equals v₂r₂. The problem with this equation is that we if you look through, we don't have any of the velocities, and we're going to be solving for one of the distances. So that gives us with 3 unknowns for this problem. There's no way we're going to be able to use that, so it can't be this. We're going to have to look for some other equations to solve for r_a. So let's go ahead and look through my equations involving r_a. So let's see. I've got this equation over here, and I've got let's see. The tsat doesn't work because there's no r_a in there. These two equations relate the semi major axis and the eccentricity. Both of those values are unknown, so I can't use either of those. So it's going to have to be this equation over here, which is that the semi major axis is equal to r_a + r_p divided by 2. Cool. So I've got what r_p is, and if I'm trying to figure out the apoapsis is, there's only one equation or one unknown I need to solve for, and that is the semi major axis. Okay? So, unfortunately, I'm going to need another equation for this. So let me go ahead and just move this over here. So how do I figure out what the semi major axis is? Okay. Let's go back to our equations. Now I can't go for this equation because this is the equation I'm just coming from. So obviously, it can't be that one. Again, I don't have any information about the eccentricities. So that means it only leaves me with one option. I'm going to have to use Kepler's 3rd law for elliptical orbits. Now this is really useful because I'm actually told some information about the orbital period, and that's the variable t. So I'm definitely going to use this equation. So I've got t t squared is equal to, 4π squared a 3 divided by g m. So all we have to do is just rearrange everything to the other side. The g m is going to go up, the 4 π squared is going to go down, and I'm going to get g m t 2 4 π 2 is equal to a cubed. So now if I rearrange, I'm just going to get that a is equal to the cube roots of that whole entire mess over there. So I've got 6.67 times 10 to the minus 11. Now I've got 2 times 10 to the 30th for the mass of the sun. Now I need t squared. Alright? The problem is that the t I'm given is in years, it's not in seconds. So I'm going to have to first convert. So we've got t is equal to 2,000. Now I've got, that's going to be in years, so I have to multiply by 365 days per year to cancel that out. Now, I've got to multiply by 86,400 because that's how many seconds there are in one day. So let me let me, sort of write that out again. So we've got seconds per day. If you didn't know that off the top of your head, that's totally fine. But the years are going to cancel with the years, the days are going to cancel with the days, and then t is just going to be 6.31 times 10 to the 10 seconds. That's how many years there are in 2000 or so that's how many seconds there are in 2000 years. Great. So we're just going to plug that into this equation over here, 6.31 times 10 to¹⁰, and now we have to square that. Don't forget that. And now we have to divide this whole entire expression by 4 π squared. Don't forget to put that in parenthesis. Anyway, so if you plug all of this huge mess into your calculator, you're going to get a semi major axis of 2.38 times 10 to¹³, and that's going to be in meters. Right? So we can actually plug this all the way back into our initial equation and then figure out what r_a is. So let's just go ahead and do that really quick. If we were trying to figure out what r_a is by isolating it so let's see. We're going to get that, 2 a. Right? We're going to multiply this up, and then we're going to have to subtract the r_p over. So 2_a minus r_p is going to be our _a. Right? That's going to be our expression, and that just comes directly from this equation right here. Great. So we've got r_a is just going to be Let's see. We've got oura is going to be 2 times we've got 2.38 times 10 to the¹³ minus the periapsis distance, which is in kilometers. So you have to be very careful because this unit right here is given in kilometers. So this is actually 3 times 10 to¹¹ in meters. Right? So this has to be in meters, and this has to be meters. So that means that our apoapsis distance over here, what I get is I get 4.73 times 10 to the¹³ in meters. And that's the answer to part a. So it goes very, very far out and then it comes in close. Usually, that's what comets do.

Alright. So for part b, now we're figuring out what is the ratio of the comet's speed at its closest point to its farthest point. What does that mean? We're looking for a ratio, so that means it's going to be something divided by something else. Now remember, the velocity at its closest point is going to be v_p, and its velocity at its farthest point is going to be v_a. So we're looking for this ratio over here. Now because we're comparing the speeds of 2 different points in the orbit, now we can act

Energy Conservation in Changing Elliptical Orbits

Hey, guys. So now we know a little bit more about elliptical orbits, we're going to see how we use energy conservation to transition from a circular to an elliptical orbit. Let's check it out. Now the idea is that we're still changing from an orbit. In this case, a circular to an elliptical one, so some work has to be done. And by the way, if you're going from an elliptical to a circular, still some work to be done. So everything I say can be done backwards here as well. Now we're still talking about changing orbits, changing works, so we're going to use energy conservation. Now when we did this for circular orbits, we were able to combine these two kinetic and potential terms into a total amount of energy. So, this simplified our process. Now, the way this worked for circular orbits was that all we needed was this distance here of little r of our circular orbit, and we use this equation, gmm/2r. Now we already know that this equation doesn't work because what happens is that as this orbit or as this object travels in its orbit, it goes from the periapsis distance, which is the closest that it gets, to the apoapsis, which is the farthest. So what happens is as it travels throughout its orbit, this distance of little r is constantly changing. Alright? So we're going to need a new equation to work with this. Now, how have we done this before? We've always taken an elliptical orbit and broken it up its axis, the minor and the major axis. And we always said that the most important variable of an elliptical orbit is the semi-major axis, which is given by the letter a. Now, this is not the same thing as the apoapsis, which is the farthest distance over here. So don't get those two things confused. But all we did was that any time we had an equation that went from circular to elliptical, we just replaced the little r with a, and it's going to work the exact same way here. So we take this equation and we replace it with negative gmm/2a. So that means that these equations over here are going to be the equations that you plug into your energy conservation here. You just have to make sure that you understand that you're going from a circular to an elliptical, which one is your initial and which one's your final, and then plug in the appropriate expressions for those. And to see how that works, we're going to go ahead and do this quick example right over here.

So we've got a spacecraft and we're told what the mass of this thing is. We're told it's initially in a circular orbit, and we're going to be calculating the energy required to change it to an elliptical orbit and we're told some characteristics about that. So let's go ahead and just draw a quick little diagram right here. So I've got the Earth and I've got the initial circular orbit like that. Now, let's try it again. There we go. That's a circular orbit. And what happens is we're going to transition to an elliptical orbit over here. So we have the initial distance of this circular orbit, little r, and then we're told what the apogee distance is. Again, remember that is this farthest distance here, not the semi-major axis. Okay? So this is the apogee and we need to know how much work is done. So or how much, how much energy is required to do that. But it's also the same thing as the amount of work. So let's go ahead and set up our energy conservation equations. So we got initial energy plus the work is equal to the final energy. Now, we're going from a circular orbit all the way to an elliptical orbit, so we just have to make sure to plug in the correct equations for those. The total energy of a circular orbit is negative gmm/2r plus the work is equal to now negative gmm/2a. So we're we're looking for the amount of work that's involved. So all I have to do is just take this term and move it to the other side and just have to, plus it because I've got a minus sign right here. So I've got the work is equal to I've got gmm/2r and it's positive, minus gmm/2a. Now let's look through my variables, make sure I've got everything. So I've got all of these letters here are just constants. I can look them up in this table over here, and I have the mass of the spacecraft. Now I'm told what the r distance is. That's this 6.87×106. Now, the only other thing I'm told is the apogee distance, But I need to find out what little a is. That's the semi-major axis. Remember, a is not the same thing as the apoapsis distance. So I need another equation to figure out what a is. Let's go ahead and look at my equations over here. So I've got, this t2=4πa2 blah blah. I've got Kepler's 3rd law for elliptical orbits. Now I've got a, I've got g and m, but I don't have any information about the period of the satellite, so I can't use that. Both of these equations over here involve a, but they also related to the eccentricity of the orbit, which I also don't have. So I can't be using these equations either. So I'm going to use this equation over here, which is my apoapsis and periapsis distance divided by 2. So I've got ra+rp/2. Now the good thing is that I have the apogee distance. I'm just told what that is as a number over here. Now what about this periapsis distance? What is that value? Well, let's take a look at my diagram. Right? So I know that I've got the I've got the semi-major axis over here, which I'm trying to find. So that's going to be this piece over here. So that's a. And the apo sorry, the periapsis, which is going to be the closest distance, is going to be this piece over here. But notice how these two distance are actually the same. So this actually becomes a really really important point that you need to know. And that when you're going from a circular orbit like we did over here to an elliptical orbit, what happens is that your little r actually becomes r p or r a, and it depends on the size of your new orbit. So let me give you a couple of rules to follow. Now just like we did with this problem, if your new orbit is larger, then that means that your little r becomes r p, just like it did for this problem over here. Right? So these two things are the same exact variable. Now, if your new orbit were smaller, then it's the opposite. Your little r, which just becomes r a. So that means that we actually know what r p is. We can just go ahead and solve that. So r a is the apogee distance 2.0×107 plus, and then we've got 6.87×106, and we're going to divide that by 2. So that means that a is equal to 1.34×107, and that's meters. We're going to plug that in there for our equation. So now we can figure out what the work is. The work is just equal to Now what we can do is we can take out this gmm/2 as a common factor. So gmm/2. Now we've got one over and the r is going to be 6.87×106 minus 1 over I've got 1.34×107. So if you work all of this out in your calculator, you're going to get the work done is equal to 1.41×1011 in joules. So this is our final answer for the amount of work. And because it's positive, that makes sense because we ended up with a larger orbit. So all of those rules for orbit still make sense, in terms of work and positives and negatives. Now the last thing I want to mention is that we went in this example from a circular to an elliptical orbit. So you might be wondering what happens when you do it backwards. Well, when you're going from elliptical to circular, all the all the rules are backwards. So we've got this elliptical orbit over here and imagine we're going sort of in the in the clockwise direction. Now if we wanted a larger orbit like for a, we'd actually have to do a burn over here, so we'd have to increase our velocity in order to basically circularize that orbit. So if you take a look at the diagram here in our new in our old elliptical orbit, this distance over here is actually our apoapsis distance, and that becomes the radius or the distance of our new circular orbit. So if you're going from an elliptical to a circular, if your new orbit is larger, then that means that r a becomes your r. And if it's smaller, then r p will become the new r. So you'd actually have to do a burner over here. You'd have some delta v, and this is your periapsis distance that would become your new orbital distance in a circular orbit. Alright. So all the rules are just backwards. Alright, guys. That's it for this one. Let me know if you guys have any questions.

A comet is in a highly elliptical orbit around the Sun with a period of 2400 years. If this comet has a mass of approximately $5\times 10^{14}$ kg, what is the total energy of its orbit?

• A. $1.24\times 10^{21}$ J
• B. $-1.24\times 10^{21}$ J
• C. $2.48\times 10^{21}$ J
• D. $-1.24\times 10^{20}$ J

Changing from a Elliptical to Circular Orbit

Hey, guys. So let's check out this problem. We've got a spacecraft that is traveling from one orbit to another. Right? We have a 2,000 kilogram spacecraft, and we're being asked to find how much energy it takes in order to go from an orbit where it has a perigee of some number and then an apogee of another number. So we're talking about some elliptical orbits right here. So let's go ahead and draw this quick little diagram. I've got the Earth that's going to be right over here, and now I've got some perigee, and then I've got some apogee out here, and just have to connect those two things with an ellipse. So it's going to look something like this. Okay? So then what happens is that now it's going to change to a circular orbit. But one of the things I notice—one of the things you should notice—is that these actually are the same numbers right here. The apogee of the elliptical orbit, so that's right over here, is actually the orbital radius of the circular orbit that it ends up in. So, basically, what happens is that it goes from this orbit of an elliptical, and then it basically just circularizes it to a circular orbit like that. So in other words, this right here is our initial orbit, and this over here is our final orbits. So we know we're going to be dealing with changing orbits. We have to use energy conservation. So we're going to go ahead and use our energy conservation equation. So we're being asked to find delta e. Let's just go ahead and write out energy conservation. So for part a, we've got the kinetic and potential, plus any work done equals final kinetic and potential. So what happens to this delta e? Where does this delta e come from? Well, remember that in energy conservation, this work is equal to the change in energy. So when they're asking for delta e, they're actually asking you to find what is the work that's done in order to change this initial elliptical orbit. So how much energy do you need to sort of bridge the gap between this elliptical orbit? Then you're going to fire the thrusters of the engine or whatever, and you're going to end up in this final circular orbit over here. Okay? So let's take a look at both sides of our energy conservation equation. We know that when we end up over here, this is going to be a circular orbit, whereas we're starting from an elliptical orbit. So let's take a look at our variables here. We've got some distances involved, and then we've got a final distance here. We're never told anything about velocities. So what we can do is we can use our shortcut equations for elliptical and circular orbit energies. So in other words, we can use these equations right here because they have the distance involved. So remember that this elliptical orbit equation just becomes negative g, big M, little m over 2 a, where that's the semi-major axis. Then we've got the work that's done is equal to the circular energy. So it's going to be negative g m m over 2 times r. Right? So we just go ahead and say, this is going to be our circular orbit equation right here, and this is going to be our elliptical orbit equation. We're just going to use those two things. So that's where these equations come from. And we're being asked to find out what is the work that's done in order to do this. So if we go ahead and move everything over to the other side, I'm going to move this term over, and what you should get is you should get the work that's done is equal to now what happens is this is going to be composite when you move it to the other side. So we're going to get gmm over 2 a minus gmm over 2 r. So we can pull out all of this stuff except for the a's and r's as greatest common factors, and what we get is that the work that's done is equal to gmm over 2, and then we've got 1 a − 1 r . So now we can start plugging everything in assuming I have all the numbers. So I've got the I'm going around the Earth, so I have the mass of the Earth and I have the gravitational constants. I have the mass of the spacecraft, and the problem is is I don't actually have what the semi-major axis is. So I never told that is specifically. I do know what this final position is, this final orbital distance. That's just this number here of 8.87. So I've got that. So the only thing I have to do is just have to figure out what a is. So what are my equations for finding out the semi-major axis? Well, hopefully, you guys realize that we're given the information about the perigee and the apogee. So we can(ss) actually use this equation right here for finding the semi-major axis. So we've got r a plus r p divided by 2. So in other words, I've got, 6.87 times 10 to the 6th plus 8.87 times 10 to the 6th. You're going to divide that by 2, and you're going to get 7.87 divided, times 10 to the 6th. Oops. Not 16. So it's just like halfway in between these two numbers right th