Hey, everyone. So we've already seen how to use Kepler's 3rd Law for circular orbits and we had this equation over here. But in some problems, you'll have an elliptical orbit, so I want to show you how to use Kepler's 3rd law in those situations. What I'm going to show you is that the equations are actually very similar. They're almost identical, but with one key difference. So I am going to show you that and we'll do a quick example. Alright, let's get started. So Kepler's third law also works for elliptical orbits. Alright, so for circular orbits, remember that we had some distance, some orbital distance or a radius of little r and we had an orbital period of T and we just use this equation, Kepler's third law, to relate those two variables. So how does it work for elliptical orbits? Well, for elliptical orbits, it's a little bit different because as this object goes around in its orbit, that distance to the star or whatever it's orbiting is constantly changing, so we can't plug that into our equations because that number is constantly different. Instead, what we're going to use is we're going to use the most important variable in elliptical orbits which is a. Remember, a is the semi-major axis. And remember, that's half the distance of the major axis which is always going to be the long one. So it turns out that we're basically going to use this equation except instead of the orbital distance of little r, we just replace it with the semi-major axis. Alright? So basically, what happens is instead of r cubed, we're going to use a cubed. Alright? That's basically all there is to it. So one thing you might remember is that we said that for circular orbits, it's really just a special case of an elliptical orbit. So for circular orbits, what happens is that r is equal to a. If I take this ellipse and I sort of squeeze it down into a circle, then basically, r just becomes or a just becomes r. It becomes that orbital distance, and it's just the same number all throughout. Alright? So, this is sort of the more generic equation and this is sort of a special case when you actually have a circle. Alright? So that's all it is for the equation. We're just gonna replace it with a cubed, but now we have some things like eccentricities and different distances to look at. So let's go ahead and take a look at an example. Alright? So we have a planet that's orbiting a star. We have the mass is 4 times 10 to the 30th. We have the eccentricity, which is, remember that the variable little e is gonna be 0.4. Now we're told the aphelion distance is this value over here. Remember, that's always gonna be the longer one. The app is always gonna be the bigger one here. So here we have got our start and then here we've got our aphelion distance. Remember that is little r, capital a or sorry, capital r and a. Right? So we want to figure out what is the orbital period. So in other words, we want to figure out T and if we have an elliptical orbit, now we're just going to use our new Kepler's Third Law equation for elliptical orbits. All right, so that's all there is to it. We're going to start out with T2=4π2a3/gm. Remember that these are just constants, the g, and we also have the mass of the star. So when I want to calculate the orbital period, the only thing I have to do is figure out what's the semi-major axis a over here. Alright? So remember that for a, we have a couple of equations in elliptical orbits, some involving the distances and some involving the eccentricity. So which one do we use? Well, the only thing we know in this problem is we know the aphelion distance which is capital r a. We know nothing about the perihelion distance over here, which is capital r p. So because of that, if you look through your equations, we can't use this one and we also can't use this one because that involves RP. So it turns out the one that we're going to use is going to be this one, which relates the aphelion distance with the eccentricity because we have both of those numbers. Okay. So this is going to be we're going to use that r a is equal to a one plus e. So I have that 1.5 times 10 to the 11th equals a, and this is gonna be 1 plus the eccentricity which is 0.4. So you can just do this really quickly here 1.5 times 10 to the 11th divided by 1.4, that's equal to a. This is gonna be 1.07 times 10 to the 11th. Okay? We're almost done here. Now we have that semi-major axis distance. Now we just pop it into this equation and finish it off, right? The last thing I have to do is just stick a square root sign because this is T squared. So you're just gonna plug in 4π squared. This is gonna be 1.07 times 10 to the 11th cubed divided by g which is 6.67 times 10 to the minus 11 times the mass of the sun which is 4 times 10 to the 30th. And when you plug that into your calculators, be very careful, put parentheses and all that stuff, what you're gonna get is 1.35 times 10 to the 7th in seconds. By the way, if you do the conversions into minutes and hours and days, what you're gonna see is this is about 156 days. And that's pretty reasonable in terms of a planet going around a star. Right? Ours is 365 days. For this one, it's 156. That's pretty reasonable. Alright? So if that's how you do these kinds of problems, let me know if you have any questions.

# Kepler's Third Law for Elliptical Orbits - Online Tutor, Practice Problems & Exam Prep

### Kepler's Third Law for Elliptical Orbits

#### Video transcript

Comet Halley has a highly elliptical orbit around the Sun, circling once every 75.6 years with its closest point to the Sun being only 0.57AU ("Astronomical Unit", where 1 AU = $1.5\times 1{0}^{11}$ m and represents the average Earth-Sun distance). How far will Comet Halley get from the Sun?

$2.64\times 1{0}^{11}$ m

$2.45\times 1{0}^{13}$ m

$5.43\times 1{0}^{12}$ m

$5.25\times 1{0}^{12}$ m

### Distant Star orbiting Black Hole

#### Video transcript

Hey, guys. Let's check out this problem together. So we have a distant star that's believed to be orbiting this black hole. So let's go ahead and draw a diagram. So we've got some orbital characteristics of this black hole. I'm told that the orbit of this star is highly elliptical. And I'm told that the period of this orbit is equal to 14 years, and it's got an eccentricity that is equal to 0.9. So it's very very very elongated. And I'm told that the period of this orbit is 14 years. Now, recent measurements have shown that the star has passed through the periapsis. So what point on the orbit is that? That's the closest distance it gets to the black hole and the variable for that is *r _{p}*. We're told that this is 1.8 times 10 to the 13th meters, and we're supposed to be figuring out what's the mass of this alleged black hole that the star orbits. So if big M is my target variable, I should go ahead and figure out which equation to use. So before, with big M's, I could use gravitational forces or accelerations. So, but we're we're not told anything about forces or accelerations. I'm going to have to use something else to figure this thing out. So we are told that there is a time, there is a distance, and we need to figure out what the mass is. So we always use Kepler's third law that relates all of those two things, all of those three things together. So we've got t2=4π2a3GM. That's the semi-major axis. Remember, that's half the distance of the long axis divided by, and that is G times big M. So that's my target variable. Oops. That's my target variable. So all I have to do is just bring this over to the other side to isolate it and then bring the t squared down so they trade places. So I've got M is equal to 4pi squared a cubed divided by G times t squared. So let's look through this equation here because, I just need to figure out if I I just need to know if I have everything to go ahead and start plugging stuff in. So 4 pi squared is just a number. Then a cubed, that's the semi-major axis. Well, I looked through the problem that I only have the periapsis distance and the eccentricity. I'm not told anything about the semimajor axis. But I do have G and I do have t squared. I do have the orbital period. So all I really need to do too is figure out what the semi-major axis is. I'm gonna go ahead and do that over here. So how do I figure out what the semi major axis is? I'm told, or I basically can look through my equations for elliptical orbits. So I've got this equation right here, that's the semi-major axis. The problem is I'm not told what the apoapsis distance is, so I can't use this. Instead, what I can do is say, which one of these equations do I know the most about? I know the eccentricity and I know the periapsis distance. So I can actually use this RP equation. So I'm gonna use that.

*R*is equal to a then one minus e. If I look through this, I have the eccentricity and I have the periapsis distance. So I can actually use this to find a. So if all I have to do is just move this one minus e to the bottom, and I've got

_{p}*r*divided by 1 minus e is equal to a. So So in other words, 1.8 times 10 to the 14th. So what you can actually do is plug this semi-major axis back into this equation and then continue solving for M. So we've got M is equal to, and I've got 4pi squared. The semi-major axis I just found was, 1.8 times 10 to the 14th. I'm gonna cube that divided by And I've got 6.67 times 10 to the minus 11. Now I've got this t. I have to be careful here because this t, this period is given to me in years. So what I have to do is actually convert over here. I'm gonna actually do that over here so I have some room. So I've got t is equal to 14 years. All I have to do is just multiply that by 365, days per year, so we can eliminate the years. And then I've got 86,400, and that's gonna be seconds per day. So again oh, okay. If you don't recognize that, you can keep working out the 24 hours, 60 minutes, 60 seconds. You'll get this number. You'll get the same number. T is equal to, if I plug all that stuff in, 44.42 times 10 to the 8th. I'm just gonna plug it back in. So I've got 44.42 times 10 to the 8th. Now I have to square that. And if you go ahead and work all of this stuff out, you're going to get a grand total for the mass of the black hole. That's 1.77 times 10 to the 37th kilograms. Now, that might just look like any normal large number. But if I can sort of express this in more familiar terms, the mass of this thing is around 8,800,000 times the mass of our sun. So it's 8,800,000 times heavier than our sun, which is pretty incredible. Anyway, so this is the answer. I hope you guys don't have any questions. If you do, just let me know. And then, I'll see you guys in the next one. Let's keep going.

_{p}