Kepler's Third Law for Elliptical Orbits - Video Tutorials & Practice Problems
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concept
Kepler's Third Law for Elliptical Orbits
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Hey, everyone. So we've already seen how to use Kepler's third 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 third 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'm gonna show you that and we'll do a quick example. All right, let's get started. So Kepler's third law also works for elliptical orbits. All right. So for circular orbits, remember that we had some distance, some orbital distance, a radius of little R and we had an orbital period of T and we just use this equation. Uh 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 of what we're going to use is we're gonna 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. All right. So basically what happens is instead of R cubed, we're gonna use a cubed. All right. 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 be or A just becomes R becomes at orbital distance. And it's just the same number all throughout. All right. So this is sort of the more generic equation and this is sort of like a special case when you actually have a circle. All right. So that's all it is for the equation you just, we're just going to replace it with a cubed. Uh 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. All right. So we have a planet that's orbiting a star. Uh we have the mass is four times 10 to the 30th, we have the eccentricity, which is, remember that variable little E is gonna be 0.4. Now, we're told the app helium distance is this value over here. Remember that's always going to be the longer one. The App is always going to be the bigger one here. So here we have got our star and then here we've got our app helium distance. Remember that is little R capital A or sorry capital R A, right? So we're going 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 T squared equals four pi squared A cubes over GM. Remember that these are just constants the G and we also have the mass of the star. So 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. All right. So remember that for A, we have a couple of equations in elliptical orbits, uh 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 app helium distance, which is capital R A, we know nothing about the perihelium distance over here, which is capital RP. 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 app hele distance with the eccentricity because we have both of those numbers. OK. So this is going to be, we're gonna 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 going to be one plus the eccentricity which is 0.4. So you can just do this really quickly here 1.5 times 10 to the 11th, uh divided by 1.4 that's equal to A. This is going to be 1.07 times 10 to the 11th. OK. So 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 only last, the last thing I have to do is just stick a square root sign because this is T squared. So you're just going to plug in four pi squared, this is going to 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 four times 10 to the 30th. And when you plug that into your calculators, be very careful parentheses and all that stuff. But you're gonna, is 1.35 times 10 to the seventh in seconds. By the way, if you do the conversions into minutes and hours and days, what you're going to see is this is about 100 and 56 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 100 and 56. That's pretty reasonable. All right. So that's how you do these kinds of problems. Uh, let me know if you have any questions.
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Problem
Problem
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 = m and represents the average Earth-Sun distance). How far will Comet Halley get from the Sun?
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m
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m
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m
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m
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example
Distant Star orbiting Black Hole
Video duration:
4m
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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. Uh 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 me measurements have shown that the star has passed through the persis. 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 RP, 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 it's 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 si could use gravitational forces are accelerations Um So, but we're, we're not told anything about forces or accelerations. I'm gonna 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 about T squared equals four pi squared a cubes, that's the semi major axis. Remember that's the 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 four pi squared A cubed divided by G times T squared. So let's look through this equation here because um I just need to figure out if I, I just need to know if I have everything, just go ahead and start plugging stuff stuff in. So four pi squared is just a number, then a cube that's the semi major axis. While I look through the problem that I only have is the per axis distance. And the eccentricity, I'm not told anything about the semi major axis, but I do have G and I do have T squared. Um 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 we figure out what the semi major axis is? I'm told? Uh 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 apple axis 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 per axis distance. So I can actually use this RP equation. So I'm gonna use that RP is equal to A then one minus E. And if I look through this, I have the eccentricity and I have the persis distance. So I can actually use this to find a. So if I, all I have to do is just move this one minus E to the bottom. And I've got RP divided by one minus A uh or sorry, R one minus E is equal to A. So in other words, 1.8 times 10 to the 13th divided by one minus 0.9. And if you work that out, you're gonna get 1.8 times 10 to the 14th. So we can actually do is plug this semi major axis back into this equation. And then continue solving four M. So we've got M is equal to, then I've got four pi squared, the semi major axis I just found was uh 1.8 times 10 to the 14th. I'm gonna cube that divided by then. 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 uh days per year. So we can eliminate the years. And then I've got 86,400 that's gonna be seconds per day. So again, ok, 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. Uh T is equal to plug all that stuff in 44.42 times 10 to the eighth. Now I'm just gonna plug it back in. So I've got 44.42 times 10 to the eighth. Now, I have to square that. And if you go ahead and work all of this stuff out, you're gonna get a grand total for the mass of the black hole that's 1.77 times 10 to the 37th kilograms. And that might just look like any normal large number. Uh But if I can sort of express this in more familiar terms, the mass of this thing is around 8.8 million times the mass of our sun. So it's 8.8 million times heavier than our sun, which is pretty incredible anyway, so this is the answer. Uh Hope you guys don't have any questions if you do, just let me know and then, uh I'll see you guys in the next one. Let's keep going.
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