Kepler's Third Law - Video Tutorials & Practice Problems

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concept

Kepler's Third Law

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8m

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alright, guys. So for this video uncovered Kepler's third Law in some detail, it's the most important of all the three Kepler's laws, and you'll definitely needed to solve orbit problems. So let's check it out. So Kepler basically said that for any circular orbit he found out that the orbital period is related to the orbital distance. So if we have some satellite here that is orbiting at some distance, little are away. It takes some time to go around in its orbit. That's capital teeth. So Kepler found out was that there was a relationship between that orbital periods squared and the cube of the orbital distance. And that's given by this equation right here. Four pi squared or cubed over G. M. Now, for those of you who have seen satellite motion equations, you've actually seen this before, so it's nothing new. So basically, what Kepler figured out was that the relationship between our anti Onley depends on the big M, the mass of the planet and not the mass of the satellite. So it only depends on the mass of the thing that you're orbiting. And so, in questions you're gonna be given, um, you might be given in orbit problems like two out of these three variables. So you can always use Kepler's third law to figure out what the other one that you're missing is. You'll be given m r T and they will be asked to figure out some combination. Of those, two are of those three, and that's basically for the equations. So how did he actually come about with this? How did you actually figure this relationship out? Well, he was looking at the characteristics of orbits in our solar system. So, for instance, I was looking at the orbital distance of the Earth, which I'm gonna call are one and the orbital distance of a different planet. This example of just using Mars are too. So if the orbital period of the earth is t one and the orbital period of the Earth 62 he noticed that because these two things orbit the same object, which is the sun. The ratios of our cubed over T squared is a constant. And because these two things over the same thing, let's see how that comes about from the equation. So t squared equals four pi squared r cubed over GM now I just want to isolate r cubed over t squared. So I'm gonna go ahead, move everything over to the left so that's gonna be g m t squared over four pi squared equals r cubed Now just gonna bring the t squared down So what happens is I get that are cubed over t squared is just gm over four pi squared. So notice how insecure Asian Everything on the right side is a constant. So if you just plugged in a number for this, you would just get a constant value Just get a number And that number depends on basically it's proportion to the mass of the sun. So if we had these two objects right here that if you took the radius the orbital distance of the earth cubed, divided by the period squared, you would just get some number here. And if you do the same exact thing for Mars, you took our two cubed over t squared, you would get the same exact number. So in problems, especially when you're asked to compare the orbital characteristics of two different um uh, two different things. You can use this ratio here to basically solve for whatever you're missing. So a lot of problems will be given to three out of these four variables, and you can use this ratio to solve for the other one. And when you are doing that comparison, thes units don't necessarily have to be an s I But it's on Lee when you're actually comparing the orbits. So that means if you were like, if you were to have this in terms of, let's see, if you were to have the our distance in terms of like a you and the t in terms of, like days or something like that, then this would also have to be an AU, and this would also have to be in days, so they have to be consistent. Otherwise, you can't do that. That's basically it. Guys, let's go ahead and take a look at an example in our own solar system. So we're gonna we're gonna be calculating the mass of the sun using the characteristics of Earth and Mars. So, in part A, we're gonna be calculating what M son is using the Kepler's third law equation. Right? So we've got Kepler's third Law and really use T Earth squared in this in part a equals four pi squared our earth cubed over g times m son. Now if I want em, son, I just need to isolate it. So I want to bring this over to the other side. But then t Earth has to come down. So they try. They just trade places. So I've got em, Son equals So I've got four pi squared our Earth Cube divided by g times t earth squared. Now I just have to go ahead and get those with those numbers are so our earth here is I'm told 450 million kilometers because I'm actually calculating something and not just comparing two things. I need that in s I So I need r e the radius of the earth, the orbital radius Not the radius of the sun or anything. The orbital radius normal distance is 1.5 times 10th, 11th and that's meters Now have a t earth were totally the earth orbits the sun once a year so that is equal to 365 days. But we need that in seconds. We need that in s I units, so I have to do is have to multiply by 24 hours in every day and then times 3600 seconds in every hour. So you got to do that to basically get what seconds are. And if you do that, you should get 13.15 times 10 to the seventh and that Z, that's in seconds. So I'm just gonna go ahead and plug everything in. So we got em, son equals four pi. Whoops. So M son equals four pi squared, then I've got 1.5 times 10 to the 11th. I've got a cube that divided by 6.67 times 10 to the minus 11 and then multiplied by 3.15 times 10 to the seventh. And then you gotta square that Just make sure that you're entering everything your calculator correctly, and you should get to 01 times 10 to the 30. And that's gonna be in kilograms because you're talking about the mass of the sun. So that's what we get when we used the Earth data. Now, how about we use Mars is okay, Well, we already have this, um, this equation right here. Let's just start with that. So now we're going to do em, son. Except instead of using Earth, we're gonna use our Mars cubes now over G, then t Mars squared. So you go and do this. Um, Now you're gonna get that the radius of the orbital radius of Mars, the orbital distance is or four pi cubes, and then this is gonna be 2.28 times 10 to the 11th. Because now we have 228 million That's gonna be cubed. Divided by that's gonna be 6.67 times 10 to the minus 11. Then we if we get the 687 days and we go through the same exact process that we did hear, what we're gonna get is we're gonna get 5.94 times 10 to the seventh, and then we've got a square. That And if you do that, you should get the mass of the sun using the Earth's or using the using marshes data, which is 1.99 times 10 to the 30. So because these two things orbit the same object the sun, we could actually get the really close. We could actually the mass of the sun, um, using both of those things and it's really close

2

example

Jupiter & Neptune's orbits

Video duration:

3m

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Hey, guys, let's take a look at this problem together. So we've got Jupiter and Neptune orbiting the sun, and I'm gonna draw a quick little diagram. What's going on? So the sons in the center, I'm gonna have the orbit of Jupiter out here like that, and I'm gonna have the orbit of Neptune. That's a little bit farther out. Obviously, it's not to scale. So I'm told that if Jupiter is some distance here that Jupiter orbits once. So in other words, the orbital period takes 11.86 years. I'm gonna write that here 11.86 it's at a distance and orbital distance, which I'll call R. J. And so we're told, if Neptune is out here, the Neptune orbits around the sun in some time, which I'll call t n. And it orbits the sun at some distance r n so were asked to find out how long it actually takes to orbit in years. In other words, our target variable is Tien. So what are we working with? We're working with the mass of the sun thing that's in the middle, the orbital distance. It's R and T. So we're gonna use Kepler's third law. Kepler's third law says that there is a relationship between our cube and T squared, and that's just a constant. And specifically, if you have two objects that are orbiting the same thing, you can set up a ratio between the two. So Kepler's Third Law and Ratio form says that if you take the RJ cubed over T j squared, that's just gonna equal a number. And it's the same number as if you were to grab our n cubed divided by T n squared. By the way, remember that these things don't have to necessarily be in S I units. As long as you have this ratio set up, then what happens is as long as these units right here years and a you are consistent, then you can set up thes ratios together. So, in other words, I'm just gonna go ahead and start isolating for T N Square. That's my target variable. So if I have this thing on the bottom, I could either cross multiply or I could just flip the fractions. Let's do that. I'm basically just gonna flip these upside down, so I got t j squared over r J Cube equals t n squared over r n cubed poops R n cubes. Now I've just gotta move this RN over to the other side And then I've isolated TN so I've got to t j squared r n cubed over r j cubed equals t n squared. So, in other words, what is the orbital period of Jupiter? Well, I'm told that that is 11.86 years, and I'm just gonna square that. And then I've got the orbital distance of Neptune, which is 30.0 point 11 a use I've got a cube that and then divided by five point to a use. And also cube that notice how all the units are consistent. And so what I'm gonna get is I am going to get to seven times 10 to the fifth. But remember, I have to take the square roots because I have t n squared. So really, what happens is, uh, the orbital period of Neptune in years, eyes expressed as 1.1 65 25. And that's in years. All right, you can actually look this up, and this is pretty close to what the actual orbit is. 165 years. Let me know if you guys have any questions with this

3

Problem

Problem

Io and Ganymede are two of Jupiter's four Galilean moons. Io orbits at an average distance of 422,000km in 1.77 days. What is Ganymede's average orbital distance (in km), if it takes 4 times longer to orbit Jupiter?

A

670,000

B

1,063,000 km

C

167,000

D

844,000

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