Kepler's Third Law: Study with Video Lessons, Practice Problems & Examples

Kepler's third law establishes a relationship between the orbital period (T) and the orbital radius (r) of celestial bodies, expressed as T^2 = 4π × r^3 / GM . This law indicates that the square of the period is proportional to the cube of the radius, allowing calculations of mass for orbiting objects. The ratio r^3 / T^2 remains constant for objects orbiting the same mass, facilitating comparative analysis of different celestial bodies.

Kepler's Third Law

Hey, everyone. So in this video, we're going to cover Kepler's third law in a little bit more detail. It's the most important of Kepler's laws, and you'll definitely need it to solve orbit problems. So let's go ahead and take a look, and we'll do a good example together. Kepler's 3rd law says that for any circular orbit, there's a relationship between the orbital period and the orbital radius or the distance. So basically what this means here is that this r, this orbital distance here, is related to the amount of time it takes for a satellite to go around. Remember, a satellite could be just basically anything that orbits something else. What he found was that the relationship of t 2 is proportional to r 3. Proportional just means there's some other constants that go in front. And so this is actually the equation. It's 4π2 r3 /(gm). Now for those of you who've seen satellite motion equations, it's okay if you haven't, but you actually should recognize this equation as the orbital period equation. And this is basically Kepler's 3rd law. So what he found was that the relationship between r and t, the radius and the orbital period, depends only on the big mass, the mass of the thing that you're orbiting, not the mass of the satellite itself. Alright? So usually, the way these problems are going to go is that you're going to be given 2 out of these 3 variables, some combination of m, r, or t, and you can use this equation to find whichever one is missing. Alright? So let's move on here because this is another type of equation, it's also pretty popular. It's when you have 2 objects that are orbiting the same mass. This is actually how Kepler sort of discovered this in the first place. So, for example, if you have the Earth that's at some radius, some orbital distance, it has some it takes some time to go around in its orbit which I'll call t one. If you look at another object, another planet in our solar system like Mars, for example, in which r two is greater than r one, what he found was that the orbital period was also longer. So t two was greater than t one. It's not exactly sort of, like, you know, double the distance, double the time. There's a little bit exponents involved, but this is basically what he found is that the farther out you go, the longer it takes for you to go around. Alright? So what also he found is that if you have 2 objects that are orbiting the same mass, so in other words, 2 objects that are orbiting something like the sun, then the ratio r3 /t 2 is going to be a constant. I can actually show you where that comes from using the equation. So we have that t2 =4π2 r3 /(gm). So if you rearrange for this, you have (gm)t2 /4π2 equals r 3. Then if you go ahead and solve, you're going to get r3 /t 2 is equal to GM / 4π2. Notice how G is a constant, This is just a number. So basically, the ratio of r3 /t 2, if you just plug in a bunch of numbers, this is just going to turn out to a big number and it only depends on the mass of the thing that you're orbiting. Alright? So what he also found is that if you have 2 objects that are orbiting the same thing, like for example, Earth and Mars both orbit the sun, then this number should also be the same. So basically this ratio is going to be equal for 2 objects that are orbiting the same mass. And so we can set up another sort of proportion or ratio here, which basically says that r3 /t 2 for one object is going to equal r3 /t 2 of another object, and that's assuming that they both orbit the same thing. Alright. So usually those problems are going to go. They're going to give you 3 out of those 4 variables and you'll be using this to solve for another. One last point here is that these units can be non SI when you're doing this sort of comparison like this as long as they're consistent with each other. And basically what this means is that if you're given something like days, or sorry, if you're given something like AU for this number and you're given, like, days for this number, both of those are non SI, that's fine. You don't have to sort of you don't have to, like, recalculate everything and turn everything into SI as long as these numbers are also given to you in AU and days. Alright? So as long as they're consistent, you can sort of do that without having to sort of, to convert everything. Alright? Let's go ahead and take a look at an example here. So we have the Earth that's orbiting the sun once a year, and we have a distance of 150,000,000 kilometers. And we're going to go ahead and calculate the mass of the sun by using the Earth's data. Alright? So in other words, we want to calculate what m sun is, and we're going to go ahead and use Kepler's third law. Alright? So the Kepler's law said that t2 =4π2 r3 /(gm), and we're going to have gmsun. But now we want to use the Earth's data and we want to use the Earth's radius. Okay? So all we have to do is to isolate this m sun. We just got to rearrange. This m sun goes up top. This t 2 goes on the bottom. So in other words, your m sun is going to be 4π2 r3 /t 2. Alright? So now these distances here are not given as SI units, 30 kilometers, and we also don't know what this once a year means. We're also given some stuff in days. Now because we're not sort of comparing these 2, we do actually have to put everything in SI. We're calculating the mass of the sun. We're not doing some sort of comparison. Right? So you have to put everything in SI. So let's go ahead and do that. Right? So this r earth cubes or sorry, r earth is going to be 150,000,000 kilometers. So that's 150 1, 2, 3, 4, 5, 6. So this is going to be 1.5 times 10 to the 8th, that's kilometers. So this is going to be 1.5 times 10 to the 11th meters. Right? You have to shift the decimal place over. Now for t Earth, this is given as what? We actually don't know, but we know sort of by everyday life that if it orbits once a year, one year for us is 365 days. So we have 365 days, but this is not SI, so I have to turn it into seconds. So how do I do this? I have to multiply by 24 hours, per one day. And then you can do 60 minutes and 60 seconds, or you can just sort of, say that there's 3600 seconds in 1 hour. Alright? So then your days cancels, your hours cancels, and you just end you end up with 3.15 times 10 to the 7th seconds. Okay? So that's what I plug into my numbers over here. So I've got 4π2, don't forget the pi is squared, times 1.5 times 10 to the 11th cubed divided by remember, g is 6.67 times 10 to the minus 11, and this is going to be 3.15 times 10 to the 7th, and you're going to square that. Alright? So there's a lot of exponents. Just be really careful when you're plugging this into your calculator. Make sure you sort of, like, put everything in parentheses and all that stuff, and what you should get is 2.01 times 10 to the 30th. That's in kilograms. Alright? And you look at this up, by the way, this is very close to what the actual mass of the sun is. Okay? So now we're going to do the same exact thing here, except now we're going to calculate the mass of the sun by using Mars' data. So I'm just going to go ahead and rewrite this same equation here, but now we're going to plug in the numbers for Mars. This is going to be r m cubed divided by, this is going to be oops. Sorry. This is g here. Yeah. This is going to be g times t Mars squared. Alright? So the only thing that's different here is I just need basically need to do to get the same exact numbers for Mars, and I'm just going to put this over here. So, your r Mars is going to be, well, 150,000,000 kilometers turned into 1.5 times 10 to the 11th. So 228,000,000 kilometers is going to be 2.28 times 10 to the 11th. Right? So the same pattern. And then for t Mars, what you're going to get is if you do the whole, you know, 687 days and then you do the conversion, what you end up with is you're going to get you're going to get 5.94 times 10 to the 7th in seconds. Okay? So that's what you plug into these numbers here. So you got 4π2. This is going to be 5 points oh, sorry. This is going to be 2.28 times 10 to the 11th cubed divided by 6.67 times 10 to the minus 11. And this is going to be 5.94 times 10 to the 7th squared. Just a lot of numbers, a lot of plugging in. What you should get here, by the way, is 1.99 times 10 to the 30th kilograms. Alright? So, basically, we've got the same exact number. We were using pretty rough numbers in here, you know, like 687, and there's also some rounding errors. But, basically, what this shows us is that using 2 objects orbital distance and periods, we get the mass of the sun, which was, by the way, the same exact number. So it's about 2 times 10 to the 30th. That's it for this one guys. Let me know if you have any questions.

Jupiter & Neptune's orbits

Hey, guys. Let's take a look at this problem together. So we've got Jupiter and Neptune orbiting the sun, and I'm going to just draw a quick little diagram of what's going on. So the sun's in the center, and I'm going to have the orbit of Jupiter out here like that. And I'm going to 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 going to write that here: 11.86. And it's at a distance, an orbital distance, which I'll call RJ. And so we're told, if Neptune is out here, then 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 we're asked to find out how long it actually takes to orbit in years. So in other words, our target variable is T_N. So what are we working with? We're working with the mass of the sun, the thing that's in the middle. The orbital distance is R and T. So we're going to use Kepler's third law. Kepler's third law says that there's a relationship between r cubed 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 in ratio form says that if you take the R_J cubed over T_J squared, that's just going to equal a number, and it's the same number as if you were to grab R_N cubed divided by T_N squared. And by the way, remember that these things don't have to necessarily be in SI units. As long as you have this ratio set up, then what happens is as long as these units right here, years and AU, are consistent, then you can set up these ratios together. So, in other words, I'm just going to go ahead and start isolating for T_N squared. That's my target variable. So if I have this thing on the bottom, I can either cross multiply or I can just flip the fractions. Let's do that. I'm basically just going to flip these upside down. So I got T_J squared over R_J cubed equals T_N squared over R_N cubed. Oops. R_N cubed. Now I've just got to move this R_N over to the other side, and then I've isolated T_N. So I've got 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 going to square that. And then I've got the orbital distance of Neptune, which is 30.11 AUs. I got to cube that and then divide it by 5.2 AUs and also cube that. Notice how all the units are consistent. And so what I'm going to get is I am going to get \(2.7 \times 10^4\). But remember, I have to take the square root because I have T_N squared. So, really, what happens is, the orbital period of Neptune in years, is expressed as 165.25, and that's in years. Alright. 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.

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

Here’s what students ask on this topic:

What is Kepler's Third Law and how is it used in orbital mechanics?

Kepler's Third Law states that the square of the orbital period (T) of a planet is directly proportional to the cube of the semi-major axis (r) of its orbit. Mathematically, it is expressed as:

$T^2=\frac{4{\pi }^2{r}^3}{GM}$

where G is the gravitational constant and M is the mass of the central object. This law is crucial in orbital mechanics as it allows us to determine the mass of celestial bodies and compare the orbital characteristics of different objects orbiting the same mass.

How do you calculate the mass of the Sun using Kepler's Third Law?

To calculate the mass of the Sun using Kepler's Third Law, you need the orbital period (T) and the orbital radius (r) of a planet. The formula is:

$M=\frac{4{\pi }^2{r}^3}{G{T}^2}$

For Earth, r = 1.5 × 10¹¹ meters and T = 3.15 × 10⁷ seconds. Plugging these values into the equation, you get:

$M=\frac{4{\pi }^2{1.5}^3\times 10{11}^3}{6.67\times 10{-11}^{}{3.15}^2\times 107^2}$

This results in approximately 2 × 10³⁰ kilograms, which is the mass of the Sun.

What is the significance of the ratio r³/T² in Kepler's Third Law?

The ratio r³/T² is significant in Kepler's Third Law because it remains constant for all objects orbiting the same central mass. This ratio is given by:

$\frac{r^3}{T^2}=\frac{G}{M}$

where G is the gravitational constant and M is the mass of the central object. This constancy allows for comparative analysis of different celestial bodies' orbits, making it easier to study and understand the dynamics of planetary systems.

How does Kepler's Third Law apply to non-circular orbits?

Kepler's Third Law applies to elliptical orbits by considering the semi-major axis (a) instead of the orbital radius (r). The law is expressed as:

$T^2=\frac{4{\pi }^2{a}^3}{GM}$

where a is the semi-major axis of the ellipse. This adaptation allows Kepler's Third Law to be used for any elliptical orbit, maintaining the relationship between the orbital period and the size of the orbit.

Can Kepler's Third Law be used for objects orbiting different masses?

Kepler's Third Law can be used for objects orbiting different masses, but the constant ratio r³/T² will differ based on the mass of the central object. For objects orbiting the same mass, this ratio remains constant. However, when comparing objects orbiting different masses, you must account for the different values of M in the equation:

$\frac{r^3}{T^2}=\frac{G}{M}$

This means that while the law is universally applicable, the specific ratio will vary depending on the mass of the central object being orbited.