ï»¿ >> Hello class, Professor Anderson here. Let's talk about satellites in orbit. We of course have tons of satellites that orbiting the earth right now. We the International Space Station. We have the global positioning satellites. We have all sorts of stuff that's up there. Kepler is up there. The Hubble. Let's talk about three in particular. One at S1 that is out at a distance 2R from the center of the earth. S2 is at the same altitude, it's also 2R from the center of the earth. And then the third, S3 is at a distance 3R from the center of the earth. Let's see if we can figure out what the period of S3 is compared to some of these others. And before we do that let's tell you the following. T1 is the period of satellite S1. What about T2? What do you think the period of satellite 2 has to be? Yeah. If it's at the same altitude it has to be exactly the same. T2 is equal to T1. It's independent of the mass of those satellites. Okay. Period of T2 is exactly equal to T1. But what about T3? How do we calculate T3? Well, to calculate T3 we need to go back to Kepler's Laws and specifically Kepler's 3rd law. And what Kepler's 3rd law said was the following. And what Kepler's 3rd law said was the following. The period squared is proportional to the semimajor axis cubed, where that proportionality constant is 4 pi squared divided by G capital M where capital M is the mass of the planet or star that you're orbiting. So in this case it would be the mass of the earth. But if it's a circular orbit then A is just the radius. That step is if it's a circular orbit. And so now look what we can say. T squared is just 4 pi squared over GM times R cubed. And I can put any subscripts on it that I want. T1 squared is this stuff, R1 cubed. T3 squared, is this stuff, R3 cubed. And now I can relate these together. How do I relate them together? You can solve the first one for 4 pi squared over GM and stick it into the second one. Or you can just divide the second equation by the first equation. Okay, which is kind of a nice little trick. So I get T3 squared over T1 squared, equals 4 pi squared over GM divided by 4 pi squared over GM. I have an R3 cubed on top. I have an R1 cubed on the bottom. All that stuff in the front cancels out. And I get T3 is equal to T1, times R3 over R1 to the 3 halves. I had a cube but I took the square root and so I ended up with that. R3 in this case is bigger than R1. And so this number is bigger than 1. So T3 is in fact bigger than T1. How much bigger? Well, let's calculate it. What we said here is R1 was equal to 2R. R3 was equal to 3R. So T3 is T1 times 3R, over 2R. To the 3 halves. The R's cancel out and we get 3 halves to the 3 halves. I have no idea what that is. Somebody punch in your calculator what is 1.5 raised to the 1.5? >> 1.84 >> Okay. Let's approximate it and then we'll confirm that you were right. This is the same as 3 cubed over 2 cubed, square rooted. Three cubed is 27. Two cubed is 8. What is 27 over 8? Well, 8 goes into 24, three times. And then I need another 3/8 on top of that. So that is about 3.3. And if I take the square root of 3.3 what do I get? Well, square root of 3 is 1.6 so it's got to be a little bit bigger than that. So we'll say 1.8. And what was the exact answer that you had? >> 1.837 >> 1.837. And this is the power of editing. If I edit this in reverse, I didn't hear his answer first, I guessed this first. 1.837 [laughter]. So this is certainly a bigger number than T1. And it has this magical value in this case. Alright?