Alright, guys. So in this video, I want to talk about a special kind of satellite orbit that you might see in some problems called a synchronous orbit. So let's go ahead and check it out.
Imagine we had a satellite above the earth and it was at some orbital distance \( r \). Right? And as it's in its orbit, it's traveling around with some orbital period, \( t_{\text{sat}} \). We've got a bunch of equations here that'll give us \( t_{\text{sat}} \). What happens is this thing is orbiting around the earth, but the earth is also spinning, at a special distance called \( r_{\text{synchronous}} \) such that the satellite's orbital period, \( t_{\text{sat}} \), will synchronize with the earth's rotation so that \( t_{\text{sat}} \) is equal to \( t_{\text{earth}} \). By the way, this equation works for any planet, not just earth. If you were to actually go outside and look at a satellite at this specific distance, then as the earth is rotating, the satellite is constantly above the same place, because the orbit period of the satellite matches the rotation period of the earth so that they are constantly synchronized. This thing would basically appear above the same place at all times. Instead of synchronous orbit, you might see stationary orbit. We use these all the time in telecommunications because we want satellites to appear above the same place at all times.
This special distance, \( r_{\text{synchronous}} \), has an equation. It's the cube root of \( \frac{GMT^{2}}{4\pi^{2}} \). I just want to remind you that this \( T \) is just the \( T \) of the planets and we can actually use it on any planet, not just the earth. We can actually see where this equation comes from pretty quickly from this \( T^{2} \) equation that we have: T 2 = 4 π r 3 G M So we just want to solve for this \( r \) right here. So move everything to the other side. We get \( GMT^{2} \) divided by \( 4\pi^{2} = r^{3} \), and all you have to do is take the cube root. So then you'll get to this equation right here. Right?
Cool. So what I want to point out is that this is the only distance where a circular geosynchronous orbit is possible. Because what happens is if you were to increase this \( r \), then your \( T \) would increase right here and it wouldn't be in sync with the earth's rotation anymore. There's only one specific distance here that's possible, and you can also make sense of that because all of these letters are capital letters. They're all big letters. They're all constants. So this is just a constant right here. Alright.
That's basically it. Let's go ahead and actually solve for what the height of earth's geosynchronous orbit is. So if we want to figure out what \( h \) is equal to, well, we know if we're ever solving for \( h \), first, we have to go ahead and solve for \( r \) first, and then we can use \( R+h \) to solve for that. What am I using? I'm using the synchronous orbit equation which is right here. So let's go ahead and write that out. I've got \( r_{\text{synchronous}} \) equals the cube root of \( G \) and I'm going to use the mass of the earth and then I've got \( T_{\text{earth}}^{2} \) divided by \( 4\pi^{2} \). Okay. The rotation period of the earth is the amount of time that it takes to spin once, which you should recognize as Earth's day. That's 24 hours. Right? It takes 24 hours for the earth to spin once. But we want that in seconds, so we just have to multiply that by 3600 and we get that that's equal to 86,400.
So now just plug everything in and we should give our synchronizes. So \( r_{\text{synchronous}} \) is equal to the cube root of a whole bunch of letters and numbers. \( 6.67 \times 10^{-11} \), and we've got \( 5.97 \times 10^{24} \), and then we've got to do \( 86400 \), but we got to square that. And now, we just divide it by \( 4\pi^{2} \). Just make sure that you plug all of this stuff carefully into your calculator. \( 4\pi^{2} \) should be in parentheses. And what you should get is you should get \( 4.22 \times 10^{7} \). But we're not quite done yet because again, we've only solved for \( r_{\text{synchronous}} \). We want the height. So now to do that last step, we just have to get \( h \) is equal to \( R - r \). We get that from this equation right here and you should get \( 4.22 \times 10^{7} \) minus the radius of the earth which is \( 6.37 \times 10^{6} \). And if you do that, you should get \( 3.59 \times 10^{7} \), and that's in meters. So if you actually put this into kilometers, you're gonna get 35,900 kilometers. You can actually Google this - the height of a stationary orbit. You're going to find out it is 35,900 kilometers.
Let me know if you guys have any questions.
