1

concept

## Geosynchronous Orbits

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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 synchronous orbit. So let's go ahead and check it out. So imagine we had a satellite above the Earth and it was at some orbital distance, are right, and as it's in its orbit, it's traveling around with some orbital period t sat. We've got a bunch of equations here that'll give us tea Sat. What? What happens is this thing is orbiting around the Earth, but the earth is also spinning, and the rotation period of the earth is called T earth. That's the amount of time that it takes for it to spin. Once the idea of a synchronous orbit is that there's a special distance called our synchronous such that the satellite's orbital period T sat will synchronize with the Earth's rotation so that T SATs is equal to t earth. Now, By the way, this equation works for any planet, not just Earth on DSO. What happens is if you would 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 off the satellite matches the rotation period of the earth. So they're constantly synchronized. So this thing would basically appear above the same place at all times. And so sometimes, instead of synchronous orbit, you might see stationary orbit. We use these all the time in telecommunications because we want satellites to peer above the same place at all times. So this special distance here are synchronous has an equation. It's the cube root of G m t squared, divided by four pi squared. Now, I just want to remind you that this t, um is just the t of the planets and we can actually use it in any planet, not just the earth. We can actually see where this equation comes from. Pretty quickly from this T squared equation that we have, we have t squared equals four pi squared R cubed, divided by gm. So we just want to solve for this are right here. So move everything to the other side We get g m t squared divided by four pi squared equals r cubed and I have to do is just take the cube root So then you'll just get to this equation right here, right? Cool. So I want to point out, is that this is the Onley distance are where circular geosynchronous orbit. It's possible because what happens is if you were to increase, this are then your tea would increase right here, and it wouldn't be in sync with the Earth's rotation anymore. So there's only one specific distance here that's possible. And you could also make sense of that because all of these letters are capital. Letters are all big letters are all constants. So this is just a constant right here, all right, It's basically it. Let's go ahead and actually solve for what the height of earth geosynchronous orbit is. So if we want to figure out what the height is, we're trying to figure out what H is equal to all right, But we know forever solving for H first, we have to go ahead and solve for little our first, and then we can use big R plus h to solve for that. Now 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 our synchronous equals the cube roots of G and I'm gonna use the mass of the earth. And then I've got t of the earth squared, divided by four pi squared. OK, well, what is that? T Earth? What is the rotation? Period of the earth? Well, we've got that. We've got all of these other letters we got G and m e. The rotation period of the earth is the amount of time that it takes to spin once once you should recognize that as Earth Day, that's 24 hours, right? Takes 24 hours for the Earth to spend once. But we want that in seconds. So we're just gonna have to multiply that by 3600 and we get that that's equal to 86,400. So now I just plug everything that we should give. Our synchronous is so our synchronous is equal to the Cube roots of a whole bunch of letters and numbers. 6.67 times 10 to the minus 11. And we got 5 97 times 10 to the 24 and then we've got to do 86,400. But we gotta square that and now we just divide it by four pi squared. Just make sure that you plug all of this stuff carefully into a calculator. Four pi squared should be in a little a parentheses. And what you should get is you should get 4.22 times 10 to the seventh. But we're not quite done yet because again, we've Onley starved for our synchronous We want the heights. So now to do that at last step, we just have to get a check is equal to r minus. Are we Get that from this equation right here. And you should get 4.22 times 10 to the seventh, minus the radius of the earth, which is 6.37 times 10 to the sixth. And if you do that, you should get 3.59 times, 10 to the seventh, and that's in meters. So if you actually put this into kilometers, you're gonna get 35,900 kilometers. You could actually google this the height of a stationary orbit. You're gonna find out it is 35,900 kilometers. That's how far it needs to be so that this satellite orbits every day, just like the Earth spins. Let me know if you guys have any questions with this.

2

example

## Find Mars' period, given synchronous satellite

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Alright, guys. So, in this problem we're asked to calculate with a period of Mars rotation is assuming that we have a satellite in synchronous orbit. So let's go ahead and draw the diagram real quick. Got Mars right here. I'm gonna have a synchronous orbit around it. So I'm told that there is a satellite here at some some distance. I'm told that there's a satellite and that satellite is in synchronous orbit and it has a velocity. The velocity is equal to 14. 50. And when I'm supposed to do is figure out what the period of Mars rotation is. So I need to figure out what T planet is, but I'm just gonna write team ours. So that's really our target variable for this problem. What is T Mars? Okay, so we've got a synchronous orbit, right? And so what you need to remember about synchronous orbits is that the period of the satellite that's orbiting it is equal to the period of the planet's rotation. So that's what we need to use for this problem. So t Mars, Where does that? Where does that variable pop up in our equations? It pops up in the synchronous orbits equations. So let's go ahead and start there. So we've got our synchronous is equal to I've got the cube root Then I've got G And then I've got the mass of Mars that I've got the period of Mars squared divided by four pi squared. So I really just solving for what this t Mars is equal to. So let's go ahead and start isolating that variable. I've got a que both sides so I've got our sink Cube is equal to g mass Mars team R squared, divided by four pi squared. And then when I move everything over to the other side, I get four pi squared are sink cubed over g times the mass of Mars and that's equal to t Mars. And that's squared, right? So let's take a look at all of these variables. This is just a constant gravitational constant I have with the mass of Mars is in this table right here. So the only thing I don't know is that orbital distance. I've never told any information about the orbital distance of this thing, so let's see how we can use something else about this problem to solve for that right, if I can figure out what that are synchronous is that I can figure out what team ours is. So how do I figure out what that little are is? Let's see, we've got our equations here and all of them involve our but it will see a lot of them involved T SATs, which is what I'm trying to find or t planet. So I can't use these equations. So let's see, I'm gonna have to use my satellite TV satellite equations and there's two basic ones I can use. I can use the square G ever gm over r. I can use this two pi r over tea. But if I take a look at this right here, one of the variables is the variable I'm trying to look for, which is t and then the other one is are so I can't use that equation because it involves two unknowns. So let's instead start with the V sat equation. So I got V sat equals the square root of G times mass over. Little are I'm trying to find out what this little our is and I actually have with the velocity of this satellite is and then these two are just constants. So let's go ahead and salt for that. Little are so I've got a square, both sides. So I've got ve sat squared equals GM over R. Now, I just gotta basically isolate our by trading places with the visa. So we got our equals. I've got g m over V. Sat squared is equal to then I've got let's see, 6.67 times, 10 to the minus 11. I've got the mass of Mars, right? The massive Mars, which is equal to I've got 6.42 times to the What is that? That is 23 and then I got divided by V Sat square. That's 14 50 that's squared. And if you do that, you should get our is equal to Let's see, I got 2.4 times 10 to the seventh, and that's in meters. So I'm gonna take this our distance. I'm gonna plug it back all the way into this equation and then solve for t Mars. So if I go in to do that and start just plugging in all of these variables, I got four pi squared. Then I've got 2.4 times. 10 to the seventh. I'm gonna cube that. Then you've got a divided by 6.67 times 10 to the minus 11 and then multiplied by six points 42 times 10 to the 23. And if you do that, you're gonna get T Mars squared, which is equal to Let's see, I got 7.83 times 10 to the ninth. But we're not done yet because again, we have that t squared. So if you take the square roots, we're gonna find that the rotation period of Mars is equal to about 88,470 seconds, which is about 24 point six hours. And this is actually the right answer. It's pretty amazing how we use two pieces of information, which is a synchronous orbit and the velocity of that synchronous orbit. We can actually figure out the rotation period of an entire planet. It's pretty cool. Let me know if you guys have any question with this. But that's the final answer. Alright, guys, we'll see the next one

3

Problem

You're on a satellite orbiting an unknown planet. The only property of this planet that you know is that days are 18 hours long. Your onboard sensors show that you're orbiting at 16,000 km above the surface, with a velocity of 3 km/s. You look down and notice that you're always above the same point on that planet as you orbit around it.

Calculate the mass of the planet.

A

5.8×10

^{14}kgB

7.4×10

^{21}kgC

4.2×10

^{24}kgD

6.3×10

^{34}kg