3Speak - YouTube - Telegram - Notes - Playlist - Vector Functions - MES Links In this video, I use the equation for Kepler's third law to calculate the required altitude above Earth's equator of a ...
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solution to problem for geosynchronous satellite. So it's possible to place the satellite into orbit about the Earth so that it remains fixed above a given location on the equator. So in others, if you have the Earth right here, and you have a satellite over here, there's the addition, so on, like that, and it's always going to be fixed with it. So as this rotates, this is also moving about it, so you're always seeing it with a telescope. Yeah, exactly the same location always. So it appears to be stationary. So that's fascinating. I'll see if I can find a video after this problem. I'll add some other contexts, as well as this Arthur C. Clark guy, is pretty famous here. So, yeah, so it's possible to place a satellite into orbit about the Earth so that it remains fixed about a given location on the equator. And now we're asked, compute the altitude that is needed for such a satellite, and the Earth's mass, is 5.98 times 10 to the power of 24 kilogram, and its radius is 6 .37 times 10 to the 6 meters. All right, and then this orbit is called the Clark Geosynchronous orbit after Arthur C. Clark, again, the 2001 Space Odyssey guy, who first proposed the idea in 1945. That's, yeah, I think it's end of World War II there. So the first such satellite, Syncombe 2 was launched in July, 1963. So he was way ahead of this time, 18 years later, that's when they launched it. First, there's a fascinating stuff. So solutions, so we adapt the equation from Problem 2C with the Earth at the center of the system. So instead of the Sun, we're doing the Earth, just because relative to the distance, this is much closer to the Earth than it is to the Sun. So this will be... and the mass is much smaller, so it's going around the Earth as opposed to around the Sun, but eventually it's going to go around the Sun as Earth goes. Alright, so we can adapt the equation for problem 2C with Earth at the center of the system, so T is the period of the satellites orbit about Earth, and M is the mass of the Earth, and A is the length of the semi-major axis of the satellites orbit measured from the Earth's center. And again, this is, the calculus book calls this semi-major axis, that's just basically the major axis divided by two. So again, remember, the major axis is 2A, that's major axis, 2A, and then the semi -major is just A. All right, so that's just A there. All right, so that's what we have. All right, so let's take a look at the problem 2C again. So scroll back up to problem 2C, and that is, well, this is problem 2C again, again, it's the same equation right here, the period. So we're going to change this up so that, well, we're just going to be looking at the mass of the earth as opposed to the mass of the sun there. All right, so. And yeah, so I just past it out there for reference. Oh, and now, since we want the satellite to remain fixed about a particular point on the Earth's equator, T, right here, the period of the satellite must coincide with the period of Earth's own rotation, so T is equal to 24 hours, which equals to 86,400 seconds. So basically, yeah, the period of the Earth's spin. Yeah, I mean, at the period of the, if you have the Earth right here and you have the satellite, I'll make it like this, something like that. This has to go around, the same amount of, that's T, has to go around the same amount of time as Earth's rotating T, so these have to be equal, so that you're always going to, if you're on here, and as Earth's rotating, you're always looking at this, so it looks stationary, so that's quite fascinating. All right. So it needs to coincide and we are given that the mass of the Earth is m equals 5.98 times 10 to power 24 kilograms. So, and also I just want to show this just to complete our work, complete, yeah, have completeness of our work. So we have a period as 24 hours. How they get this is basically times it by 60 minutes in an hour, and then times it by, well, 60 seconds in a minute. So basically times 60 squared, and this equals to 86,000 400 seconds. And I'll do a calculation check soon. So what we have here, yeah, what do we have here? Well, we have, this is the period that we need to have. So in other words, we need to find the length of the semi-major axis, basically the major axis divided by 2. So we'll solve this. That's going to be, well, the cube root of that. So this is going to be a. which is going to equal to move everything over to this size, so GM go on top, this one's at the bottom, so we'll have a, uh, I'll put this, uh, t squared, or actually instead of the t square, I put the GMT as we did before, so g mt squared, divide this by, uh, yeah, divided by four pi squared. And then we're going to do the cube, uh, root right here. So in other words, 1 over 3. So you could do that, or you could also write it like this. You could also write, I think, like this, 3, like that. I'll put both just for a completeness. And then this equals to GMT squared divided by 4 pi squared, and then this is 1 over 3, like that. Just where we wanted to write both just for a... mathematical, uh, all right, so now this is going to be approximately equal to when we put in the values. So let's put these values all in here. And, uh, the top part is going to be, uh, well, we have the G. G is going to be, again, the same thing as, uh, as before, 6.67. that is this right here, G, 6.67 times 10 to the negative 11, Newton's meters per second. I'm a Newton meters squared, divided by a kilogram square. So, 6.67, I'll put this nicer. All right, 6.67 times 10 to the negative 11, Newton's meters squared over kilograms, squared, all right, and then this is going to be like that, GM, and then mass of the Earth is 5.98 times 10 to the 24 kilograms. All right, and then we have the last one is going to be, yeah, is the period squared, so that's just going to be 86,000, 86,400. 2nd, 86,400 seconds, and then all squared, and then this is going to be, well, put a close bracket, divide this by 4 pi squared, and then this is going to be power of 1 cubed. All right, and now we plug this into the calculator, and then again all the units, just they all cancel, we're going to get the meters, as we did over above in Problem 3, so we just get the meters. like that, and then a number of Newton's is kilogram per meters, times meters a second squared, and then what we get is kilograms, kilograms, kilograms divided by kilogram squared, cancels, and we get a meter squared, and then the seconds cancels with this second squared, and all we're left with is, yeah, meters cubed, and then we get this cube root, so we get meters. So this is going to be approximately equals two, when you plug in the calculator, this is going to be four. I'll put the A there. So thus A is the semi-major axis, 2.3 times 10 to the 7 meters like that. All right, so now that we have this length of the semi-major axis, or basically major axis divided by 2, let's do a quick calculation check, and then we'll go further. So calculation check, so again, we have this part here 24 times 60 times 60. plug that in, this per space equals 86, 400. And then over here, the 6.67 times 10 to the negative 11 times 5.98 times 10 to the 24 times 86, yeah, 1,400 squared, and then all times it by, yeah, that's this part squared, and then all times it by, or divided by 4 pi squared, and then a cube root, and again, you can just use this giant formula, and you get the answer. all nice and tidy there, equals to, yeah, 4.2250 and then, et cetera, e7. In other words, 4.23, rounded up, round is 5, 3, and then times 10 to the negative, times 10 to the 7. So again, fascinating stuff. Alright, so now going further, so if we assume a circular orbit, the radius of the orbit is A, so again, assuming it's not an ellipse, it's a circle, again, just because, uh, relative to the Earth, the orbit isn't that much bigger. than the Earth, it's still bigger but not as big. So the radius, and also for simplicity for this video, I'm pretty sure the actual satellites will be more accurate. So anyways, if we assume a circular orbit, the radius of the Earth, radius of the orbit is A, and since the radius of the Earth is... Yeah, radius of the Earth is 6.37 times 10 to power of 6 meters. The required altitude above the Earth's surface for the satellite is, and basically what I'll do, I'm going to draw this out now because it's a circle. And yeah, it's a circle. So what I'm going to do, I'm going to try to make it more accurate, just because this is 10 to the 7 is the semi-major axis, in other words, the radius, because we're assuming a circle. And again, just for simplicity's sake, assuming a circle, and also because Earth is pretty big, and the satellite isn't that far away, I'm assuming. But regardless, I think the actual astronomers would have the satellite at more accurate elliptical orbit. What I'm going to do now is I'm going to insert a circle. I'm trying to make this accurate. Alright, so we have that, and I'm going to go right here and try to make this as accurate as possible. So that is a circle. I'm going to make it a bit more circularly set up. Okay, so it's going to be like this. I'll make it bigger. Anyways, it's not fully around, but I just wanted to make it looks decent enough. So now, and the reason is this is times 10 to the 7. This is 10 to the 6. So basically 10 times bigger is the semi -major axis. So Earth is going to be something like this. It's going to be pretty small, actually. I'm just going to draw it a bit. It's not going to be fully accurate, but there's the Earth. It has radius R. I'll go like this, this says, yeah, it all has radius R, and now the length, the major axis of the satellite's going to be like this, this is A, like that, and then the satellite's going to be somewhere here, it's again, not the scale, but it's going to be something like this, and then it goes all the way around, yeah, so it goes all the way around, and then the period is going to be T has to equal to, well, 24 hours, I'll put hours like that, and this Earth is rotating. Yeah, I'll put this, how am I going to do this? Yeah, I'll just move this R out of there. I'll make this a bit bigger, just so we can fit this in. So this is going to be our R, and now this is going to be rotating, well, rotating the Earth's rotating just at 24 hours. So again, that's Earth's a Earth's spin. So these have to be equal, and then basically, yeah, that's the to scale, roughly to scale, it's going to be a bit, actually even further out, because this is, it's not going to be 10 lengths, yeah, it's going to be around 10 lengths, but anyways, I've made it like that, and we know the Earth, and so on, so the required altitude is, so we know A, so basically, the record altitude is going to be from here to here. And, or I'll make it from here, I'll make it here, so this is going to be, and move this here. So that's going to be the required altitude, and that's going to be A minus the radius. All right, so required altitude is going to be A minus R, which equals 2, and we know what A is, A is above here, 4.2 3 times 10 times 10 power of 7. 4.23 times 10 to the 7 meters, subtracted by the radius that are 6.37 times 10 to the 6.6.37 times 10 to the power of 6 meters. And also this is going to be approximate because we're using rounding up numbers and all that. So this is going to be approximately equal to, well, when you plug this to the calculator, we've got 3.59 times 10 to the... power of 7 meters. All right, so it becomes like that, or, and then divided by, again, by 1,000, so this is going to be basically divided by 1,000, we're going to get 3 .59 times 10 to the power of 4 kilometers. Again, it's the standard units right there in metric system. Yeah, like that, or this is going to be three, I'll make this in actual full numbers, so it basically four decimal place, so we're going to get three, five, and then we have nine zero, kilometers. So 35,000 kilometers, 35,900 kilometers altitude, and this is going to be altitude. Yeah, altitude above Earth, all right, just above. I'll say above equator. All right, let's circle this. Alright, and now let's do a cool calculation check, so we have, basically, calculation check 4 .23 times 10 to the power of 7, and this part right here, minus 6.3 -7 times 10 to power of 6, and then usable built-in one-note calculator space, yeah, equals 3.593, or 3.59, just rounded down, times 10 to power of 7. There's E7 to 7, and then divided by 1,000. We just put that in, divided by 1 ,000. equals that, and just put a space here like that. All right here, I quickly just added an animation from Wikipedia, so here's animation, not the scale of a geosynchronous satellite orbiting the Earth. So if I click that over there, I already have it open, so we just go over here, geosynchronous orbit on Wikipedia, and here's animation, not the scale, showing geosynchronous satellite orbiting the Earth. So there it is, it looks like that, and it's going around it, and as it's basically following along the Earth's rotation. And it's always going to be pointing around this point on the equator there. But again, we use 24 hours, it's actually Earth's rotation is about 23 hours, 56 minutes, and 4 seconds, and that's one sidereal day there. That's basically a fixed point like that. And if you make it solar day, basically, we're accounting for the sun movement. it's going to be, I think you add, yeah, you add another minute there. That's why it's 24 hours, there, this is 23 hours, 56 minutes. All right, and then I just want to make a point right here. So over the course of a day, the object's position in the sky may remain still or trace out a path, typically in a figure 8 form. whose precise characteristics depend on the orbit's inclination and eccentricity. Again, that's with the whole ellipse, because we just assume the circular orbit. So a circular geosynchronous orbit has a constant altitude of 35,786 kilometers, and that's pretty close to ours. We had 35,900. So again, pretty close there. And then next thing I'm going to do is I want to click over here. So if I click here, Figure 8, the reason it's Figure 8, is, again, as Earth's rotating around, you're going to have a shape like that. This is a geosynchronous shape like a figure 8. And again, these are called an alema. Let's go back up to here. So you also have that for the sun here, so a diagram of an anolema, and I'll hear this pronunciation of it. Let's say this. Anolema. Yeah, so that's an analema, all right. And, yeah, so this one's diagram of an an analema looking east and a northern hemisphere. The dates of the sun's position are shown. This analamas is calculated for 9 a.m. not photographed. Oh, yeah, so here's the dates right there. This would be, I think, this is June 21st, and then this would be, yeah, this is December right here. Okay, this is December 22nd. Yeah, so notice of December 22nd, it's at the lowest, and these are 9 a.m. photos, so the sun's lowest in the winter, and then it's highest in the summer, and notice that's why it goes up, and then because Earth's moving around and tilting and all that, you're going to have this weird shape like up. All right, and you also have an analema of the moon, and you could, if you do a search right here, see where it shows up on the moon. So there's an alamma of the moon. Apparently, more people have walked on the moon than have captured the an ala. Interesting. So if I go here, this is credit Rich Richens. All right, that's quite interesting name there. So anyways, this is July 20, 2005. Yeah, I mean, July 13th, 2005. So you can see right here. So there's a full, I mean, a new moon there, it's a hiding, and then you see the crescent, it gets bigger and bigger, and this is going behind the Earth. And you can see that different shape, like that's, like, that's pretty amazing there. And I mean, behind the Earth, I mean, the sun is going to be opposite of this. So as the sun's opposite, and it, you're going to have this refraction and so on, that's red. more the red light is showing on there. So it gets bigger, so it notice to figure it. So again, absolutely fascinating stuff. So let's go back over here. All right, so now I'm going to end this video with just some info on Arthur C. Clark, again, this guy is everywhere. So according to Wikipedia, first of all, so more info on Arthur C. Clark. So according to Wikipedia, Herman Podknick, described both geosynchronous orbits in general in 1929. The first popular literature to mention that geosynchronous orbit was in 1948 by George O. Smith, but Arthur C. Clark was, yeah, he popularized and expanded on the concept in 1945, hence it is sometimes called the Clark orbit or the Clark Geosynchronous orbit. So there he is with this epic shirt there. And also has this New York Times Magazine walked on the moon there. Interesting. And there's a nice Mac. All right. So I want to go over some info about this guy. It's pretty fascinating stuff here. So, Arthur C. Clark, born on December 16th, 1917 and died on March 19, 2008, is the famous author of the 1968 novel-turned movie, 2001, A Space Odyssey. Interestingly, it was written concurrently or at the same time as Stanley Kubrick's film version and published one month on June 1968 after the movie was released, which was April 2nd, 1968. I'm not sure the exact day in June, but... So here is the cover of the novel there. So 2001, Space Odyssey, a novel by Arthur C. Clark, and based on a screenplay by Stanley Kubrick and Arthur C. Clark. And notice how they're in space here, they got a space spacecraft there. And also notice this giant circular thingy that, and that's basically, it creates a centrifugal force to mimic gravity. So you can exist in there while you're basically rotating constantly. That's quite fascinating there. Now some interesting info on it. So the 2001 Space Odyssey movie features a black monolith, which looks just like the Millennium Hilton Hotel building that stood facing the Twin Towers. And the hotel phone number is, you can make it up, is 2112, 693, 2001. So there's a black monolith there. And yeah, so you hear the monolith there, there's the monkeys there, or gorillas. And now you have, interestingly, so there's the Millennium. Hilton Hotel standing right next to the Twin Towers on 9-11 there, and again, the contact number is 2112, 693, 2001. Again, it's called Millennium. and we'll get to that naming as well. So again, it looks like just a black monolith there, and then it has 2001 in there. And also, there's actually someone on 9-11, her name was Tammy Michaels, and she actually called, called someone and said her, she was at the Millennium Hotel, and her phone numbers, 2112, 693, 2001. So we'll open this up, actually. So I posted on Hive blockchain slash, this is on ThreeSpeak. It's also on YouTube, on my other channel, Mias Truth, but I'm going to play this beginning. It's only a few seconds. Just play this. Blames are coming out the window, we've been told to stay our hotel, and I'm watching this right in front of me, I want to call it to tell you, and then at the morning hotel, my phone unit was 2-on-2, since 940, 2001. Alright, so that's quite amazing. All right, I'm just going to close that, and let's go back here. Yeah, so the film, yeah, so that's pretty interesting. So, the film was also interestingly released on April 2nd, 1968, which is the same year as... as when the World Trade Center WTC Construction began, and a difference, interestingly, between 1968 and 2001 is 33 years, and it's a pretty interesting number. So, April 2nd, 1968, Stanley Kubrick, and Arthur C. Clark's movie, Space Odyssey is released, and then you have August 6, 1968 World Trade Center construction began then. All right, and here I just put a link to that article as well. Again, and this is all inside this one article I wrote on the Hive blockchain. here, so you can go and see that article. But going further, and since year one is the start of…