 Hi, I'm Zor. Welcome to a new Zor education. The previous lecture was dedicated to ideal rocket equation or formula of Tsilkovsky. Basically, it's a very simple equation which combines together the increment of the speed of the rocket and the ratio of the mass of the rocket before with the propellant and after certain amount of time when the propellant is exhausted. Now, today I would like to use this formula to make a couple of very very simple calculations. Now, this lecture is part of the course Physics for Teens presented on Unisor.com. I suggest you to watch this lecture from the website because there are other lectures. First of all, there is a course of mass for Teens which is prerequisite for this course, especially as far as vectors and calculus is concerned. Secondly, the course is presented in some logical sequence. So I would definitely recommend you to basically start from the beginning and go with the way how the course actually goes. It gives you some logical view onto an entire subject. Plus, every lecture has very detailed notes. There are even some exams for those people who would like to challenge themselves and well, the site is free and it has no advertisement, so no financial strings attached at all. You don't even have to register. I mean, if you do, it's good. All right, so the calculations we would like to make today are related to basically launching the rocket. In space. So we will be calculating what actually is necessary and let me tell you, I was surprised myself with the results of this. Basically, how much fuel, how much propellant you need to launch the rocket. So let's use this Tsilkovsky's formula, the formula which is basically the ideal rocket equation. So let me just specify it again and we will discuss it a little bit. So that's how it looks. Now, delta V is an increment in the speed of the rocket in some inertial system of coordinates, some inertial reference frame. Let's say it's related to stars. Doesn't really matter. And we can always assume that the rocket goes along the positive direction of the x-axis. Now, to basically make this particular motion, it's supposed to exhaust certain propellant. So the very important characteristic of the rocket movement versus something which we have started before is that rocket is losing its mass as it moves forward. I mean if it wants to accelerate, of course. I mean if it works, I mean if it moves with certain speed, constant speed along certain trajectory, basically it moves without any loss of mass, but if it wants to accelerate, it throws away the propellant and that's what makes it to go faster. Or if the rocket wants to slow down, it turns with its back towards the direction where it's moving and again throws propellant and that slows down. So if propellant is thrown in the opposite direction to the movement relative to the movement of the rocket, it accelerates. If it's in the same direction, we can assume it's positive basically because it all depends on us. So if it moves, if propellant is exhausted towards the same direction, the positive direction of the x-axis where the rocket is moving, then it decelerates. And that's actually meaning of this sine minuses. So delta V is increment of the speed of the rocket. Now V E is effective exhaust velocity. Now again, this is negative here, negative because in its vector form V E is directed either towards the negative direction of the x-axis, which means my delta V should be positive to accelerate, right? And to make it positive, I need this minus sign. If, however, V E is positive, which means it goes into the same direction as the rocket itself is moving, then this would be negative and this would be negative 2, which means my speed of the rocket is decreasing. It decelerates. And this is the logarithm of the ratio of the mass of the rocket in the beginning and at the end of a certain time period during which we are the rocket exhausts its fuel, its propellant, right? Now, and I was telling you that I was a little bit surprised. Now, I was surprised of this ratio. Apparently, we need a lot of fuel, a lot of propellant to launch our rocket to an orbit. And that's exactly what we are going to calculate. Now, let's consider that we would like to launch our rocket on the orbit around the Earth. Now, in some other lecture which precedes this one, I think a couple of lectures before it, I have calculated that the speed which is necessary to circulate around the Earth somewhere were international space station right now, it's it's circulating. You need about 7.8 kilometers per second speed of the rocket. If it's less than that, it will fall down onto the Earth. The Earth's gravity will pull it. If it's this, then it will be on an orbit. So, it will fly away and then fall down exactly by the same distance to keep it on the same distance from the Earth. Okay, now, so we know this which means we know delta V because initial speed of the rocket is obviously zero. So, starting from the zero, we have to reach the speed 7.8 kilometers per second during certain amount of time. So, my question is what kind of, how much fuel do I need? Now, this is something which must be specified. Now, this is a relative to the rocket effective speed of the propellant. Now, contemporary engines, rocket engines, and I didn't know it, I just looked at Wikipedia or something, have a speed of exhaust about 4 kilometers per second. That's pretty fast, actually. So, when the fuel burns, it actually throws with this speed whatever the results of the burn is. So, this is your propellant. So, the propellant speed VE is this. Now, obviously, using these two factors, I can find this logarithm. Okay. So, VE, well, obviously, I should put minus and then this minus would be plus. So, it would be 7.8 is equal to 4 times logarithm of M in the beginning divided by M at the end. By the way, this is always greater than 1 because in the beginning, with the fuel, with the propellant, the rocket mass is greater than at the end when everything is exhausted. So, which means my logarithm of M in the beginning divided by M at the end is equal approximately 7.8 divided by 4, which is about 195. So, logarithm of this ratio is 195. Okay. So, the ratio itself equals e to the power of 1.95, which is approximately 7. So, here is very important ratio. The mass of the rocket in the beginning of the movement is 7 times greater than the mass at the end. So, which means that if I have a rocket where this is the useful part and this is my fuel and let's say useful part is about 1000 kilogram, 1 ton metric ton, all right, which is reasonable, 1 metric ton because you have equipment, you have the budget of the rocket itself. I mean, it's reasonable amount of weight or mass rather. Then this thing is supposed to be 6000 kilograms. So, for every ton you would like to launch on the orbit, you have to burn 6 tons metric tons of fuel. Now, that's what I was surprised with frankly. I didn't kind of feel that this is so big. Six times greater is the mass of the fuel than the mass of the rocket itself, the useful mass of the rocket. And that was my first problem, my first calculation rather. Now, my second calculation is related to trip a little bit further. So, instead of going around the Earth, where let's say International Space Station or all these satellites are rotating, I would like to go to let's say Mars, which means I have to go away from the gravity of the Earth. Now, we can very easily calculate how much speed I need for this, but in any case I just ask you to trust me that there are calculations which result in the speed necessary of 11.2 kilometers per second. So, as you see, we need more. It was 7.8 kilometers per second to get around the Earth in orbit, but if you would like to go away from Earth, you need more. And this is how much you need. Okay, now let's see how much fuel you need for this. That would be even more surprising, quite frankly. Well, let's again, let's just calculate. So, V, the effective speed of the fuel of the propellant is 4 kilometers per second. So, my logarithm of m from the beginning to the end is equal to 11.2 divided by 4, which is 2.8. So, the ratio of the mass is e to the point 2.8, which is what? Approximately 16. So, again, the same picture. This is the rocket. This is the useful part. Let's say this is 1,000 kilograms. Now, this should be 15,000 kilograms. So, you have to have 15 times more weight of the fuel relative to the useful weight of the rocket with all its equipment and maybe people, I don't know. But in any case, that's how much fuel you need. And that's a problem, obviously. I mean, it's not such an easy thing to have 15 tons of fuel to launch one ton of useful weight. Now, these are obviously very, I would say, teaching kind of calculations. The real calculations, the real rocket science is much more complex, obviously. Because, for instance, to launch the satellite on the orbit, if this is the Earth, you basically launch it like this. So, you have to calculate the speed a little bit more, a little bit more involved in detail conditions. But in any case, it's still something around the same calculations with obvious corrections on difference in, let's say, Earth's gravity on the Earth's level and the Earth's surface and on the orbit. I mean, you have to really take everything into calculation. Whenever you're launching the real rocket, it's really a complex thing, very complex thing. I mean, besides the technical achievement, whatever it is, it's a lot just on calculation purposes. So, the calculations are involved. And by the way, when you're launching the rocket, you know, Earth is rotating by itself. So, it helps if you're launching the rocket in such a way that the rotation of the Earth helps you. Which means, since rotation is towards the East, it means that it's easier to launch the rocket towards East, because the Earth's rotation helps you move a little bit forward. And it's substantial speed as well. I think we did calculate the speed of the rotation. So, that's basically it. There are very simple two problems which we have solved using this rocket equation. But again, you really have to feel how difficult it is to deal with all these issues related to the launching because it requires a lot of fuel to launch anything significant on the orbit. And obviously, these were the first calculations scientists really were going through before they started to, you know, to complicate the picture with the rotation of the Earth and stuff like this. But anyway, this is the beginning, relatively simple, but with somewhat surprising results, I would say. Okay. Why don't you read the notes for this lecture? If you go to unizor.com, go to physics for teens. This is the mechanics part of the course. And then you have to go to dynamics. And it's one of the dynamics topics. That's it. Thank you very much and good luck.