 Hi, I'm Zor. Welcome to Unizor education. We continue talking about rocket science. It's kind of a nice education term. Today, we will talk about how this rocket calculations are related to gravity. So basically we are talking about a major component of space exploration, how the rocket should be calculated, how all these parameters that will speed, etc. If we are, let's say, leaving the Earth and are trying to launch the rocket on the orbit. Now before, when we were talking about ideal rocket equation, we were talking about space without any kind of external forces, external gravitation fields, for instance. Now we will introduce this component into the equation and we will see how it changes the equation. So that's the subject of this lecture. Now, this lecture is part of the physics for teens course of physics, which is presented on Unizor.com. I do suggest you to watch this lecture from the website. You just go through a directory. You go to physics and then from physics to mechanics and then dynamics and it will be part of the last topic on the dynamics. Now why? Because these lectures are not only on YouTube, because every lecture on the website has a very detailed notes, practically whatever I'm talking about is written. Plus the most important thing, I think, the website is logically structured. So all the lectures are follow each other in a logical sequence. Now the prerequisite for this course is mass for teens on the same website and you really have to be familiar with calculus and with vector algebra. So that's that's prerequisite. So I will be using relatively freely concepts of derivative, differential, etc. These are all parts of these rocket calculations. And the site is free, by the way. Completely free. There are no advertisement, no financial strings attached. Okay. Now, let me start from something which we already started in the previous lectures. Namely, the equation of how the speed of the rocket is changing. This is basically v of t at the end minus v at time the beginning. So these are speeds of the rocket in the beginning when it starts moving or its engine actually starts working. It may be on the surface of the Earth, which means this will be zero. Or maybe it's somewhere in space when engine starts working. And this is at the end of the period during which we are measuring the speed as a result of working engine. So this is what we are going to investigate. We did already use this and receive this formula. Now this is effective speed of the exhaust. So when propellant is exhausted, let's say backwards, then the rocket goes forward. Now we are always assuming that the rocket moves along the positive direction of some axis, let's say x-axis, which means we are disregarding y and z-coordinates. And the propeller, propellant is exhausted towards the negative end of the x. So that's why this speed, if we interpret it as a vector, is actually negative. And minus this is positive times logarithm of mass of t in the beginning divided by mass t at the end. Now the mass at the end of the period of acceleration is obviously less than the mass in the beginning of this period. That's why this ratio is greater than 1, logarithm is positive. Effective exhaustion speed is negative, but this is a minus, so it's also positive. And that's why we have this delta V of t positive. Right? Now this formula in many textbooks is not really written with this minus sign, which means they actually consider the E as positive. And that's why basically this is the result. Written in this way in a vector form, the formula is more, I think, flexible because sometimes we want to decelerate the rocket. For instance, rocket goes to some point and then it wants to decelerate. So it turns with its back towards its movement. And then starts working. I mean engine starts working. And obviously because we are exhausting towards positive direction of the movement, we are decelerating the rocket. And it's written here, actually, because if VE is positive towards the direction of the rocket, with minus it will be negative. And that's why the delta would be negative, which means we are reducing the speed with deceleration. So that's why I'd prefer to have this minus sign. But you should understand that it's not really that very important. Besides, in most cases we are considering when the rocket starts and accelerates. We don't think about deceleration for whatever reason. Okay, now I would like to introduce the gravitation. Now this is an equation in empty space. There is absolutely no gravitation fields, no drag from the air or anything like that. No resistance to the moving because the rocket moves in an empty space, basically. So this is this equation. Now I'm introducing the gravitation. And I will do exactly the same. I will use exactly the same logic as I was using when I derived this particular equation. So my first tool which I'm using is the conservation of momentum. So I will check what's the momentum of the rocket in the beginning of its movement and at the end of the movement. And they're supposed to be equal to each other. Except in this particular case, we have to consider that there is a force acting on the rocket. The force of gravity. Now, so we're assuming that the rocket starts from the surface of the earth and it starts usually with zero speed. But in any case, we will see that that's not really very important. But what is important is that there is a force which is directed against the rocket movement and the force, if the rocket works, if the rocket engine works during certain amount of time, this force has some impulse. And this impulse is supposed to be added to the entire momentum. Well, in this case, it's subtracted because it's opposite to the movement. Now, you remember this formula. This is the second law of Newton. Now, instead of A, we can have dV by dt. Because what is acceleration A? It's the first derivative of the speed, right? Or dMv of dt. So that was the formula when the mass was actually constant and we just put it inside differential. But now, if mass is not the constant, this is the correct form of this equation, which means that f times dt equals d of mv. And f might be actually even the force which depends on the time, too. Maybe. But it doesn't really matter. But anyway, this is the equation which equates impulse of the force during the moment, during the time, infinitesimal time, dt, and increment of the momentum of the motion. So this force, if it's gravity force, should also be added at the end of my equation, which combines my momentum before and momentum after. So before, force did not act. But during the time of my acceleration, my force was the gravity force and it was actually acting. And we have to take into consideration additional momentum, which it gives. Well, additional means with a minus sign in this case, because it goes against the movement. But anyway, this is something which we have to take into consideration. Everything else is exactly as before. So what was before? Now, before, I had my momentum being this. Now, in most cases, we can probably consider that at the time of the start, t equals t beginning. The rocket is on the launch pad on the earth. And basically, its speed is zero. And that means that my momentum is zero. Now, what happens if my engines are working during the time dt? This is an infinitesimal time from t plus dt. This is interval of time, which I am considering right now. So in the beginning of this, it was m of t times v of t. Mass is variable because we are exhausting propellant. V is variable because we are accelerating. But in the beginning of this time, I had certain mass and certain speed. Now, what happens during the time dt? Infinitesimal period of time. Well, we have exhausted certain amount of mass. Now, what kind of mass we have exhausted? Well, we have exhausted mass, which was at the moment t minus mass at the moment t plus dt. That's what we have exhausted, right? Now, what is it actually? Well, this is minus differential of m of t. Why? Well, because differential of function is the value of this function at the moment t plus dt minus the value of the function at the moment t. So right now, I would like my mass to be positive. So if I'm asking how much mass I have exhausted, it's supposed to be with a minus sign because differential is obviously negative in this case. This is smaller. This is bigger. This is after the exhaustion. This is before. So this is negative. So my differential is negative. And that's why the amount of mass which has been exhausted is supposed to be positive. And that's how I wrote it here. Now, this mass, which was exhausted, was exhausted with certain speed, right? Now, what is the speed of exhaustion? Well, my effective speed is ve. That's a speed of the propellant relative to the rocket. But rocket is also moving with the speed v of t, right? So we have to add basically the rocket speed with exhaust speed to get the real speed of the propellant. And this is the mass. So my momentum of exhausted propellant is its mass times its speed. So this is momentum of the propellant. This is its mass, positive mass, because it's minus and d is negative. And this is the speed with which we exhaust the propellant, the gases, whatever is burning. Now, the next component is the rocket itself. So gases are going one way and they have this momentum. By the way, in the beginning, if v of t is equal to zero, my ve is negative. So the whole thing is negative. So this is negative momentum, right? In the very beginning. When v of t is zero and my initial momentum, if we start from the moment zero, actually, when the rocket is in the launch pad, this is obviously zero. So this is negative. But the rocket has a positive momentum because the rocket goes to the positive direction of the x. So what's the momentum of the rocket? Well, the rocket has a mass m of t plus dt. That's its final mass, right? Which is equal to, look at this, mt of dt goes here, it's minus, it will be plus, d goes here. So instead of this, I can put m of t plus dm of t. That's the mass of the rocket. And what's the speed of the rocket? Well, speed of the rocket is v of t plus dt. But again, it's absolutely the same formula for differential. Instead of v at time of v plus dt, I can put v of t plus differential of v of dt. Again, if you feel some kind of uncomfortable about this, go back to the calculus. It's all there. It's very, very simple thing. So this is the momentum of the rocket itself, whatever remains of the rocket without the propellant, which has been exhausted. So this is, let's say in the beginning, when v of t is equal to 0, this is negative and this is positive. And that's why their sum is equal to 0, right? But now, not quite. Now we have to really overcome an impulse of the force of gravity, which is equal to what is the force of gravity? The force of gravity is m of t times g, right? Mass times acceleration of the free fall. And I have to multiply by the time during which this force was acting, right? So that's what basically the whole thing is. Now, if I will combine these three components together, I will basically have to, I have to get exactly the same thing as my initial momentum, because this is, in a sense, this is what actually the conservation of momentum is. This is momentum in the beginning of this period, from t to t plus, from t to t plus dt. This is the beginning momentum. And the combination of these three is the ending momentum. And the conservation of momentum says that they must be the same, right? So let's make this equation. Okay, so what is really the equation? First, I have to have this component minus dm of t times v of t plus ve. That's one component. Right? Now, second component is this one. This is the momentum of the rocket. New mass times new speed. And I have to add the impulse from the gravity. Force, mass times acceleration times time. So during the time dt, infinitesimal interval of time from some point t to some point t plus dt. So during this interval of time dt, this was in the beginning of this period. This was at the end of this period. And basically, that's it. So the only thing which I have added is this one, this component, relative to flying the rocket in a free space without any kind of forces. So let's just simplify that a little bit. Now, what follows is this. If I will open these parentheses, what will I have? I will have mt times vt. I will have vt times d of mt. I will have mt times dvt. And I will have differential times differential. And I will completely throw it away because this is an infinitesimal of higher order because I will be doing integration obviously after that. And this is infinitesimal of the higher order. And one integration will give me infinitesimal, which can be dropped at the limit. So instead of this, I can just put this. And now I can basically do the following. I can do this and oops, sorry, not this. This stays. And this. Right? Then I can do this minus dm of tvt and plus dm of tvt. So I can do this and this. So what remains, what remains is the following. I will have this one. I will have this one. And I can, and I have this one without this. So the final equation looks like zero. This is zero equals minus ve times dm of t plus m of t dv and plus this. So this is my final equation, which I can integrate right now to get the formula for total increment of my speed from t begin to t end on this time interval. So what will I do? I will obviously divide by m t and I will have this goes to the plus ve dm of t divided by m of t equals, no, I will put this also to the left, minus g, minus g times dt equals to dv of t. So this goes to the left. This goes to the left. This is the only thing which remains on the right. And I divide everything by m of t, by m of t, by m of t here, here and divide it here. So let me reformulate it a little bit more. dv of t is equal to, now what is this? This is obviously differential of logarithm, right? So the first derivative of logarithm of something is one over that something, that's m of t. And then you have to multiply it by derivative of the inner function, which is m derivative of t. And that's what gives you differential. And minus g dt. Now I can integrate it very simply. Integrate from t begin to t end, t begin, t end, t begin, t end. So what will be my result? Well, if I'm integrating differential of some function, the result will be function, right? So it will be v of t, v of t end, minus v of t begin. This is exactly delta v, which we are interested in. How much speed we will gain, right? Equals to, now this is the constant, so it goes outside. And then I will have again the function, logarithm of m of t end, minus logarithm of m of t begin. And minus integral of this is obviously g times t end, minus t begin. Okay, fine, that's done. Let me just simplify it maybe a little bit and that will be the answer. So delta v is equal to, now the difference between logarithm is logarithm of their ratio, right? Now ratio t end divided by mass of t end divided by mass at t begin. Well, the problem is mass of t end divided by mass of t begin. Well, t end is smaller, mass of t end at the end of the period is smaller than mass in the beginning, right? Because we are exhausting the propeller, which means that my ratio is less than one. When ratio is less than one, logarithm is negative, right? And I don't like to have, to deal with negative logarithm. So I will inverse it and I will put here beginning and this is end and put minus sign in front of it, right? So as a result, I will have minus v e, logarithm of mass at t begin divided by mass of t end and minus g t end minus t begin. So that's my final formula, which is almost the same as my ideal rocket equation except this member. So this is the only thing which my gravitation added to the equation. My old equation was from here to here and now it's this way. Now it's kind of obvious that this is supposed to be this way. Why? Well, because as the rocket comes up, it also falls down because of the gravity. Now the gravity has acceleration g and the interval of time during which the rocket goes down and at the same time falls down is from t end to t begin. And that's why you have this minus g times the difference in time. So we could have come up with this without all these calculations, but I just wanted to make again the point that we're talking about conservation of momentum at the gravitation force has an impulse which is supposed to be taken into equation and that's basically why we have the same thing plus this particular member which is kind of obvious. Now what's important actually here? Well, important is that v e is again, v e is negative when the rocket starts, right? Because it's going against the direction of the rocket which we assume to be positive. So let's say this is a z axis and the rocket goes up along the z axis and obviously its speed vector of speed its velocity is positive and this one should be negative then with minus sign it will be positive. And unless by absolute value this thing exceeds this one, the rocket will not fly. It will just stay still on the launching pad. So you have to have a pretty substantial effective exhaust by absolute value to overcome this obviously. Now what else is important here? What is important is that there are complications. You see, you might think about this way, okay, we have taken into consideration the gravity and that's probably the biggest factor which influence all our calculations and must be taken into account. Now what has not been taken into account in these calculations? Well, number one, when the rocket starts from the surface of the earth there is an air resistance which drags it down, right? So that's, we would not take it into consideration at all. And the air resistance obviously depends on the rocket form shape itself but also on the density of the air and density of the air is changing the higher we go the less dense the air is. So we have to really take it into consideration which is not easy. Another thing which we did not take into consideration is that the gravity as we are moving off the earth higher and higher, the gravity itself is also changing because we are increasing our distance from the planet. So G is not actually constant in the real life situation. Which kind of complicates obviously our equation? I mean we have integrated equation basically, right? Now if we will take into account all these factors, the equation differential equation from which we started, which is what's happening during the time dt, this differential equation would be a little bit more complicated and sufficiently complicated not to be able to solve it in integrals. So we will not be able to come up with a formula like this one. Instead people basically do it numerically and taking into consideration all these factors actually does complicate the whole picture sufficiently to call rocket science in quotes as something which is really very very complicated. Yes you definitely can say so. There are many different factors. Another factor which I would like to mention is if you launch the rocket towards the east the earth's rotation actually helps to start from certain speed which is not really zero. So the speed in the very beginning V of T begin is not really zero. It's actually the speed of the rotation of the earth if you point it to the east. Now if it's at some angle maybe that's another story again. I mean you have to really take into consideration lots of factors. Okay now I do suggest you to read the notes for this lecture. It's in this website Unisor.com and just try to familiarize yourself and basically that's it. That probably completes my theoretical explanation of what rocket science is about. Again not covering completely all the complexity but giving you some flavor of basically this is the probably the most important part of it because all these additional things like the fact that the gravity is changing as we are going up or the density of the air etc. They are definitely contributing factors which the real calculations performed for launching the rockets must be taken. However for our purposes this is the major component because whatever we are adding is significantly smaller in magnitude and impact on the movement of the rocket. But still important and we have to take it into consideration when we are dealing with real rockets to launch. Okay that's it for today. Thank you very much and good luck.