 Hi, I'm Zor. Welcome to Unizor education. We'll talk about rocket science today. Well, basically more precisely it will be ideal rocket equation. That's the the real theme. Rocket science is the name of the whole topic which probably will have some problems related to this and it's a nice catchy name anyway for this topic. So this lecture is part of the course called physics for team presented on Unizor.com. If you found this lecture on uni on YouTube Where all my lectures actually are stored I do suggest you still to go through the Unizor.com website because it contains basically the course which means logically dependent on each other Topics presented not only in the video format, but also in a textual format as well like like a textbook Plus this site contains mass for teens which is a prerequisite for this course, especially the calculus wise Calculus is definitely needed for any kind of physics, which I'm talking about. So you have to be proficient In calculus and the site is completely free by the way It does not contain any advertisement. So I do recommend you to basically take the course and This particular physics for team course is being created as we speak. So right now I am ending the Dynamics portion of the mechanics and then whatever other Aspects of the course will be I will continue basically recording the lectures and providing cold the textual material All right back to rocket science Now we are talking about very ideal situation You see ideal rocket equation, which means the rocket is considered to be flying completely freely in the space and there is only stars there only stars to provide us with inertial reference frame Which we will use to To check the speed of the of the rocket Now so we are assuming ideal situation. There are no Gravitational fields completely empty space. The rocket is moving and what's most important with the rocket movement versus movements of other Objects which we were considering before The rocket moves or propels forward accelerates because It actually takes the piece of its own mass the propellant and throws it If it throws in this direction, it goes to this direction. So basically that's what's happening and If it's constantly throwing propellant towards this direction It will increase the speed in this if it's moving now if it's already moving in this direction and Turns with its back with its engines Towards the movement and then starts throwing the propellant. It will obviously decelerates So that's how we understand the whole physics of this, right? so what I'm doing right now, I will try to derive certain equation which basically connects together certain characteristics of this movement, so important thing is that our Mass of the rocket is changing because it throws away the propellant which is part of its own mass So mass is a function of the time Now I will consider the time interval from T begin to T end So I consider my rocket to be completely free. Nothing is working up until the moment T begin So there is certain speed and certain mass of the rocket at moment T begin certain speed then it turns on the engines and starts throwing the propellant with certain speed relative to Rocket the speed of the propellant is constant because that's basically characteristic of the engine how it works so there is a constant speed of It's called effective exhaust speed. This is the speed of the of these gases or whatever it throws away Relative to the rocket itself and its constant obviously It's not constant relative to inertial frame obviously because the rocket itself is moving, right? So the speed of these gases Which we are throwing away is actually Vt plus Ve So this is the speed in the inertial reference frame relative to which we will put all our equations But this is the constant relative to the rocket now at the moment T end We basically turn off the engine and then we see that our mass has diminished by certain amount Because we have exhausted the propellant, but our speed is increased if we Direct the exhaustion back in a negative direction if we are moving to the positive direction where propelling this propellant backwards is negative direction or if we are propelling towards our Movement so we are moving in the positive direction and we are directing this also In the same direction then we will slow down obviously So that's how the rocket for instance can sit on a can land on the on the planet It turns with its engine Down towards the planet and it goes towards the planet But the engines if they work they slow down up until the moment when the landing actually occurs So this is what we have now Why the rocket let's consider we are exhausting our Propellant backwards, so the rocket goes this way and propellant goes the opposite direction Why the rocket increases this its speed? well There is a very simple law called law of conservation of momentum now what happens before and After certain period of time and I will use the infinitesimal period of time from T to T plus DT so DT is infinitesimal increment Again, I refer you to calculus about all these infinitesimal things how they are working, etc. If you if you are not familiar with this I don't think it will be really well understood whatever I'm going to do next So this is a differential of time which is infinitesimal increment of the time and I would like to know what happens During this infinitesimal period of time if my engines are working So before this period of time at moment T I have certain momentum of the entire system and the entire system contains The rocket and certain amount of propellant in it that was before What happened after at the moment T plus DT well my entire rocket is This is V of T. This is V of T plus DT and This is propellant which we have exhausted and its speed was V of T plus V E So that's what happens before and after now from the from the law of conservation of momentum This momentum should be equal to some of these two Momentums and that's where we get the increase of this because we are moving this in this direction momentum is the the vector basically, right, so if If this vector is going to this direction now this vector should be greater than this one by absolute value To some be equal to this one, right? It's kind of obvious, right? So some of these is equal to this similarly if I have 10 here and I have minus 2 here. That's supposed to be 12, right? So the sum of these should be equal to this. So this is how we increase the speed and This is minus because it's this direction, right? Okay This is a nice picture and we will basically use this picture to To construct our law of conservation of momentum as an equation Now before my total mass, this is a moment T My total mass was this and our speed was this. So this is momentum of the entire system at moment T Now we exhausted certain amount of fuel What happens afterward? Well, obviously, we will have M of T plus DT that's the new mass, right mass at the moment T plus DT times V of T plus DT That's the speed of this thing So a new mass which is actually smaller than this one, right and New speed which is supposed to be bigger, right? So plus This thing now, what is the mass of this what what we have lost well, we have lost M of T minus M of T plus DT That's what we have lost This is what was before The mass, this is what was after so the difference between them This is bigger than this one. So this difference is exactly the mass of the Exhausted propellant and the speed we know the speed V of T plus VE and this is supposed to be equal. So this is my equation Basically, that's it the whole Work is done and whatever remains is just pure technicality Now, let's just think about it. What is M of T plus DT? This is the differential of the function. It's equal to M of T plus differential of the M of T Again back to calculus Why because differential of the of the function is defined as the value of the function at moment T plus DT Minus the value of the function at moment T. That's what differential of the function at moment T is Similarly V of T plus DT is equal to V of T plus differential of function V of T We got that and what is this? Well, this is minus D M of T right M of T plus DT minus M of T is DT here We have it in reverse because this is something which was before and this is after exhaust So that's smaller and the difference is really the mass now why minus here well because differential is negative So to make it positive. I have to make it with a minus sign obviously, right? But whatever the logic is the definition of this is exactly minus DmT it follows from here and this is a different definition of differential. So there is nothing, you know Very arbitrary here. This is just from the definition of the differential of the function So let's use this and see what we have So M of T Times V of T is equal to Now this is this M of T plus DT plus Plus differential of M of T times V of T plus differential V of T That's this piece plus this piece Now this is as I was saying it's minus differential of M T times V of T plus VE So this is our equation Which we are going to Open the parenthesis So M of T Times V of T is equal to M of T times V of T plus M of T times differential of V of T plus V of T times differential M of T That this and two differentials M of T V of T Now minus minus V of T times this and minus VE times differential M of T Okay, this is a little bit better and Here is what we will do this cancels with this This cancels with this Now what is this? Back to calculus The product of two differentials one differential is an infinitesimal and another is infinitesimal So their product is infinitesimal of the higher order Which we can ignore because we will integrate the whole thing to get rid of the Differentials of the first order and this is differential of the second order It will be infinitesimal which we can ignore So what's the result the result is? M of T times D V of T minus this equals to zero or We can put it this way or We can put it this way Okay Now this is now very very simple because this is VE is a constant, right? What is differential of the function divided by function? Well, it's differential of logarithm of this function again if if you Don't know why back to calculus The the derivative of logarithm is one over whatever is under logarithm So it's which is one over empty and then we have to derive Make a derivative from the inner function, which is in this case differential All right. So this is the final formula Which we are going to integrate on this period because this is what happens during this interval of time infinitesimal infinite infinitesimal in interval of time from T to T plus DT so if I will integrate it on any time interval From beginning to end So I assume that during this period of time My engines are working They start working at this moment They end working at this moment and they are always exhausting with a constant relative to the rocket speed The fuel the total amount of fuel which we have exhausted basically is during the infinitesimal Time period is is a differential of m of t but now we are integrating and we will get the total amount of fuel Which we have exhausted Okay, so what's the differential? Integral of differential that's the function itself v of t and that should be v of t at the end Minus v of t in the beginning So that's on the left on the right. We have now v is a constant now we have logarithm of m of t plus Sorry plus t and Minus logarithm of m t beginning Almost done These are all simplifications technical part only Right now the the logic was finished a long time ago when I derived the first equation from the law of Conservation of momentum this is just technicality So what is this? This is increment of the speed We have gained the new speed And I will put it delta v During this period from beginning to end Now what is this? V e Now the difference of logarithm is logarithm of the ratio. Remember this So it's logarithm of m of t and Divided by m of t beginning But I Don't like it a little bit. I'm telling you why So you see mass at the end is smaller than mass at the beginning right because we are exhausting certain propellants So this ratio is less than one which means logarithm of this ratio is Negative I kind of like it to be positive. So I will change the order it would be Mass in the beginning divided by mass at the end, but I will put minus in front. I Like it better mass in the beginning which is bigger and That's why my logarithm is positive and this is the final formula Now let me talk a little bit about this formula. This minus is actually very important Why we came up with the minus? Actually in many textbooks, they don't really specify this because they do not Consider the speed as a vector, but speed is a vector. So it's a velocity actually. That's the right term speed is just absolute value of the velocity now if our Exhaust is directed To the opposite direction then the movement and Let's assume that the movement is towards the positive And of some axis of this inertial frame. So we have this inertial frame and let's say our rocket is moving along the x-axis Towards the positive direction right then if I'm exhausting back Which means our DE as a vector is negative since it's directed to the negative part of the x-axis, right? So this is negative With a minus it will be positive now. This is positive. So this is positive now What does it mean that delta V is positive? It means our speed is increasing, right? We have the positive increase and that's our acceleration comes from so if we are exhausting to the back and Moving to the front then we are increasing our speed of the rocket as we are exhausting the propellant But as I was saying before sometimes the rocket wants to decelerate for instance, the rocket goes down to the earth or planet whatever and it starts and turns instead of Knows first it goes the back first and The engine start working and the rocket moves in one direction and the exhaust also in this direction So exhaust is positive because our rocket is always moving towards the positive value of the x-axis So this is positive and now this is negative and that's what makes our delta negative that's logarithm is always positive, right and Delta negative means we are decelerating. So that's why I prefer this minus to be here It signifies actually the vector character of the movement of the rocket itself and the exhaust of the propellant as another object All right, this formula is named after Russian mathematician and astronomer Tselkovsky Although he was not the first one who derived it There were other people you can actually go to Wikipedia and and check who else But he derived it independently and for whatever historical reason the formula is called by his name Tselkovsky Well, that's basically it. I do recommend you to go to the Unisor to Physics 14. It's a dynamics mechanics dynamics and the rocket science. That's how you go to this topic Examine the text for this lecture. I think it's very important if you will read it again It's like a textbook, but now you're prepared for Whatever you will be reading and I think it's a very useful Reading because maybe I did not did not explain in all the details certain aspects and you will see it in front of you That probably would be really very beneficial That's it. Thank you very much and good luck