 Hi, I'm Zor. Welcome to Unisort Education. I would like to continue talking about gravity and solve a couple of problems, very practical problems, by the way, like measuring the mass of the Earth. Now, this lecture is part of the course of Physics 14 presented on Unisort.com, as well as there are some other courses of this Unisort.com website. Everything is free, no advertisement. I do suggest you to watch this lecture from the website, just go into the website and menu for Physics 14. This is Mechanics, because if you will find it somewhere else, like on YouTube, etc., it will be only the video presentation. On the website there is also a textual part, like notes, which basically is like a textbook. And there are some problems for those people who would like to challenge themselves. Okay, so, continuation of gravity. Today we will talk about free-falling. Now, what is actually free-falling? Well, obviously, we are talking about some planet, like planet Earth, where we live. And an object, which is subjected only to the force of gravity, and we just drop it to the surface of the Earth and see how it behaves. Well, it's a purely mechanical problem, so there are such typical mechanical variables like mass, force, acceleration, right? And we know that the only force we are dealing with right now is the force of gravity. There is nothing else. So, the object is gravitated towards Earth, and that's why it's going with some kind of an acceleration. Okay, now we are talking about ideal situation, and what is ideal? First of all, we are talking about a planet being an ideal sphere that the mass of this planet, let's call it M, is evenly distributed within this sphere uniformly. Now, our object we are dealing with has also a certain mass, M. And then we would like to know what is the acceleration of this object free-falling onto the surface of the Earth if it's right at the Earth. So, we are just lifting it on infinitesimally small height and let it go and then measure the acceleration. Now, we know the force between them, right? The force of gravitation is some kind of a constant, gravitational constant, is proportional to each of the mass and reverse proportional to the radius of the planet. Now, here we actually assume that, and this is a very important actual assumption, so instead of considering a sphere with the mass evenly distributed within this sphere, we consider this to be a point object with all the mass concentrated in its center. So, we are talking about two point objects now. One is the probe object which we are dropping onto the surface. Another is a point object with mass M and the distance between them is the radius of the planet. Now, you might actually have some doubts that this particular replacement of the sphere with evenly distributed mass. Replacement of this with a point object with the whole mass concentrated in the center, it's kind of stretching. Well, actually it's not. I mean, there are certain relatively sophisticated calculations which prove that this is exactly the same. So, that's what we are basically, we'll skip this particular proof and we'll just assume that this is a valid assumption. And it is. So, we know how the force is expressed. The force of attraction between the planet, mass M, capital M and the object lowercase M. So, what's the acceleration? Well, obviously, acceleration is equal to force divided by mass. This is the second law of Newton, right? From which we can derive that this is equal to gM divided by r square. Great, so now we can determine acceleration of the object which is free-falling on the surface of, let's say, Earth. Wow, not so fast. I can actually imagine how we can calculate the radius of the Earth. Well, let's say we can travel around the equator and divide it by 2 pi, something like this. But the mass of the Earth, nobody knows the mass of the Earth, right? So, we are using this formula for a different purpose. In as much as we can determine the radius of the Earth, it's much easier even to determine experimentally the acceleration of the free-falling object. Now, first of all, as we see, it does not depend on the mass of the object. This acceleration, which means big objects, small objects, they are all falling with the same exactly acceleration. And if we don't observe it, if we are comparing, let's say, a metal ball and some kind of a feather, but that's because of the air resistance, etc., which resists falling of the feather in the environment without the air. If we will pump out the air, let's say, from a tube, the feather will fall exactly the same way as a metal ball. So, we can actually measure this, and knowing the constant G, because again, we can measure the force of attraction between two different objects, not necessarily the Earth, just any two objects, because G is a universal constant, right? So, we know G, we know R, well, we can come up with R after certain manipulations, and we measure acceleration. That's how we can measure the mass of the Earth. So, let's resolve this for the mass. Mass is equal to A times R squared divided by G. Now, we measure the acceleration on the surface of the Earth as being 9.8 meters by second square. Now, we know the radius of the Earth, again, using whatever geography we know about. It's equal to 6.4 times 10 to the 6 of meters. We know the constant G, so this is R, this is A. We know the constant G is 6.674 times 10 to the minus 11, and the dimension of this is kind of sophisticated, it's Newton. We have to have the Newton, so we have to divide by kilogram square and multiply by meter square, right? So, this is the dimension. So, if we will calculate this based on these numbers, we will have the mass of the Earth. And I did it before, and it's approximately 10 to the 24 kilogram. This is the mass of the Earth calculated using the gravity, the laws of gravity and certain experimental data. Well, that's useful. Now, we know the mass of the Earth. Well, obviously, we can do some more precise calculation. These are very approximate numbers, and this is as well. Something in series like 5.9, blah, blah, blah, I don't remember. Okay, so the purpose of this part of the lecture is basically to determine the mass of the Earth using the concept of a free fall and the laws of gravity and certain experimental data. Okay, now I would like to switch to another problem. Also very interesting, by the way. Now, we are launching satellites in certain orbits. So, let's do some satellite launching. What I would like to know is, if I would like to launch a satellite at certain height above the level of surface of the Earth, and I would like to launch it and basically let it go by itself, by inertia. I need to know the linear speed I should really develop for this particular satellite so it will circulate around the Earth on a circular orbit. So that's my task, actually. I have to find out what speed do I need. Well, which obviously means what kind of a rocket I need to launch the satellite on the orbit, and the rocket will propel it forward with certain speeds. So I need to know that speed so the satellite will then by inertia itself, only by inertia, it will go around the Earth. Well, not exactly by inertia of its own movement, combined with the gravitation of the Earth, obviously, because by inertia itself it will move along the straight line. But since the Earth, so this is Earth, and this is how I launch my satellite. So the satellite goes this way, but Earth gravitates it this way, and that's why it's going this way. So that's what I need. I need to know the speed, which obviously depends on the height. The further from the Earth, probably the speed should be less, because the gravity is less, right? So let's just talk about this, how it can be done. All right. Well, first of all, from kinematics, and by the way, if you don't remember it, obviously you can go back to the Unisor.com to kinematics section, and there is a whole topic related to rotation. Whenever you are rotating, then your acceleration A towards the center of rotation, so if you are rotating, there is acceleration going directed towards the center of rotation, always. Now, it's equal to radius times angular speed squared. Now, this is derived in one of the lectures on rotation in kinematics, and again I invite you to go to this lecture if you don't remember this formula. Well, in using the linear speed instead of angular speed, it would look like v squared divided by r, because v is equal to r times omega. Linear speed is related to angular speed using this particular formula, where r is the radius of rotation. Now, in our case, radius of rotation is not really radius of the planet, it's radius of the planet plus the height above the surface, which we actually would like to use as a parameter. Depending on this height, I have to know the speed. So, instead of this formula, I will use a very similar formula. A is equal to v squared divided by r plus h, where r is the radius of the planet, and h is the height above the surface where my satellite is supposed to fly. Alright, so that's my formula. Now, at the same time, I know from my previous calculations that acceleration should be equal to g times mass of the Earth divided by, now, again, r squared but instead of r, which is acceleration on the surface, it doesn't really matter what radius is. So, in this case, I can put the radius, which is basically a distance from the center of the Earth, where r is the radius of the Earth and h is the height above the surface. So, this is the formula which I have just calculated before, derived before in the previous half of this lecture. Now, since the gravity and the force of gravity is the only force which keeps this on the orbit, this is the same acceleration must be. So, it's the same force and the same acceleration. So, if my object has a mass a, then the force would be equal m a in this case, which is m v squared divided by r plus h. And this is the same force as this one, if I will multiply m times a, this will be the force of acceleration, the force of gravitation. So, these force are the same. So, mass actually, the lower case mass is obviously reduced and I have the quality between these acceleration, acceleration which is caused by gravity and acceleration which must be if I am on that orbit, otherwise I'm going off the orbit. Okay, so let's just equate them to each other, r plus h squared equals to v squared r plus h. Now, m, we already know, we had just calculated m. We know the m based on the same kind of a formula and certain observations, certain experiments where I really observe my free falling acceleration. So, from this I can derive basically v which is equal to square root of g times m divided by r plus h. So, again, now I know from the previous half of the lecture mass, now I know the radius of the Earth. Now, this is dependency between the height I would like my satellite to be above the Earth and speed necessary for it to go by itself without any additional rockets around the circle, around the Earth in the circular orbit. Now, if I will put, let's say we have an international space station which I know is above the Earth at 400 kilometers. So, this is h. So, if I will substitute all the numbers and get the speed, I did these calculations and the v would be exact, would be equal to 0.78 times 10 to the fourth meters per second. Or about 28,000 kilometers per hour which is kind of more habitual way of expressing the speed. Now, in miles it would be something like 16,000 miles per hour. That's a lot. I mean, considering the speed of the car, let's say it's whatever, 100 kilometers per hour. So, this is 280 times faster than the speed of the car. All right, well, basically I wanted to show you how the laws of gravity with certain experimental help allows us to calculate, first of all, mass of the Earth. Now, it also allows us to calculate certain parameters of the satellites if we want these satellites to be used for certain purposes, which means going to a certain height and circulate the Earth. So, this international space station, it just goes by itself around the Earth, approximately at this height, which means its linear speed on the orbit is this one. Okay, that's it for today. I do recommend you to read the notes for this lecture on Unisor.com. It's a nice kind of textbook style explanation of basically the same concepts which I have just talked about. That's it. Thanks a lot and good luck.