 In this video, I want to calculate how long it takes the moon to go around the Earth once. Now, how are we going to do that? We're going to be using several laws. Newton's second law of motion, F equals ma. And we're going to be using that when we can assume that the moon goes around the Earth in a circular path, we would have a centripetal acceleration of a equals velocity squared over the radius. And then we're also going to be using the universal law of gravity where we have the force of gravity is g times mass of the Earth in this case, times mass of the moon over radius squared, radius in this case being the distance between the center of the Earth and the moon. And we're also going to be using that the circumference of the path to go around once 2 pi r will be traveled at time t if we travel at a speed v. Therefore, we can conclude that v equals 2 pi r over the time it takes to go around. Now, first thing we have to do is, we have to do the free biograph of the moon. What forces are acting on the moon and there is actually only one that's acting on the moon, which is the force of gravity, which points towards the center of the Earth. We also know that the centripetal acceleration is pointing to a point close to the center of the Earth, so you're just going to assume your centripetal acceleration goes that way. It's actually pointing to the center of the circle of the moon, which is not exactly the center of the Earth, but we're going to see when we do our calculations how close we get. Also, you're assuming a circular path, you're not sure if it actually is a circular path, so we're just going to see. So, now let's work this out in, let's assume we go in left direction, where we have that the sum of all forces is equal to ma. In my case, I have the force of gravity, which is equal to the mass of the moon times the acceleration of the moon. The force of gravity and the acceleration point is the same direction, so we can plug in that g times the mass of the moon times the mass of the Earth over distance squared is mass of the moon times the acceleration, which is v squared over r for the centripetal case. Now, first simplification, the mass of the moon times the south, so one thing that we do not need to know. So, let me rewrite gravity constant times mass of the Earth over distance squared between moon and Earth is equal to v squared. We already determined that v squared is 2 pi r over time, so we have 2 2 pi r over time squared over r, which leads me to, if I solve this for t here, I get that the time that it takes the moon to go around the Earth once, which is the period, is the square root of 4 pi squared times radius power of 3 divided by capital G times mass of the Earth. Now, before we plug in numbers, let's see if that equation actually works out if I look at the units. So, for the units I have on the top, I had square root of 4 pi squared will have no units, then I have the distance, which is much in meters, so meters cube, and I'm going to have to divide that by the universal gravitational constant, which is in Newton square meters per kilograms squared. And then I multiply this with the mass of the Earth is kilograms. So a couple of things fall out, the square makes me just going into meters, then I have one kilogram that falls out, so I have meters over. So what is a Newton? A Newton is a kilogram meter per second square, which I had divided by the kilogram. So Newton kilogram is per square second kilograms cancels out, the meter cancels out. One over second square is square root of second square, which gives me seconds. So I'm quite confident that my result for the period is actually correct. If I use my distance between Earth and Moon being 3.84 times 10 to the 8 meters, I know the universal gravitational constant is 6.67 times 10 to the minus 11 Newton square meters per kilogram squared. And I have my mass of the Earth, which I determined in the previous video as 5.97 times 10 to the 24 kilograms. When I plug all of this in, I get the time of 2.36 times 10 to the 6 seconds. Now that answer probably doesn't tell you anything, like who knows what 2.36 times 10 to the 6 seconds are. So let's convert this into this. So we divide this by 60, then we get minutes, we divide it by 60, then we get hours, we divide it by 24. And what we get is 27.4 days. Well, isn't that almost a month? Yeah, it is, because the name month actually comes from Moon. A month is the time it takes to go the Moon around once. Now why is this not exactly what we usually consider a month? This is just until the Moon gets back to its original position. What we consider a month is until the Moon looks the same again, which is also influenced by the fact that the Earth itself is turning slightly. So this is the time it takes the Moon to go back to its exact same position as measured in reference to the stars. So 27.4 days, approximately one month.