The big idea here, is that the gravitational force we feel into the surface of the earth that just makes a constant acceleration, is the same as this general expression that we just talked about. This was Newton's big insight, is to this earthbound force and the force that causes the motion of the planets and the moon were the same. Let's go ahead and use it for something astronomical or well, not quite astronomical, let's run some numbers for the Earth and the Moon and see what we get. It'll be fun. Let's see. Let's get the force between them, and just to see what the magnitude is, who knows? Let's see. Here's the Earth, and the mass of the earth is 6 times 10_24 kilograms. Somewhere out here, not to scale, is the moon, an the mass of the moon is 7.3 times 10_22 kilograms. Less massive. We run the numbers real quick. The magnitude of the force, we don't really want to think about directions right now is g 6.674 times 10_-11 I don't know why I can't remember that number, times the two masses, 6 times 10_24, 7.3 times 10_22 over the separation squared. How far is it from the Earth to the Moon? It's a long way, 3.85 times 10_8 meters squared. All in MKS units. You do all that and you get an easier number to remember. 2 times 10_20 Newton's. That's a big force. Well, it takes a lot of force to pull the moon and the circle, the moon is big. Force is big, this number is small. Gravitational forces tend to be weak. It's pretty far apart, but they're really heavy. These are big, massive objects that we're talking about. So 2 times 10_20 Newton's. Let's see if we can use that to think about the orbit of the moon a little bit. Now that we've talked about this force that always pulls to the center, we can already think about how that influences circular motion. Because remember in circular motion, such as an orbit and we have the moon here, we know circular motion you've got some speed that you're going v, and it happens because you have a centripetal force giving you a centripetal acceleration. So for simple orbit calculation, you basically just say this. The GMm over r squared force is the centripetal force and it goes around. So let's see if we figure out how fast the moon goes and see if we get the right answer. We would say 2 times 10_20 Newton's equals MV squared over R, mass of the moon times the speed of the moon over R, which is this radius here, 3.85 times 10_8. But what I want to do is I don't really know the speed of the Moon. Does anybody know how fast the moon is going? Now they don't know. What we're going to do is change it real quick to the period, because another period of the moon, it's about a month. What is it? 28 days, 27 days. So the period is 2 Pi R, the circumference of the orbit and the v was squared over the period squared. So we changed the formula like that. Then we can solve for the period and see if we get what you're supposed to get. We can simplify, just square the two Pi and that becomes R just like that, bring the T squared over here and we got our mass of the moon, right? Yeah, so let's see. It would have the mass of the moon 73 times 10_22. There would have 4 Pi squared. There would have the radius of the moon, or the distance there, 3.85 times 10_8. This is getting a little unnecessary because we already used all these numbers here we can't cancel them over nothing. There's nothing left, all over the force 2 times 10_20 Newtons. Anyway, if you do that and take the square root, you get that T is 2.3 times 10_6 seconds. I don't really think of my life in seconds.I think of it in hours and days. If you divide that by 3600 to get it to hours and you divided by 24 to get to days, you get 27.2 days, which is about right for the orbital period of the moon. So you can see this F equals G Mm over r squared. You already know how to use it. You've already thought a lot about forces in kinematics, pretty basic force and you can apply it in a lot of places.
