Welcome back to the Cosmic Classroom. So I had some questions this week about the relationship between Newton's version of Kepler's third law and Kepler's third law. So the book shows you that Kepler's third law is P square equals A cube the, the period of orbit around the Sun depends on the average distance from the planet to the sun. Newton, Newton's version looks similar, but for some, it may not be similar enough. So Newton's version show that P square over 2 pi square equals A cube over G M plus, plus big M plus little m. Those two relationships say exactly the same in the case of a planet orbiting the sun. The case of a planet orbiting the sun big M, the mass of the big object, and little m are pretty much equivalent to big M. Alright? And if you plug in all the constants here, G and 2? and if you use A in this at the SI with meters per second, excuse me, A meters and if you used the period in seconds and if you do all the math you end up with P? = A?, if you do the unit transformation. They are the same thing, but you're still not convinced and you still don't understand why is it that there're two versions. Kepler's look much better. The problem is that Kepler's is only useful for the solar system, it's only useful when there, there're two bodies interacting and one of them is much more massive than the other, so that we can discard the mass of the second one. However, in many cases the the bodies may have similar masses, may even have equal masses, and the problem becomes more complex. Also Kepler as I mentioned in, in a previous video derive this equation empirically, he just observed it to be an equation. While Newton really derived it from physical principles, and that's what I'm going to show you how to do. So, Newton's version becomes important when we have two bodies and we can't ignore the mass of either one of them, we can't say that "all that's very small." So here I have a big body, that I'll call big M, and I have a little one, that I'll call little m. And they are at a certain distance apart, that I'll creatively call D. Alright? The center of mass of the system, the place that this is this, this objects will orbit, its neither the center of big M nor the center of little m, it's somewhere in between and it's called the center of mass. That's why little, that's where they'll orbit. So this object here,big M will orbit, like this. Alright? Oops... that should be a circle. While the object little m will orbit around the center in a circle here, so they orbit this way. The distance from the center of big M to the center of mass I will call D_big M the distance from M to the center of mass. The center, the distance from this, the center of mass to the little mass are creatively called D little m. Now the, the position of the center of mass depends on the mass of these two bodies, in such a way that D M, the distance from the little body to the center of mass,is equal. The big distance, the total distance from one center to the other, that multiplies M divided by little m plus big M. And you could do, you know an elect,an (allegros) ANALAGOUS relationship you can also say that the distance from big M to little m is also equal D and now here we'll have little m,and again m plus big M. So things are getting more complicated. The reason the objects, the objects are orbiting the center of mass is because they attracted to each other gravitationally. So the little m is attracted to big M with a certain force, F, just as,as much as the big M is attracted to the little m with the same force. Alright? Due to gravity. So the force of attraction depends on the constant of grave,grave,Gravitational Constant, the product of the mass of the two bodies, and the distance between the center square. Now, each one of this objects is orbiting in a circular orbit in this case. So pick one in look at what's happening with that one. So let me pick little m. So little m is being attracted by beginning with this force and this force is then acting as a Centripetal force. So this force is also centripetal force that depends on the mass of little m. On the velocity square little m and the distance, but now it's not the distance from big M to little m instead is from, the distance from little m to the center of mass, which is D, M that's it. Right? So the two forces, this force is the force that is creating a Centripetal circle of motion. So those forces are equal. So now I can say that G M M / D? = M V? / D M This M goes away with that M and now let's remember the definition of velocity. So the definition of velocity right here is the distance travel, in this case 2? D M divided by the time that it takes to go around, which is the period, P. So go back here and I'll substitute V 4 2? D M / P. So G M / D? = 2? D M all of this square over D M P?. G M D? = 2?? that multiplies D M divided by P?. I hope you followed that much. Now we ARE almost there. We need to remember that D M is D that multiplies big M over little m plus big M. So now what we need to do is substituted and M right there and things will cancel out, so let's do it. G M / D? = 2?? that multiplies D big M little m plus big M, right there just brought this instead of the M, over P?. Alright? So we rearrange this,we bring the P? this way,the D,the D? that away with this will become a D? and you already start to see that you have the P? you have the D?, this M will go away with that M and look you suddenly got Newton's version of Kepler's third Law. G that multiplies M plus big M over D?, cubed because this one went there equals 2?? over P?. Cross multiply and you have what you really want P? over 2?? all of it squared equals D? that multiplies G little m plus M and remember that D is the average distance, is the distance between the two centers. So this is your A, that's you used to,you saw it as an A. Pretty easy,boring but easy. This is it. I hope it helped those of you who wanted to see what's the relationship between Newton's version of Kepler's Law and Kepler's Law.