All right today we're solving this interesting thought experiment where if you have a hole that goes straight through the earth from Point A all the way to the opposite side, through the diameter through the center um You know if you drop a mass M what would it do? So the first question we've got we've got a radius X. Of some point throughout the middle of the earth and the full radius of the earth are calculate the mass of that smaller sphere if they are concentric circles. So we're going to use to relations we're going to use the fact that the density of the earth is the same throughout the density of the small, spherical is the density of the full earth and density is equal to mass over volume. So we've got the mass of that little small sphere divided by the volume of a sphere which is four thirds pi x cubed. Where X is the radius of the smaller sphere is equal to the mass of the entire earth. Capital M divided by four thirds pi capital r squared. Now we see we have four thirds pi on each side, we divide them out. If we saw for the little mass M, we get little M is equal to big M times the ratio X over are cute from there. We can solve for the force of gravity using our classic Newton's equation of universal gravity. So we've got the gravity constant G times the little mass, M times the bigger mass, which is the still the small concentric circle with a dime or radius of X divided by the distance we are, which is X squared. Now we can sub in capital M. With our expression from part A. So we get g little M mass of the Earth times X cubed divided by the X squared from before times where it is cubed. Now we can see we have X appearing twice. We can cancel it out and we get the force of gravity is equal to G. M. M. X. Over our cute remember Little M. Is the mass of the small object. Capital M. Is the mass of the earth. Now we're tasked with figuring out how this is simple harmonic motion. So if recall at the center, if you plug in X is equal to zero, you're going to get a force of zero. But you know that the kinetic energy will be at its max because it's sped up from the force felt all the way to point A If you look at pointe au of a maximum force but zero velocity as a result of the object will oscillate along that whole through the Earth passing through the center and then filling an opposite force until it reaches the opposite side and then it will go back and go back and forth. Now we're gonna go to part D. Were tasked with finding the period of that oscillation. We're going to take the same period equation from her spring mass damper so the period T. Is equal to two pi times the square root of mass over stiffness. K. Now we don't have a spring here. So it's kind of weird. We don't know what the stiffness would be if we can figure out an equivalent stiffness by looking at our force equation and part beat. So we know that F. Is equal to G. M. M. X. Over our well Gm argued Gmm over. R cubed is roughly Kay, if we look at Hook's law which is F. Is equal to K. X. So we get a stiffness of G M. M over R cubed. Thus we substitute these two into our period equation. We get to pie times. The mass is still M. To buy the by G M M. And then the rQ goes to the top. We can cancel little M. Thus we get a period of two pi radius cubed, gravitational constant G times mass of the Earth M. Now in part E were given numbers for all three of these in addition to the constant to pie and you can solve to get a period of 85 minutes Now, part F asks if there is a satellite that is orbiting the Earth at a radius of our, what would be the period here. Now, if you look in our period equation, everything is a constant about Earth radius R, mass M of the Earth and gravitational constant G. In addition to the two pi. So if it's orbiting around the Earth, you'll actually have The same period since it's a function of the same three constants. So that's the period is the same at 85 minutes, I don't know.