(II) A sphere of radius r₁ has a concentric spherical cavity o...

Question

(II) A sphere of radius r₁ has a concentric spherical cavity of radius r₂ (Fig. 8–42). Assume this spherical shell of thickness r₁ - r₂ is uniform and has a total mass M. Show that the gravitational potential energy of a mass m at a distance r from the center of the shell ( r > r₁ ) is given by
U = - (GmM/r).
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U = - (GmM/r).

Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem in a large hollow ball of outer radius. R one was put into space. Its inner radius is R two. The shell of the hollow ball of thickness R one minus R two is uniform in mass distribution and has a total mass of M. If another mass M prime was placed at a distance of R R is greater than R one away from the center of the hollow ball, calculate the gravitational potential energy of this mass M prime. So that's our end goal is ultimately, we're trying to figure out what the gravitational potential energy of the mass M prime is. That's our final answer that we're ultimately trying to solve for. So as we could see in our diagram that's provided to us by, by the problem itself, we have a large hollow ball of radius R one which is denoted by this doughnut shape, that's yellow, this yellow donut shape and its inner radius is R two. So inside this yellow donut, we have our white circle and that denotes R two and the shell of the hollow ball of thickness R one minus R two is uniform in mass distribution and has a total mass of M. And another mass M prime is placed at a distance of R. So our distance of R going from the left uh sorry. So on the left is our donut shape and then going to the right is our radius or thickness R and our mass M. So M prime is gonna be denoted by this pink red circle where the arrow is pointing to. And that's for R. So from the center of our inner space two RM prime mass on the outside is a distance R denoted by our, our arrow here. Awesome. So with that in mind, let's look at our multiple choice answers to see what our final answer might be. A is negative G multiplied by M all divided by RB is negative G multiplied by M prime, divided by R C is negative G multiplied by M prime plus M all divided by R. And finally D is negative G multiplied by M multiplied by M prime all divided by R. OK. Moving right along here. So first off, we must determine the density of the shelf by dividing its mass by its volume. We will also then use the density value to help us determine the potential of mass M prime. Due to a solid sphere of radius R one instead of a hollow ball with all else the same. So then in a similar fashion, we will find the potential of the mass M prime due to a solid sphere of radius R two with all else the same. And after that, we will then subtract the potential of M prime due to the solid sphere of radius R two from this potential due to the solid sphere of radius R one. And this will give us the value required for the gravitational potential energy. So now at this stage, we must recall and use the density equation of a sphere which for the case of this problem, since we are dealing with two different radius values, we can write this equation as Ro where Roe is usually the Greek symbol that denotes density is equal to M divided by four thirds multiplied by Pi multiplied by R one to the power of three minus R two to the power of three Awesome, which we can also write as capital M solid R one. So we're gonna denote S is solid divided by capital V. One is equal to lowercase M divided by four thirds multiplied by pi multiplied by capital R one to the power of three minus capital R two to the power of three Awesome. So we can write this equation because the solid sphere of radius R one has the same density. So now we can simplify this equation to write that capital M solid R one. So solid R one is equal to lower case M divided by four thirds multiply by pi multiplied by R one to the power of three minus R two to the power of three multiplied by V one. Awesome. And then we can simplify even further to write that capital MS. So M solid R one is equal to M divided by four thirds multiplied by pi multiplied by R one to the power of three minus R two to the power of three. And this is all multiplied by four thirds multiplied by pi multiplied by R one to the power of three. Awesome. And similarly, we can write the equation for the solid sphere of radius R two as so capital MS R two. So the mass of the salt sphere at R two is equal to M divided by four divided by three, multiplied by pi multiplied by R one to the power of three minus R two to the power of three, which we can multiply all of this by like like above four, divided by three, multiplied by pi multiplied by R two to the power of three. Awesome. So now at this stage, we must recall and use the equation for the gravitational potential which we can write as capital U hollow, which I'm gonna denote hollow as H. So you subscript H so U hollow is equal to ur one where U denotes the potential. Awesome. So Ur one minus Ur two. So once again, you is the potential. Awesome. And we can write this as you hollow is equal to negative G. So capital G multiplied by M solid comma R one multiplied by M prime, all divided by R minus negative capital G multiplied by capital M solid comma two multiplied by M prime, all divided by R which we can simplify and write that U hollow is equal to negative capital G multiplied by M prime divided by R multiplied by capital M solid, R one minus capital M solid comma R two, which we can simplify and write or we can expand it. I should say rather we can expand it. And we can say that you hollow is equal to negative capital G multiplied by M prime divided by R multiplied by lowercase M divided by four thirds multiplied by pi multiplied by R one to the power of three minus R two to the power of three, multiplied by four thirds multiplied by R one to the power of three minus M divided by four thirds multiplied by pi multiplied by R one to the power of three minus R two to the power of three. Awesome. And this is all multiplied by and I'm running out of room here. So I'll draw a little arrow. So this is all multiplied by four thirds multiplied by pi multiplied by R two to the power of three So moving right along here. So now we could start to simplify. So U hollow is equal to negative capital G multiplied by M multiplied by M prime, all divided by R multiplied by R one to the power of three, divided by R one to the power of three minus R two to the power of three minus R two to the power of three divided by R one to the power of three minus R two to the power of three. Fantastic. So then we'll simplify one last time and we will find that U hollow is equal to negative capital G multiplied by M multiplied by M prime, all divided by capital R. And this is our final answer. Hooray, we did it, we found our final final answer. Hooray. That was a lot of work, but it was really rewarding when we found that final answer. So looking at our multiple choice answers, the correct answer has to be the letter D which states that negative capital G multiplied by M multiplied by M prime all divided by R and that's it. We found our final answer. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.

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