OpenStax College Physics for AP® Courses, Chapter 13, Problem 36 (Problems & Exercises)

This is College Physics Answers with Shaun Dychko. We're going to assume that many Ping-Pong balls are going to be placed on the surface of the Earth and there are going to be as many Ping-Pong balls as Avogadro's number. And we're given a diameter for each ball three and three-quarter centimeters which we convert into meters and then divide by two because we want to know what the radius of each ball is because our volume formula for a ball is 4/3 pi R cubed. And now I have a feeling that this answer for the depth of this layer of balls is going to be in units of kilometers. So we're going to take this radius and convert it into kilometers so that we have kilometers everywhere in our formulas. So we divide this 1.875 times 10 to the minus 2 meters by 1000 and we get 1.875 times 10 to the minus five kilometers is the radius of one Ping-Pong ball. I looked up the radius of the Earth 6,371 kilometers. So to find the depth of this layer of Ping-Pong balls we'll find the difference in volume between this big outer sphere including the Earth and the layer of Ping-Pong balls and then subtract from that volume the volume of the Earth which is 4/3 pi times radius of the Earth cubed. And then that'll give us the volume of this shell. And then we know that the volume of the shell is going to be the number of Ping-Pong balls which is Avogadro's number multiplied by the volume per ball. Now the volume for a single ball is going to be 4/3 pi times the radius of a ball cubed, but we're told that this space between the balls will also add 25% to this volume. So we add 0.25 times the volume of a single ball. This works out to 1.25 times the volume of a single ball times Avogadro's number - is the total volume of this shell of Ping-Pong balls. And the shell is just this section here consisting only of the balls. So we can equate this volume to this volume and figure out what D is and that'll be the answer to our question - the depth of Ping-Pong balls needed to make this volume of shell. So we have this copied here in place of V. So we have 4/3 pi times radius of the Earth plus this depth all cubed minus 4/3 pi times radius of the Earth cubed equals 1 and 1/4 times Avogadro's number times 4/3 pi times radius of a ball cubed. And then we add 4/3 pi R E cubed to both sides. And we get this line here. And then the 4/3 pi is a common factor in every term. So we can divide both sides by 4/3 pi and so it cancels everywhere. And we end up with this line. And then we take the cube root of both sides which is the same as raising both sides to the exponent one-third. So we have radius of the earth plus the depth of the Ping-Pong ball layer equals the cube root of 1 and 1/4 times Avogadro's number times radius of the ball cubed plus radius of the Earth cubed. And then subtract radius of the Earth from both sides and we have an expression for the depth. So the depth is the cube root of 1 and a quarter times Avogadro's number which is 6.02 times 10 to the 23 times the radius of a single ball cubed which is 1.875 times ten to the minus five kilometers cubed, plus the radius of the Earth 6,371 kilometers cubed minus the radius of the Earth. And this is 40.4 kilometers. So this layer of Ping-Pong balls will be 40.4 kilometers deep in order to have an Avogadro number of Ping-Pong balls on the Earth.