Under some circumstances, a star can collapse into an extremely dense object made mostly of neutrons and called a neutron star. The density of a neutron star is roughly 10^14 times as great as that of ordinary solid matter. Suppose we represent the star as a uniform, solid, rigid sphere, both before and after the collapse. The star's initial radius was 7.0 * 10^5 km (comparable to our sun); its final radius is 16 km. If the original star rotated once in 30 days, find the angular speed of the neutron star.

Video transcript

Okay. Hey, everyone in this problem, we're told that neutron stars are believed to form when a massive star depletes its fuel and collapses the collapse, crushes together all protons and electrons into neutrons? The density of a neutron star is about 10 to the times greater than the density of the sun. We can imagine that the sun, the radius of 6.96 times 10 to the five kilometers, collapses into a neutron star of radius 15 kilometers. Using the Sun's average rotational speed of one rotation every days. What would be the rotational speed of the neutron star formed? They were told to model the sun and the neutron star formed as uniform solid spheres. All right, so let's start just by writing out some of this information we're given. Okay, we're given kind of an initial situation. And a final situation where the initial situation is when we have the sun. And the final situation is when we have the neutron star, Okay, the sun collapses into a neutron star. So we go from our initial state to our final state. Alright, so initially, what are we told about the initial Situation while we're told the radius of the sun is our initial radius ri is 6.96 times 10 to the five km. Now we want to convert this into meters. So this is going to be 6.96 times 10 to the five kilometers times 1000 m per kilometer. Okay, The unit of kilometer cancels. And we're left with 6.96 times 10 to the exponents. eight m. Okay, so to go from kilometers. two m, we multiply by 1000. Alright, what else are we told while we're told it's average rotational speed. Okay, So that means that our angular speed or angular velocity omega i it's going to be equal to what? Well, this is one rotation every 27 days. Okay, well one rotation we know is two pi ingredients. Okay, if you go all the way around the circle back to where you started, that's two pi radiance. So we have two pi radiance Per 27 days. Okay, now we want to convert this into radiance per second. So let's multiply. We have one day And then one day we have 86,400 seconds. So the unit of day cancels just like the customers did. And we're left with an angular velocity of 2.69 3, 4 times 10 To the - radiance per second. Alright, so that's all the information we're told about the sun. Okay, so let's move on to the neutron star. Okay, so the neutron star we're told has a radius of 15 km. And again, multiplying by 1000 m/km2. This gives us a radius of 15,000 m. Now it's angular velocity or angular speed. That's what we're trying to find. We're trying to find that rotational speed. So we don't know this. And that is what we're trying to find. Okay. And the other information we're told is that the final density or the density of a neutron star. Okay. So the density will call it D. F. Is equal to 10 to the 14 times greater the density of the sun. Okay. So that's going to be equal to 10 to the exponent. 14 times the density initially or the density of the sun. Alright. So we have all of this information that we're given. Now. We want to find this rotational speed. This omega F. We have information about the radius and the angular velocity. So let's recall our conservation of angular momentum. Okay. In this case we have no net external torques. We have conservation of angular momentum. And what that tells us is that the angular momentum L initially is going to be equal to the angular momentum L. F. Final Okay, now let's recall what is angular momentum, angular momentum is given by I the moment of inertia times omega the angular velocity. Mhm. Just like with the regular momentum, we have mass times velocity. In this case we have a moment of inertia times angular velocity. Okay. Alright. And same for the final case I f the moment of inertial final times omega F. That final rotational speed. All right. So what about the moment of inertia I how can we calculate this? Well, you can look at a table in your text work or that your professor provided. We're gonna be looking at the moment of inertia for a solid sphere, uniform solid sphere. Okay. We're told that is a shape to consider these planets or the sun in the neutron star. Okay. And so if we do that, we find that the moment of inertia I for uniform solid sphere is going to be 2/ M the mass are the radius squared. So on the left hand side we get 2/5 the initial mass. The initial radius squared times the initial angular velocity and same thing on the right side. Okay, It's still a uniform solid sphere. So we get to fifth the final mass, the final radius squared times that final angular velocity, rotational speed. Okay. And again, we're looking for omega. Alright, so we know omega I we know the radius R. I. And R. F. We don't necessarily know these masses M. I. And M F. Okay, we're told some information about the density and we know that we can relate the density to the mass. So let's go ahead and check there to figure out the relationship between M. I. And M. F. And we don't want to make any wrong assumptions here. So let's look at the density. Um and see what we can find out. Okay. All right, so, we know that the final density D. F. And before we move on, let me label this as equation one. Because we're gonna come back to that. So the final density is equal to 10 to the 14 times the initial density. Okay, well what is density recall that the density D. Is equal to the mass M divided by the volume. V Putting these two together. This tells us that the final mass divided by the final volume is going to be equal to tend to the exponent times the initial mass divided by the initial volume. Okay, so now we have this relationship between these two masses that we're interested in. The only thing in here that we don't know yet is these volumes? Okay. But remember we have solid spheres, we know the radius so we can go ahead and find the volumes. Okay, so let's start with the one on the left. Okay. The final volume V f we're talking about the volume of a solid sphere. So we get four thirds pi r cubed. This is 4/3 pi times now. The final radius we found was 15,000 m. K. We were given 15 km. We converted to meters. All cubed. Okay, and doing the same for the initial volume V. I again, solid sphere. So we get four thirds pi r cute. And in this case the radius was 6.96 times 10 to the eight m. All cute. Okay, all right, so we have our two volumes. Let's go back to our equation and substitute it. So we get the mass M f. Final mass divided by the volume four thirds pi times 15,000 m. All cute. Is equal to 10 to the 14. The initial mass M. I divided by the initial volume 4/ pi times 6.96 times 10 to the eight m occupied. Alright, so this 4/3 pi in both terms we can divide out and it will cancel. And then if we multiply we get M. F. Times 6.96 times 10 to the exponents eight m. All cute. Is equal to 10 to the exponent. 14 times M. I times 15,000 m. All cute. Alright. So if we divide by this 6.96 times 10 to the eight and work this out, we're going to find that the final mass M F is equal to 1.1 times the initial mass. M. I. Okay. And so these masses are pretty much equal at least significant digits. They're equal. Okay, so we could have assumed that the masses were equal and divide that out in our equation. But it's good that we checked. Okay. We weren't given the information about the mass. We were given the information about the density and we used that to kind of figure out this relationship between those masses. Okay. Alright. So if we go back to equation one what do we have? Well our equation is that 2/5 the initial mass. The initial radius squared? Omega I is equal to 2/5. Final mass. Final radius squared. Oh my God. Okay and now we know this. M F is equal to 1.1 M I. Okay, so we can go ahead and substitute in our values And we get 2/5 times am I? We don't know the value of M. I times the initial radius 6.96 times 10 to the eight m squared times the initial angular speed, which we calculated to be 2.69, 3, 4 times 10 to the -6 radiance per second. And on the right hand side we get 2/5 times M. F. Which we just found to be 1.1 times the initial mass. M I times the radius of 15, m. All squared many times omega the angular speed we're looking for. Alright, so now, instead of having M I. On one side, M F on the other side and not knowing if they're equivalent, we have M I on both sides and we can divide out by that mass. Okay, so we didn't need to know the exact mass but we did need to know the relationship between those two masses in order to work out this calculation. Alright, so on the left hand side we're going to have 5. 18, 9 times 10 to the exponent 11. And on the right hand side we get 9. Times 10 to the exponents seven times are angular velocity omega f. We can divide to isolate omega F. We get an omega F value and we missed our units here. Okay. We have meters squared times of radiant per second. So we have meters squared times, radian per second. On the left hand side, on the right hand side we have meters squared. So when we divide we're left with the unit of radiance per second. Okay, so we get And 92.98 radiance per second. Okay. And if we kind of approximate we're gonna round this and write it to two significant digits in scientific notation. We get 5.8 times 10 to the three radiance per second for that final rotational speed. The speed of that neutron star. Alright, so if we go back up to our answer choices, we see that we have answer choice C. Okay, that rotational speed of the neutron star is gonna be 5.8 times 10 to the three radiance per second. Thanks everyone for watching. I hope this video helped see you in the next one.