Suppose that an asteroid traveling straight toward the center ...

Question

Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bury itself just below the surface. What would have to be the mass of this asteroid, in terms of the earth's mass M, for the day to become 25.0% longer than it presently is as a result of the collision? Assume that the asteroid is very small compared to the earth and that the earth is uniform throughout.

Accepted Answer

Hey everyone welcome back in this problem. We have a planet and it's going to be orbiting too slowly around its star. Okay. It's going to be attracted by gravity and fall into the star. Okay, So we have the planet moving radial E. That means not orbiting the sun or spinning on its axis and it's going to embed into the sun's equator and were told to take the mass of the sun to be M. And what we're asked to find is what is the mass of the planet In terms of M. That increases the Sun's average period of 27 days by 10%. And we can assume that the planet is much smaller and that the sun is consistent at all points. And when we're reading problems like this and we hear the sun is consistent all points we think of is okay, that's a solid sphere. Okay. And what that does is that allows us to calculate an angular momentum. Okay. We'll need to know the shape. And so we think about that. When you're reading these questions, what is that piece of information telling us? How can we relate that to the math? All right, So let's draw what's going on here. We have our son and then we have this little planet here and it's going to be moving radiantly towards the sun and then it is going to embed itself in the sun's equator. All right now because this is much smaller. This planet is much smaller than the sun. We're going to assume that it embeds on the surface of that of that, son. Alright, so let's start with our conservation of angular momentum. We know we have conservation of angular momentum so we can write that L not the initial angular momentum is equal to L. F. The final angular momentum. The initial angular momentum. We have 22 things. We have the little planet and we have the sun. So we're gonna have contributions from each. We're going to have the angular momentum of the sun plus the angular momentum of the planet. And after this planet embeds into the sun we just have one thing. Okay, so they're moving together one velocity. And so we're going to have the angular momentum final which is gonna be the angular momentum of those together. Alright, so working out some more details here, let's recall. The angular momentum is given by I omega. Okay. Where i is the moment of inertia and omega is the angular velocity. So for the sun we have I. S. And we're gonna call it omega. Not the initial angular velocity of the sun. We have the angular momentum of the planet. P. Okay, well let's pay attention P. We're told is moving radial E. Okay, it's not orbiting the sun, it's not spinning on its axis. And so that is going to be a linear motion. When we have linear motion. We're writing angular momentum recall. We use our um VI Okay, we're just gonna put the subscript P2 mean planet for each of these times sine of the angle Theta on the right hand side. We're going to have similarly to with the sun. We have the moment of inertia. Oops. Not w moment of inertia final times. Omega final the final angular philosophy. Alright, let's work from left to right and see what kind of information we can fill in. We have the moment of inertia of this fear and this is where that information about the sun comes back. The sun we're told is this um consistent at all points? We've said that means it's a solid sphere. Okay. And if we look at our moment of inertia table, okay, you can look in your textbook or the table that your professor provided for you and you're gonna see at the moment of inertia for solid sphere is two fists big. M. Okay. The mass of the sun, big r the radius of the sun squared. And then we have omega. Not still. Now when we're looking at this planet moving radial E. Planet moving radial E. And then embedding in the equator while that motion is parallel. Okay, so that motion is parallel, that means our angle is zero. Well, if our angle is zero, sine of theta is also going to be a zero. So this entire middle term is going to go to zero. Okay. Alright, well that's nice, we don't have to worry about that. And then we can do the right hand side on the right hand side. We have I. F. Well let's work out over here, off to the side, what if is gonna look like? So I. F. Is going to be the final moment of inertia and that's gonna be the moment of inertia of this sun with the planet embedded in it. Okay, so we're gonna have that, it's going to be made up of the moment of inertia of the sun plus the moment of inertia of the planet that's embedded. We use subscript P. E. There for the embedded planet while the moment of inertia of the sun we know it's two half or 2/5 sorry, big M big R squared. Now the moment of inertia of the planet we mentioned because it's much smaller, we're gonna assume that embeds on the surface because it's much smaller. We can take it as a point mass. So if we're looking at the moment of inertia of a point mass, looking at our table, we see that it's going to be M. R. Squared. Okay, where r is the radius or the distance to the axis of rotation? Okay. And again in this problem we're assuming it's embedding at the surface, so the radius is going to be the radius of the sun. Okay, so here we're gonna have plus m the mass of the planet, little mm P. And then the radius is going to be the radius of the sun squared. Big R squared. Okay. Alright, so that's I. F. So we can go ahead and put that quantity into our equation right here, We're gonna have to over five um R squared plus little mp big R squared times omega F. Alright, so we've gotten to this point in the equation. Let's stop and think about what it is. We're trying to find we're trying to find the mass of the planet in terms of m. Okay, so let's start there. The mass of the planet is this little mp. So that's what we're looking for and we want to in terms of N. So that we increase the Sun's period by 10%. Okay, Alright, so let's go and deal with that piece of the question. Okay, we want the final period to be the initial period plus 10%. Okay, so that's gonna be plus 0.1, tina. Okay, so we can just write this as one plus one or 1.1. Sorry, tina. Okay, so we want the final period T. F. To be 1.1 times the initial period. All right, so we don't have period in our equation. We don't have period. And let's just go ahead and call this equation red star. Okay, so in that red start equation we don't have period but we do have angular momentum. Okay, well let's recall how can we relate angular momentum to period. Well, we know that the period is going to be two pi divided by the corresponding angular momentum. Okay, so we can write that for the final period and similarly for the initial period. T not that's gonna be two pi over. Oh my God. Alright, so let's combine what we've written about the periods. Okay, so T. F. Well, we know that we can write, so let's use these guys into this formula. Okay, we can write to pi over omega F. And let me just scroll down so we can have a little bit more room to work here. All right, two pi over my F. Okay, and then on the right hand side we have 1.1 times T not well, what's T not two pi over omega? Not. Alright, so now we've got an equation in terms of omega, which is good because that's the variable we have in that red star equation that we're working with. Okay, two pi we can divide on both sides and then multiplying that's gonna give us omega nought is equal to 1.1. Oh my gosh, okay, so it turns out that If we want the final period to be 1.1 times the initial period, 10% more than we need our initial angular velocity to be 1.1 time bigger than the final velocity. Okay, so that's kind of neat. Alright, so let's work with that. We're gonna go ahead and we're gonna substitute that omega not in our red star equation. Okay, so everywhere in our red star equation where we have omega naught, we're going to write it in terms of oh my God F. Instead. So we have two fists Big M. Big R squared. Omega not which is now going to be 1.1 times omega F. All right, And on the right hand side we have the same thing. Two fists, Big M. Big R squared plus little mp. Which is that mass that we're looking for of the planet Big R squared? Oh my God. Alright, so we have omega F on both sides so we can divide those terms are going to go away on the left hand side. Just simplifying. We're going to get 0.44 big M. Big R squared on the right hand side. But we see that we have R squared in both terms. So let's go ahead and factor that out. So we get R squared and that's going to be all times simplifying. Again, we get 0.4 M in the second term we're left with. Just this little mp. Well, R squared on both sides so we can go ahead and divide And then moving the 0.4 M to the left. We're going to have a little M. P is equal to 0.04. Mhm. And that's exactly what we were looking for. Now. We know the mass of the planet, we need is going to be 0.04 times the mass of the sun. Okay, in order to have the final period increased by 10%. Okay. So if we scroll back up to the options we have here, that's going to correspond with solution a. Alright. I hope that helped. Thanks everyone for watching.

Key Concepts

Conservation of Angular Momentum

Moment of Inertia

Rotational Kinematics

Watch next