A particle is released at a height r_E (radius of Earth) abov...

Question

A particle is released at a height r_E (radius of Earth) above the Earth’s surface. Determine its velocity when it hits the Earth. Ignore air resistance. [Hint: Use Newton’s second law, the law of universal gravitation, the chain rule, and integrate.]

Accepted Answer

Welcome back. Everyone in this problem. A space probe is initially hovering A two R above Mars surface suddenly its thrusters dysfunction and it starts to fall freely toward the Mars surface. Given that R is Mars radius and M is the mass of Mars, calculate the probe's velocity just before it touches the surface. Assume that there is no air on Mars and take the outward direction from the center of Mars along its radius. As positive. A hint for this problem is to apply Newton's second law, the law of universal gravitation. And then utilizing the chain rule integrate to find the result for our answer choices. A says it's the square root of GM divided by RB, the negative value of the square root of GM divided by two RC, two multiplied by the square root of two GM divided by three R and D negative two multiplied by the square root of GM divided by three R. Now to solve this problem, our hints can be pretty useful because what are we trying to figure out here? Well, if we, if we were to draw, let's say that this is Mars. OK. And this is our probe just above Mars. And like our problem says, it's hovering at a distance of two are above Mars surface. OK. So that means then that here if we can represent on our diagram R as the surface of Mars, OK, as the radius sorry of Mars, then we want to find the velocity just as it touches the surface just before it touches the surface. In other words, we want to find the velocity at this point. No. How are we going to find that? Well, by Newton's second law, recall that it tells us that our gravitational force is going to be equal to negative GMM divided by R squared. OK. Where M capital M is the mass of Mars common M is the mass of the satellite and R is the distance of the satellite from the center of Mars while G is the gravitational constant. Now, what else do we know? Well, we know that this gravitational force based on Newton's second law is also going to be equal to the mass of the probe multiplied by its acceleration. So that means then that ma is going to be equal to negative GMM divided by R squared. If we can M from both sides, then we find that the acceleration equals negative GM. OK. Divided by R squared. Now, here's the question, what do we know about acceleration? Well, we know that our acceleration is our change in velocity with respect to time. And remember we're trying to figure out our speed, our velocity just before it touches the surface. So we can say then that our change in velocity with respect to time is equal to negative GM divided by R squared. Now again, what do we know about our change in velocity with respect to time? Well, we can relate it to our distance R by using the chain rule because by the chain rule, we can rewrite DVD T as DVD R or change in velocity with respect to distance multiplied by DRDT or change in distance with respect to time. Now recall that that's the very definition of velocity or a change in distance with respect to time. So then we can rewrite our problem as DVD R multiplied by VK are changing distant with respect to time and that equals negative GM divided by R squared. Now, what if we were to put our similar terms on one side? OK. Then now we find that our changing VNVDV multiplied by V equals negative GMK divided by R squared. Dr. Now, if you notice here, we can rewrite this as negative GM multiplied by D divided by R squared. And now based on how our problem is set up, we can integrate both sides of our equation. We can integrate. So let me write that here, we can integrate our left hand side with respect to the OK. Let me add that here and we can integrate on the right. Take out our constant negative GM and integrate with respect to R for one divided by R squared. Now, what are going to be our limits? Well, for our speed, remember it started to fall. So it started from rest down to our final speed that's called that VF And for our distance R, it's going to go from three R all the way to R. And why is that so? Well, again, our diagram can help us remember on our diagram, it's two RM. Its distance is two RM above Mars surface. So overall if we look at its entire distance from the center, because remember we, we, we are taking the outward direction from the center of Mars along its radius as positive then from the center up to the satellite is going to be three RM. And we want it to where it touches the surface, which would be RM. OK. So that's why we're integrating, let me move this here between RM and three RM. OK. That distance. So now let's go back to our problem then then we can start to integrate. And on our left hand side, that means we're going to add one to the power divide by the new power. Since it starts from zero, we don't need to find a bounder by substituting zero. So it's just going to be a half of VF squared. OK. On our left hand side and on the right now, when we integrate one divided by R squared. That's like integrating R to the negative two where we add one to the power OK. And divide by the new power and negative two plus one is negative one. So it's negative GM OK multiplied by negative R because that's negative uh that's R to the negative one divided by negative one between the boundaries of R and three R. We can write this even further. Let me make a note here, we can write this even further to then make our GM positive, negative multiplied by negative is positive. So that's gonna be GM. OK. Divided by R because R is raised to the negative one, OK? Between RM and three RM. Now let's substitute to our values for are here. Now, when we do that, OK? So that's a half VF screen is going to be equal to GM divided by RM minus GM divided by three RM. OK. Now both of these are the same term. So in a sense here, uh we can, we can, we, well, they are similar terms, OK. Notice that they are very similar. So we can start to simplify and when we do that, we're going to get two GM divided by three RM. OK? Because we could express our first fraction in terms of three RM, which would be three GM divided by three RM, same denominators and three GM minus GM equals two GM. OK. Now, since that's the case, we can multiply both sides by two because remember we're trying to solve for speed or velocity. So that tells us then that VF squared equals four GM divided by three IM and no, if we square root both sides, that means VF OK is equal to the square root of four GM, which is plus or minus two GM. OK. Well, you know what, let me write it out for you to see it. So that's the square root of four GM divided by three IM and now the square root of four GM, we can find the square root of four by itself. So that's plus or minus two, multiplied by the square root of GM divided by the, well, the square root of the entire term here. So that's the square root of GM divided by three RM because this term is still under the square root, we only took out the four. So now we have our speed. The only thing we need to ask ourselves, is it positive or is it negative? Well, since the outward direction along the radius is positive and the velocity of the probe is towards the center of miles, then we would take the negative result. Thus, our final speed is going to be equal to negative two, multiplied by the square root of GM divided by three RM. If we go back to our answer choices that tells us, then that D is the correct answer. Thanks for watching everyone. I hope this video helped

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A particle is released at a height r_E (radius of Earth) above the Earth’s surface. Determine its velocity when it hits the Earth. Ignore air resistance. [Hint: Use Newton’s second law, the law of universal gravitation, the chain rule, and integrate.]

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