 Hello. This is a video about work, an example of propulsion, an application of integration. So in this example, there's a space module that weighs 15 metric tons on the surface of Earth. How much work is done in propelling the module to a height of 800 miles above the Earth? And we're going to assume a radius of 4,000 miles for planet Earth. We'll ignore air resistance and propellant weight. So here's what we have. We have a little image here of planet Earth. I want to draw a coordinate plane using this. So the center of the Earth will be my origin, and then you'll have your x and y axis. And you know that the distance, the radius of the Earth is 4,000, 4,000 miles. And you know that there's a space module, so I'll draw a little space capsule thing. That's my space capsule. And it's going to be propelled 800 miles above the Earth. So the distance between planet Earth and the space capsule will be, or space module, will be 800 miles. So I have 4,000 miles, then I have 800 miles. So together that's, well, 4,800 miles. The capsule is going to be 48 miles from the center of the Earth once it's propelled. So one thing we need here is we need one of our physics laws that says the weight of a body varies inversely with the square of its distance. We'll use x to represent distance from the center of Earth. So we know work equals force times distance. Well, we can find the force component of this by saying, okay, force is equal to k over x squared. That's how you write that weight, which is force of a body, varies inversely with the square of its distance, x, f equals k over x squared. k is the constant or the relationship constant. So we need to actually find k as it does exist in this situation. We know that f equals 15 and x equals 4,000. So the force is 15 because it's 15 metric tons for the space module. And then initially you are 4,000 miles from the center of the Earth before you begin propelling upward. So you get 15 equals k over 4,000 squared. You get 15 equals k over 16, followed by 6 zeros. That's 16 million. And you actually do end up getting that k equals 240 million. So force as a function of distance x is 240 million over x squared. So to calculate work, we're going to integrate work equals force times distance. We're going to integrate this 240 million next to the negative second. I'll go ahead and put it in the format ready to integrate. And then your distance increment will be your delta x represented by dx. And I'm propelling from 4,000 miles from the Earth center all the way up to 4,800. I'm going from on the y-axis here. I'm going from 4,000 all the way up to a distance of 4,800. All right. So in this situation, I can actually evaluate this integral. So yes, you do get 240 million x to the negative first over negative one. And you're going to evaluate from the upper bound is x equals 4,800. And the lower bound is x equals 4,000. Then you can work with this little more, clean it up a little bit. You get negative 240 million over x. And you're evaluating lower bound x equals 4,000 all the way up to x equals 4,800. Anyway, you plug in 4,800, you plug in 4,000, you'll end up getting negative 50,000 plus the 60,000. So this actually becomes 10,000. So the work done is 10,000. And your distance is miles. And then your force here is tons. So it would be mile tons. If you convert this to foot pounds, miles to feet, tons to pounds. 1.164 times 10 to the 11 foot pounds. If you want to convert it to Joules, well, you know that one foot pound is equal to 1.35582 Joules. 1.35582 Joules. So what this means is that when you convert 1.164 times 10 to 11 foot pounds to Joules, you actually get 1.578 times 10 to the 11th Joules. And that's the amount of work that was done. So that's a propulsion example. Thanks for watching.