[i]Sorry about the one day delay on this week's posting. Posting this completely slipped my mind yesterday![/i] Welcome to week 11 of Science Friday! This week, we are going to continue our discussion of classical gravity. However, this week’s post is going to focus solely on the idea of escape velocity. Before reading this post, you should read last week’s SciFri which establishes the fundamentals of classical gravity. I will be assuming here that the readers have this background. On to the actual science! When we throw a ball into the air, it, very expectedly, comes back to the ground. The faster we throw the ball upward, the longer it takes to return to us. Taking this reasoning to its natural conclusion, we recognize that, at a certain upward velocity, the ball will not return to us and will effectively escape the gravitational field. How do we calculate this escape velocity? From a first year physics class, the following formula is presented v = √(2GM/r) where G is the gravitational constant, M is the mass of the object producing the gravitational field (usually a planet/moon), and r is the distance from the center of the planet/moon to the object we wish to escape. But where does this formula come from? It is my view that formulas are useless if you do not know how they were developed. We can derive this equation using some very basic integral calculus. If you are not familiar with this level of mathematics, I will try to clarify as I go along. We start with the universal law of gravitation F = GMm / r^2 From basic mechanics, we know that the [i]work[/i] (energy) needed to move an object a distance “r” by a force “F” is described by the equation W = Fr If we wanted to be more mathematically formal, we would say that the work necessary to move an object a distance r by a force F is equal to the dot product of the force vector and displacement vector. If you do not understand this, don’t sweat it. This means that the [i]integral of force with respect to distance[/i] is equal to [i]work[/i]. Mathematically, the integration operation is essentially finding the area of infinitely many rectangles with infinitely small lengths. When we integrate force with respect to distance, we are finding the force times distance (area of rectangles) over an entire distance interval. Knowing this, let’s integrate the universal law of gravitation. ∫F = ∫(GMm / r^2) dr Our limits of integration for the definite integral will be from r to infinity, with r being the radius of the planet. The reason why the lower limit is r is because we want to know the escape velocity from the surface of the planet. The upper limit is infinity because the escape velocity implies an object completely leaving a planet’s gravitational field. Since G, M, and m are constants, we can factor them out of the integrand. GMm ∫(1 / r^2) dr Integrating, we get GMm(-1/r) Since we are evaluating this as a definite integral from r to infinity, our expression becomes limit as b approaches infinity of GMm(-1/b + 1/r), which equals GMm / r Since the above expression is equal to work, we can set it equal to the kinetic energy expression, (1/2)mv² (1/2)mv² = GMm / r Multiplying both sides by 2 and dividing both sides by m, we get v² = 2GM / r Therefore v = √(2GM/r) And there we are. Let’s quickly use this formula to calculate the escape velocity for planet Earth (notice that it does not depend on the object’s mass, only on the planet’s mass, similar to orbital speed). radius of Earth (r) = 6,371,000 meters mass of Earth (M) = 5.972e24 kilograms Calculating v, we get roughly 11.2 km per second. This assumes no air resistance, of course. Note that this is not the velocity a rocket needs to leave the Earth. For a rocket, as long as there is continuous propulsion, it will be able to leave the Earth’s effective gravitational field. I hope you enjoyed this week’s SciFri. While I have completed the topic of classical gravity, we are by no means done. Einstein’s famous general theory of relativity provides a slightly more accurate framework for assessing the gravitational force and also poses an explanation of how it works (notice that all of our discussion on gravity thus far has described what happens but not exactly why it happens). I will not be covering general relativity as it currently exceeds my mathematical understanding (i.e., it requires calculus of several variables to adequately understand). As a result, I will be moving on to a new topic next week. Please leave comments and questions below. I look forward to reading reactions to my posts and participating in discussions. As always, you can check out previous weeks of SciFri by clicking the #sciencefriday tag and clicking “all time” next to “created.”