JavaScript is required to use Bungie.net

OffTopic

Surf a Flood of random discussion.
5/10/2013 10:19:31 PM
5

Science Friday, Week 10: Introduction to Classical Gravity

Welcome to the tenth week anniversary of Science Friday! It has been a truly wonderful experience writing SciFri for the past two-and-a-half months. At first, I thought I would just be relaying information without much consequence. However, my actual experience has been very different; I’ve learned a lot myself by researching and articulating scientific and mathematical ideas in a cohesive manner. Today, I want to talk about a phenomenon that is perhaps the most pervasive in the entire known Universe. A force of nature that is so well understood ... and isn’t at the same time. I am, of course, talking about gravity. This week, I will cover the classical understanding of gravity and expand upon it in the following weeks. This topic is far too expansive to cover in one week. From ancient times, mankind instinctually knew about the existence of gravity. After all, something had to explain why we fall back to the ground when in the air or why a ball sails through the sky only for a brief period of time. However, this concept was not well understood—that is, mathematically—until the seventeenth century. The work of Galileo and Newton after him truly elucidated the nature of this force that seemingly nails us to the Earth’s surface. In week 2 of SciFri, I discussed the instrumental work of Galileo in describing the kinematics of free falling objects. I would urge those who have not read this SciFri or are not familiar with Galileo’s work to do so. [url=http://www.bungie.net/7_Science-Friday-Week-2-The-Beginning-of-Science-wit/en-us/Forum/Post?id=59985172]Here is the link.[/url]. While Galileo’s work was invaluable to the understanding of gravity as we know it, the true hero of our classical understanding of gravity is none other than Sir Isaac Newton. While Newton did not [i]discover[/i] gravity per se, he recognized that the force that gives objects with [i]mass[/i] weight is the same force that permits the orbits of the planets and moons. From this stroke of insight, Newton developed the [b]universal law of gravitation[/b], which is described a very simple and elegant formula: F = GMm / r^2 (equation 1) F = force of gravity (often describes as “weight”) G = universal gravitation constant (6.67e-11 N*m^2 / kg^2 ) M = mass of object 1 (usually the larger body or planet) m = mass of object 2 (usually the smaller body or object) r = distance between the two objects’ center of gravity (center of planet to another center of gravity) Newton’s second law of motion also describes force, but slightly differently. F = ma (equation 2) If we call F in equation 2 the force due to gravity, then little a becomes g, the acceleration due to gravity. F = mg Now that we have two equations describing the force due to gravity, we can combine them. mg = GMm / r^2 g = GM / r^2 (equation 3) From equations 1 and 3 alone, we draw enormous insight into the kinematics planetary motion and weight. Using equation 1, we can calculate with what force the sun pulls on the Earth (and therefore the force the Earth pulls on the Sun by Newton’s third law). While equation 1 assumes the two masses are “point” masses (i.e., all of their mass is contained within a single point), this approximation is remarkably accurate in normalized settings. Of course, to obtain exact answers, calculus is used to infinitely sum up infinite radii of mass (integrate). Another equation worth mentioning that appears from Newton’s universal law describes orbiting kinematics of objects. From classical circular motion, we know that the centripetal force (inward force of circular motion) is described by F = mv^2 / r (equation 4) F = centripetal force v = tangential speed r = radius of circular path If the centripetal force is the force due to gravity, then we can set equation 4 equal to equation 1. mv^2 / r = GMm / r^2 Simplifying, we get v = sqrt(GM / r) (equation 5) In equation 5, v represents the tangential speed of an orbiting object. As we can see, for a planet with a given mass M and given distance from orbiting object, r, the minimal speed to maintain orbit is constant. In other words, it does not matter how massive the object orbiting a planet is: whether it be another planet or a golf ball, the minimum speed to maintain orbit is exactly the same. Next week, I will wrap up this discussion of classical gravity by using the information established here (and a bit of integral calculus!) to derive the formula of the escape speed of an object leaving a planet. Beyond that, we may venture into the modern understanding of gravity (i.e., general relativity, courtesy of Einstein) and even foray into the realm of particle physics. Looks like we will be talking about gravity for a while! I hope you enjoyed this intro segment into gravity. Tune in next week for more!

Posting in language:

 

Play nice. Take a minute to review our Code of Conduct before submitting your post. Cancel Edit Create Fireteam Post

  • I think b.next permits the use of special characters which might help to make your wonderful posts a bit prettier. test: ² , √ , ∫ , ∑ , → , ≈

    Posting in language:

     

    Play nice. Take a minute to review our Code of Conduct before submitting your post. Cancel Edit Create Fireteam Post

    1 Reply
    • Edited by Winy: 5/11/2013 2:53:34 AM
      What's also interesting is that the Universal Law of Gravitation's equation is remarkably similar to Coulomb's Law: Coulomb's Law: F(En) = K * [q(1) * q(2)] / d^2 (Just an example problem): K = 9.0E^9 N m^2/c^2 q(e^-) = -1.6E^-19 q(p^+) = 1.6E^-19 Assume d = 3.7E^-11 m F(En) = 9.0E^9 * [-.16E^-19 * 1.6E^-19) / 1.369E^-21 F(En) = 1.7E^-7 N Compared to a problem using Newton's Law, you can understand the massive spike in how much weaker gravity is than the electromagnetic force. F(g) = G * [m(1) * m(2)] / d^2 G = 6.7E^-11 N m^2/kg^2 m(e^-) = 9.1E^-31 kg m(p^+) = 1.7E^-27 kg d = 3.7E^-11 m (Same distance as before) F(g) = 6.7E^-11 = (9.1E^-31 * 1.7E^-27) / 1.369E^-21 F(g) = 7.6E^-47 N Electromagnetic force: 1.7E^-7 N Gravitational force: 7.6E^-47 N (<-- Holy balls, that's tiny) The Electromagnetic Force is 2.2E^39 times stronger than the Gravitational Force.

      Posting in language:

       

      Play nice. Take a minute to review our Code of Conduct before submitting your post. Cancel Edit Create Fireteam Post

      6 Replies
      • [url=http://research.microsoft.com/apps/tools/tuva/#data=3|||]I'll just leave this here.[/url]

        Posting in language:

         

        Play nice. Take a minute to review our Code of Conduct before submitting your post. Cancel Edit Create Fireteam Post

      • I recognise all of these equations... <:/

        Posting in language:

         

        Play nice. Take a minute to review our Code of Conduct before submitting your post. Cancel Edit Create Fireteam Post

      • Love these treads, great job again!

        Posting in language:

         

        Play nice. Take a minute to review our Code of Conduct before submitting your post. Cancel Edit Create Fireteam Post

        1 Reply
        You are not allowed to view this content.
        ;
        preload icon
        preload icon
        preload icon