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Unformatted text preview: MA 36600 LECTURE NOTES: MONDAY, JANUARY 26 Modeling with First Order Equations (cont’d) Escape Velocity. Say that we have a rocket of mass m which blasts off from the surface of the Earth at an initial velocity v . Gravity pulls back on the rocket no matter how high the rocket goes, so is it possible that the rocket can ever leave the Earth and go to another planet? The escape velocity v e of a rocket is that velocity such that the pull of gravity is too weak to force the rocket to return. That is, if the rocket has an initial velocity v > v e then it will never return! We derive a formula for the escape velocity v e . Recall Newton’s Law of Gravity and Newton’s Second Law of Motion: m d 2 x dt 2 =- G mM r 2 where we define the variables m = mass of the rocket M = mass of the Earth r = distance between the center of masses of the rocket and the Earth R = radius of the Earth = 6 . 38 × 10 6 meters = 3963 . 11miles x = r- R = height of the rocket from the surface of the Earth G = universal gravitational constant g = GM R 2 = 9 . 81meters / sec 2 = 6 . 10 × 10- 3 miles / sec 2 The differential equation above reduces to the equation m d 2 x dt 2 =- mg R 2 ( x + R ) 2 which can also be written as d 2 x dt 2 =- g 1 + x R 2 . (Compare with Computer Lab #2.) We wish to solve this differential equation. Unfortunately, we do not yet know how to solve second order differential equations which are nonlinear, so we perform a trick. Weyet know how to solve second order differential equations which are nonlinear, so we perform a trick....
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This note was uploaded on 05/24/2011 for the course MA 36600 taught by Professor Ma during the Spring '09 term at Purdue.
- Spring '09