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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 v0 . 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 ve 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 v0 > ve then it will never return! We derive a formula for the escape velocity ve . Recall Newton’s Law of Gravity and Newton’s Second Law of Motion: d2 x mM m 2 = −G 2 dt r 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 × 106 meters = 3963.11 miles x = r − R = height of the rocket from the surface of the Earth G = universal gravitational constant GM g= = 9.81 meters/sec2 = 6.10 × 10−3 miles/sec2 R2 The differential equation above reduces to the equation m d2 x m g R2 =− 2 dt2 (x + R) which can also be written as d2 x =−￿ dt2 g . x ￿2 R (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. We will write a differential equation which involves x and v instead of x and t. To this end, we note the following using the Chain Rule: 1+ d2 x dv dv dx dv = = · =v . dt2 dt dx dt dx Hence we find the initial value problem mv dv m g R2 =− 2, dx (x + R) x(0) = 0, v (0) = v0 . The left-hand side of this differential equation is the exact derivative of a function: ￿ ￿ dv d1 2 mv = mv . dx dx 2 Upon integrating both sides we find that can solve this differential equation implicitly: 1 m g R2 m v2 = +C 2 x+R for some constant C . This is an integral curve for the differential equation. This constant C can be found from the initial condition: 1 1 1 m g R2 2 2 C = m v0 − m g R =⇒ m v 2 − m v0 = − m g R. 2 2 2 x+R 1 ...
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This note was uploaded on 11/30/2011 for the course MATH 366 taught by Professor Edraygoins during the Spring '09 term at Purdue University-West Lafayette.

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