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# lecture_6 (dragged) 1 - 2 MA 36600 LECTURE NOTES: MONDAY,...

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2 MA 36600 LECTURE NOTES: MONDAY, JANUARY 26 This is the implicit solution for the initial value problem. (The left-hand side is the change in kinetic energy, whereas the right-hand side is the change in potential energy.) This expression states how the distance x = x ( t ) is related to the velocity v = v ( t ). As t →∞ , we want x ( t ) →∞ and v ( t ) 0; this means the rocket can travel inFnitely far away from the Earth and never reverse its velocity. This gives the inequality 1 2 mv 2 0 lim t →∞ ° 1 2 mv ( t ) 2 1 2 mv 2 0 ± =l im t →∞ ° mgR 2 x ( t )+ R mgR ± = mgR which reduces to 1 2 mv 2 0 mgR = v 0 ² 2 gR. The expression v e = ² 2 gR = 11 . 19 km / sec = 6 . 95 miles/sec is the escape velocity. Is it possible to reach the escape velocity in a commercial jet plane, or even in the space shuttle? The
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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.

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