140_pdfsam_math 54 differential equation solutions odd

# 140_pdfsam_math 54 differential equation solutions odd - v...

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Chapter 3 25. (a) From Newton’s second law we have m dv dt = GMm r 2 . Dividing both sides by m , the mass of the rocket, and letting g = GM/R 2 we get dv dt = gR 2 r 2 , where g is the gravitational force of Earth, R is the radius of Earth and r is the distance between Earth and the projectile. (b) Using the equation found in part (a), letting dv/dt =( dv/dr )( dr/dt ) and knowing that dr/dt = v ,weget v dv dr = gR 2 r 2 . (c) The di±erential equation found in part (b) is separable and can be written in the form vdv = gR 2 r 2 dr. If the projectile leaves Earth with a velocity of v 0 we have the initial value problem vdv = gR 2 r 2 dr , v ± ± ± r = R = v 0 . Integrating we get
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Unformatted text preview: v 2 2 = gR 2 r + K, where K is an arbitrary constant. We can ²nd the constant K by using the initial value as follows: K = v 2 2 − gR 2 R = v 2 2 − gR. Substituting this formula for K and solving for the velocity we obtain v 2 = 2 gR 2 r + v 2 − 2 gR. (d) In order for the velocity of the projectile to always remain positive, (2 gR 2 /r ) + v 2 must be greater than 2 gR as r approaches in²nity. This means lim r →∞ ² 2 gR 2 r + v 2 ³ > 2 gR ⇒ v 2 > 2 gR. Therefore, v 2 − 2 gR > 0. 136...
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## This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at Berkeley.

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