sol7 - Massachusetts Institute of Technology Physics...

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Massachusetts Institute of Technology - Physics Department Physics-8.01 Fall 1999 Solutions for Assignment # 7 by Dru Renner [email protected] November 7, 1999 9:50 pm Problem 7.1 (Ohanian, page 271, problem 55) Throughout this problem we will use the rocket equation: v f v i = u ln à M i M f ! Consider two rockets: (1) a multiple-stage rocket with initial mass M 0 and with a ±nal mass of the last stage M ; (2) a single-stage rocket with initial mass M 0 and ±nal mass M . Suppose all stages of (1) and the single stage of (2) have the same exhaust speed u . We know that the terminal speed of the single-stage rocket is v 2 = u ln µ M 0 M Rocket (1) has less fuel than rocket (2) because each jettisoned stage has some mass; we should expect that rocket (1) will have a smaller terminal speed than rocket (2). To see this, consider a two stage rocket. Suppose the mass of the fuel for the ±rst stage is M F 1 , the mass of the ±rst jettisoned stage is M 1 , and the mass of the fuel for the second stage is M F 2 . Then the ±nal mass will be M = M 0 M F 1 M 1 M F 2 . After burning MF 1, the speed of the rocket is v ± = u ln µ M 0 M 0 M F 1 Then the ±rst stage is jettisoned, so the mass is reduced to M 0 M F 1 M 1 . If we assume the ±rst stage moves away at zero velocity relative to the remainder of the rocket then the remainder of the rocket will not speed up due to the jettisoned stage. Then M F 2 is burned, and the speed of the rocket is v ±± = u ln µ M 0 M F 1 M 1 M 0 M F 1 M 1 M 2 = u ln µ M 0 M F 1 M 1 M Thus the terminal speed for rocket (1) is 1
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v 1 = v ± + v ±± = u ln µ M 0 M 0 M F 1 + u ln µ M 0 M F 1 M 1 M = u ln µ M 0 M 0 M F 1 · M 0 M F 1 M 1 M = u ln µ M 0 M u ln µ M 0 M F 1 M 0 M F 1 M 1 = v 2 u ln µ M 0 M F 1 M 0 M F 1 M 1 Since M 1 > 0 the second piece in the above expression is negative, so v 1 <v 2 For each stage that is jettisoned there will be similar terms which further reduce v 1 . Problem 7.2 Consider velocities positive if along the initial direction of the spacecraft. Let the ±nal velocity of the spacecraft be v ± and the ±nal velocity of the planet be V ± . Conservation of momentum gives mv + MV = mv ± + ± = V ± V = m M · ( v v ± )( 1 ) Mechanical energy is conserved. If we consider the initial and ±nal moments of the system to occur when there is sufficient separation between the planets, then the gravitational potential energy can be ignored. Therefore kinetic energy is conserved. 1 2 mv 2 + 1 2 2 = 1 2 mv ± 2 + 1 2 ± 2 = m M · ( v v ± ) · ( v + v ± )=( V ± V ) · ( V ± + V ) Using Equation ( ?? ) we can rewrite the above equation as m M · ( v v ± ) · ( v + v ± )= m M · ( v v ± ) · ( V ± + V V ± = v + v ± V Substituting this expression back into Equation ( ??
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This note was uploaded on 11/23/2009 for the course 640 373 taught by Professor Staff during the Spring '08 term at Rutgers.

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sol7 - Massachusetts Institute of Technology Physics...

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