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 M i M f Consider two rockets: (1) a multiple-stage rocket with initial mass M 0 and with a final mass of the last stage M ; (2) a single-stage rocket with initial mass M 0 and final 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 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 first stage is M F 1 , the mass of the first jettisoned stage is M 1 , and the mass of the fuel for the second stage is M F 2 . Then the final 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 M 0 M 0 M F 1 Then the first stage is jettisoned, so the mass is reduced to M 0 M F 1 M 1 . If we assume the first 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 M 0 M F 1 M 1 M 0 M F 1 M 1 M 2 = u ln 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 M 0 M 0 M F 1 + u ln M 0 M F 1 M 1 M = u ln M 0 M 0 M F 1 · M 0 M F 1 M 1 M = u ln M 0 M u ln M 0 M F 1 M 0 M F 1 M 1 = v 2 u ln 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 final velocity of the spacecraft be v and the final velocity of the planet be V . Conservation of momentum gives mv + MV = mv + MV = V V = m M · ( v v ) (1) Mechanical energy is conserved. If we consider the initial and final 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 MV 2 = 1 2 mv 2 + 1 2 MV 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 ( ?? ) we obtain v + v 2 V = m M · ( v v ) = v = m M 1 1 + m M · v + 2 1 + m M · V 2
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(b) The mass of a spacecraft is very small compared to the mass of a planet, i.e.
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