Aerospace - N icole Maistrow ENGR 110 Aerospace 1 To solve...

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Nicole Maistrow ENGR 110 Aerospace 1) To solve this problem, I first manipulated the rocket equation given to find V. I found that V = -U e ln(M final /M initial ). Next, I know that since 90% of the initial mass is propellant, that the final mass would be 10% of the initial mass. I then plugged in the numbers given for U e to find the maximum V. a. V = U e ln(M final /M initial ) U e = 4000 m/s M initial = 100 M final = 100-90 = 10 V = 4000 m/s (ln(100/10)) = 4000(ln10) = 9210.340372 m/s 1 m/s = 2.23693629 mph 9210.340372 m/s * 2.23693629 mph = 20602.94462 mph b. V = U e ln(M final /M initial ) U e = 40000 m/s M initial = 100 M final = 100-90 = 10 V = 4000 m/s (ln(100/10)) = 40000(ln10) = 92103.40372 m/s 1 m/s = 2.23693629 mph 92103.40372 m/s * 2.23693629 mph = 206029.4462 mph c. I know that specific energy is equal to kinetic energy/mass, and that the equation for kinetic energy = ½mv 2 . The mass’s cancel out, so I am left with ½v 2 . For v, I used the velocity I found in part a for the chemical propulsion option and the velocity from part b for the advanced propulsion option. Specific energy = KE/m
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This note was uploaded on 02/17/2011 for the course ENGR 110 taught by Professor None during the Fall '08 term at University of Michigan.

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Aerospace - N icole Maistrow ENGR 110 Aerospace 1 To solve...

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