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Jules Verne's crazy idea

by Andrew
(New York)










































    1. In 1865 Jules Verne proposed sending men to the Moon by firing a space capsule from a 220 m long canon with final speed of 11 km/s.

          (a) What would have been the unrealistically large acceleration experienced by the space travelers during their launch?

          (b) Compare your answer with the free-fall acceleration. Why is this “unreasonably large?”

          (c) How long would the cannon have to be in order to launch the explorers to 11 km/s with an acceleration that people can survive?

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Sep 14, 2012
Acceleration from cannon
by: Staff


Answer:


Part I


Definitions:

         vinitial = initial velocity

         vfinal = final velocity

         a = acceleration

         Δd = change in distance (distance traveled)


Initial Values:


         vinitial = 0

         vfinal = 11 km/s

         a = unknown

         Δd = 220 m


Equation:

         final = v²initial + 2a(Δd)



Calculation of acceleration:


         (11 km/s)² = 0² + 2a(220 m)

         (11 km/s)² = 2a(220 m)

         (11 km/s)² / (220 m) = 2a(220 m)/ (220 m)

         (11 km/s)² / (220 m) = 2a(1)

         (11 km/s)² / (220 m) = 2a

         [(11 km/s)² / (220 m)] * (1 / 2) = 2a * (1 / 2)

         [(11 km/s)² / (220 m)] * (1 / 2) = a * ( 2 / 2)

         [(11 km/s)² / (220 m)] * (1 / 2) = a * ( 1)

         [(11 km/s)² / (220 m)] * (1 / 2) = a

         a = [(11 km/s)² / (220 m)] * (1 / 2)

         a = [(11 km/s)² / (2*220 m)]

         a = [(11 km/s)² / (440 m)]



     before completing the final calculation, convert km/s to m/s

         1 km = 1000 m

         Multiply 11 km/s by 1000 m/km

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Sep 14, 2012
Acceleration from cannon
by: Staff


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Part II

         a = (11 km/s * 1000 m/km)² / (440 m)

         a = (11 km/ km * 1000 m/s)² / (440 m)

         a = (11 km / km * 1000 m/s)² / (440 m)

         a = (11 * 1000 m/s)² / (440 m)

         a = (11000 m/s)² / (440 m)

         a = (11000² m²/s²) / (440 m)

         a = (11000² / 440 ) * (m²/s²) * (1 / m)

         a = (11000² / 440 ) * (m/s²) * (m / m)

         a = (11000² / 440 ) * (m/s²) * (m / m)

         a = (11000² / 440 ) * (m/s²) * 1

         a = (11000² / 440 ) * (m/s²)

         a = (121000000 / 440 ) * (m/s²)

         a = 275000 m/s²

         a = 275000 m/s²




     convert the answer to g’s

         1 g = 9.8 m/s²

         Multiply the acceleration “a” in m/s² by 1 g per 9.8 m/s²



         a = (275000 m/s²) * (1 g / 9.8 m/s²)

         a = (275000 / 9.8) * g * [(m/s²) / (m/s²)]

         a = (275000 / 9.8) * g * [((m/s²) / (m/s²)]

         a = (275000 / 9.8) * g * 1

         a = (275000 / 9.8) * g

         a = 28061.224489795917 * g

         a = 28061.22 g’s


         a = 28061.22 g’s


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Sep 14, 2012
Acceleration from cannon
by: Staff


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Part III



Final Answers


     (a) What would have been the unrealistically large acceleration experienced by the space travelers during their launch?

         A human can stand an acceleration of 15g for a short time. This is fifteen times the earth’s gravitational pull.

         The acceleration which would be experienced by a human being shot out of the cannon described in the problem statement is 28061.22 g’s. This is over twenty-eight thousand times the earth’s gravitational pull.



     (b) Compare your answer with the free-fall acceleration. Why is this “unreasonably large?”

         Free fall acceleration is 1 g. See part (a) for an explanation.



     (c) How long would the cannon have to be in order to launch the explorers to 11 km/s with an acceleration that people can survive?


         final = v²initial + 2a(Δd)


         vinitial = 0

         vfinal = 11 km/s * 1000 m/km = 11000 m/s

         a = 15 g (max acceleration humans can withstand)

         Δd = unknown


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Sep 14, 2012
Acceleration from cannon
by: Staff


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Part IV

         (11000 m/s)² = 0² + 2 * (15 g’s) * (9.8 m/s² per g) * (Δd)

         (11000² m²/s²) = 0² + 2 * (15 g’s ) * (9.8 m/s² / g) * (Δd)

         (11000² m²/s²) = 0² + 2 * (15 g/g ) * (9.8 m/s²) * (Δd)

         (11000² m²/s²) = 0² + 2 * (15 * [(g/g) * (9.8 m/s²) * (Δd)

         (11000² m²/s²) = 0² + 2 * (15 * 1 ) * (9.8 m/s²) * (Δd)

         (11000² m²/s²) = 0² + 2 * (15) * (9.8 m/s²) * (Δd)

         (11000² m²/s²) = 2 * (15) * (9.8 m/s²) * (Δd)

         (11000² m²/s²) = (294 m/s²) * (Δd)

         (11000² m²/s²) / (294 m/s²) = (294 m/s²) * (Δd) / (294 m/s²)

         (11000² m²/s²) / (294 m/s²) = (Δd)* [(294 m/s²) / (294 m/s²)]

         (11000² m²/s²) / (294 m/s²) = (Δd)* [(294 m/s²)/ (294 m/s²)]

         (11000² m²/s²) / (294 m/s²) = (Δd)* 1

         (11000² m²/s²) / (294 m/s²) = (Δd)

         Δd = (11000² m²/s²) / (294 m/s²)

         Δd = [(11000² m) / (294)] * [(m/s²) / (m/s²)]

         Δd = [(11000² m) / (294)] * [(m/s²)/ (m/s²)]

         Δd = [(11000² m) / (294)] * 1

         Δd = (11000² m) / (294)

         Δd = (121000000 m) / (294)

         Δd = 411564.62585034012 m

         Δd = 411564.6 m



         The cannon would require a barrel 411564.6 meters long to limit the maximum acceleration in the barrel to 15 g.







Thanks for writing.

Staff
www.solving-math-problems.com



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