# 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?

### Comments for Jules Verne's crazy idea

 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:          v²final = 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 --------------------------------------------------

 Sep 14, 2012 Acceleration from cannon by: Staff -------------------------------------------------- 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 --------------------------------------------------

 Sep 14, 2012 Acceleration from cannon by: Staff -------------------------------------------------- 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?          v²final = 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 --------------------------------------------------

 Sep 14, 2012 Acceleration from cannon by: Staff -------------------------------------------------- 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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