icole N Maistrow ENGR 110 Aerospace 1) To solve this problem, I fi rst manipulated the rocket equation given to f ind V. I found that V = -U e l n(M final /M ini tial ). Next, I know that since 90% of the ini tial 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 t he maximum V. a. V = U e l n(M final /M ini tial ) U e = 4000 m/s M ini tial = 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 l n(M final /M ini tial ) U e = 40000 m/s M ini tial = 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 = mv2. The masss cancel out, so I am left with v2. 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 KE = mv2 ; specific energy = mv2/m = v2 Chemical Propulsion: V = 9210.340372 m/s Specific energy = v2 = (9210.340372)2 M J/kg = 4,605.17186 M J/kg Advanced Propulsion: V = 92103.40372 m/s
Specific energy = v2 = (92103.40372)2 M J/kg = 46,051.7186 M J/kg d. Energy of combustion of hydrogen = 141.4 kJ/g 141.4 kJ/g * 1g/10-3kg * 10-3 M J/1kJ = 141.4 M J/kg http://www.ausetute.com.au/fuelsdef.html E nergy of combustion of natural gas = 11,000 kcal/kg 11,000 kcal/kg * .004184 MJ/kcal = 46.024 M J/kg http://www.engineeringtoolbox.com/gross-net-heating-values-d_420.html E nergy of combustion of automotive gas = 10,444 Btu/lb 10,444 Btu/lb * 236.1 J/kg/1 Btu/lb * 1 MJ/106 J = 24.2937884 M J/kg 2) a. To find the amount of antimatter needed to create such an explosion, I used the equation given in the hint in the problem. However, I read that antimatter is identical to matter but with the opposite electrical charge. T herefore, to find antimatter, I used the equation +/- E = mc2 , k nowing that t here is no such thing as negative mass, even with We antimatter. were also g iven the hint that the blast is about 1000x the H i roshima atomic bomb blast, so I know that for E in this equation, or the energy, I can multiply the amount of energy released in the atomic bomb blast by 1000. In this equation, c is the speed of light, or about 3 x 108 m/s, so I have that constant, and I know that I am looking for the mass of the antimatter in ounces. I then solved the equation for mass by dividing E/c2 t o find m. E nergy in bomb at H i roshima = 15,000 tons of TNT 1 ton of TNT = 4,184,000,000 joules E = 6.276 x 1013 joules C = 3 x 108 m/s E = +/- mc2 E /c2 = m 6.276 x 1013 joules/(3 x 108 m/s)2 = m m = 6.973 x 10-4 k g 1 kilogram = 35.2739619 ounces 6.973 x 10-4 k g * 35.2739619 ounces = .0245977094 ounces
h ttp://metricsystemconversion.info/ton-of-TNT-energy-equivalent-to-jouleJ.html?func=detail (convert ton of TNT to joules) http://www.btinternet.com/~j.doyle/SR/Emc2/Equation.htm (units used in E = mc2/speed of light constant) http://www.cfo.doe.gov/me70/manhattan/hiroshima.htm (energy in bomb at H i roshima) http://www.aolnews.com/2010/11/18/what-is-antimatter-four-essentialquestions-and-answers/ (equation and charge for antimatter) b. To find the speed that this much antimatter would propel the shuttle orbiter to, I used the equation E = mv2. I used the energy that I used in part a and derived from the amount of energy in the bomb at H i roshima (6.276 x 1013 joules), and used the mass found in part a (6.973 x 10-4 k g). I then isolated v, and plugged in these numbers to find the speed in meters/second. T hen, since the problem asked for the speed in miles per hour, I converted meters/second to miles per hour. E = mv2 E = 6.276 x 1013 joules m = 6.973 x 10-4 k g v= v = 424274209.2 m/s 1 m/s = 2.23693629 mph 424274209.2 m/s * 2.23693629 mph = 949,074,375.5 mph To find how long i t would take to reach the Moon from Earth, I used distance/ rate = time. I found that the distance between Earth and the Moon is 238,855 m iles, and I used the velocity I found in the last question (949,074,375.5 mph) as the rate. Distance/Rate = t ime 238,855 miles/949,074,375.5 miles/hr = 2.516715298 x 10-4 h ours