Lecture 21 — March 3, 2000 “Rocket Propulsion”

Physics 116a — Lecture 21 — March 3, 2000 “Rocket Propulsion”

Reading Serway, Chapter 8.7. See also Fishbane et al., Physics for …, pp. 249 ff. Problems None due. Just catch up on any outstanding homework.. Objectives To understand conservation of energy and momentum in propulsion. Understand the concept of escape velocity. Be able to calculate rocket velocity from fuel parameters. Demo Apollo XI film clips (from www.nasa.gov), and rocket scooter.

1. Minutiae bureaucratiae 1.1. Re WebAssign problems: We will extend due dates so that you can catch up. No credit for anything not turned in by Monday, March 13th. 1.2. See the course Web site for some statistical information on Exam 2, the an- swer sheet for Exam 2, and grades. 2. A brief history of space flight. Rockets are now in more or less routine use for both scientific and technological applications. Rocket propulsion furnishes a good example of the conservation laws for momentum and energy. 2.1. The father of American rocketry was Robert Goddard, who was on the fac- ulty at Clark University. (See www.clarku.edu for more information.) Ridi- culed by the New York Times in 1920 for thinking that a rocket could work in space, Goddard responded that “every vision is a joke until the first man ac- complishes it.” [The Times printed a retraction the day the Apollo astronauts left for the moon in 1969!] 2.2. The era of “working in space” really began with the work of Werner von Braun and his colleagues at Huntsville, AL after World War II. The Saturn V launch vehicle which powered the Apollo missions was designed by von Braun at age sixteen. 3. The “rocket equation.” What the New York Times did not realize is that a rocket needs nothing to “push against,” as we can show from conservation of momentum. If there are no external forces on the rocket, and noting that the backward velocity of the gases is (v-uexhaust) in the rest frame of the earth,

Dp = Dprocket + Dpfuel = D(mv)rocket - (v - uexhaust)Dm æD v æD m Dp = Dm ×v + m×Dv- v×Dm+u Dm = 0 Þ m ö = - u ö exhaust è Dt ø exhaust è Dt ø In the limit of arbitrarily small time intervals Dt, we get the “rocket equation” æ dv dm dm mi ö m = - uexhaust Þ dv = - uexhaust Þ v f - vi = uexhaustlnç ÷ (1) dt dt m è m f ø

Physics 116a — Lecture 17 —Non-Conservative Forces — February 21, 2000

4. Example: the Saturn V. To understand the application of this rocket, consider the parameters for the booster rocket that powered the Apollo astronauts on the first stage of their journey. The parameters of the rocket are: initial mass, 6 4 mI = 2.5·10 kg, burn rate 1.6·10 kg/s (that’s ten metric tons/s!), burn time is 2 min, and exhaust speed is 3.0·103 m/s. 4.1. To calculate the final mass and the logarithmic ratio in Eq. (1), we use the burn rate and burn time: æ dm 6 4 6 mi ö mf = mi - tburn = 2.5 ×10 kg- 1.6×10 kg/s×120s = 0.6 ×10 kg Þ lnç ÷ = 1.5 dt è mf ø

4.2. We also need to take account of the effects of gravity: during the entire burn time, there is a net reduction of velocity by –gt (that’s just the defini- tion of the change in average velocity), so that æ mi ö 3 2 3 v f - vi = uexlnç ÷ - gt = 3 ×10 m /s ×1.5 - 9.8m/s×120s = 3.2 ×10 m/s (2) è mf ø

4.3. This gives a final velocity of 3.2·103 m/s — about triple the velocity of sound at sea level. 5. Escape velocity from earth. It is useful to compare the velocity we just com- puted with the velocity required to escape from the earth’s gravitational pull, the so-called “escape velocity.” To just escape from the earth’s field would mean that an object would have zero total energy, and knowing that the zero of potential energy is very far away, we get (using conservation of energy): 2 mvesc GMem GMem K(re) + U(re) = K(¥) + U(¥ ) Þ - = 0 - = 0 (3) 2 re r¥ Here we have used the fact that the gravitational potential energy can be set equal to zero at infinity. But this means that - 11 24 2GM 2(6.67 ×10 )(5.97 ×10 ) (4) vesc = = 6 m/s=11.2 m/s re (6.37×10 )

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