Problem Solving Assignment 1. Rocket launching: In this problem we are looking at some of the aspects that need
to be considered in designing and describing the launching a single- stage rocket with a payload. In what follows we will be ignoring any eects of cross-winds, rotation of Earth or air resistance. We begin by assuming that the rocket is launched from Earth's surface in such a way that it follows a simple vertical trajectory. As a single-stage rocket, immediately after launch there will be a period of rapid acceleration as the fuel consumed provides thrust for the upward motion. One of the complex issues to be considered is that as the fuel is used, the mass of the rocket (i.e. fuel + engine + payload) decreases, and hence the acceleration will increase for the same amount of thrust. The period of time over which the rocket engine is ring is known as the `boost phase' of the trajectory, hence when the fuel runs out, the `boost phase' is at an end. Representing the rocket's total mass with the un- known function m(t), if we assume that one of the design parameters for the rocket engine is to set a constant fuel burn rate, then we will have the rela- tionship dm=dt = ô€€€b where b is the fuel burn rate constant measured in kg sô€€€1. The initial total mass of the rocket can be represented by the constant m0, and we can assume that the fuel is a signicant frac- tion of this mass. In this case, the quantity of fuel can again be a design parameter, so we will take the mass of fuel to be X m0, where 0 < X < 1. If we further assume that the `boost phase' is short enough, and we assign the function h(t) to be the rocket's height above ground-level, then we can de- scribe the acceleration experienced by the rocket during this period as d2h dt2 = b m0 ô€€€ b t ô€€€ g ; where g is the acceleration due to gravity, and b is the thrust provided by the rocket engine, being an experimentally determined constant. If we launch the rocket from a stationary position, we can jus- tiably include the initial conditions h(0) = 0 and dh=dt = 0 when t = 0. (a) Show that the expression for the height h(t) of the rocket when it just runs out of fuel is given by m0 b (1 ô€€€ X) ln(1 ô€€€ X) + X 1 ô€€€ g 2b Xm0 : (b) What is the velocity of the rocket at the end of the `boost phase'? (c) Astronauts aboard the `vomit comet' or its current equivalent1, experience a rapid loss of weight as the acceleration due to gravity is counteracted by the acceleration provided by the aircraft. The rate of weight-loss expe- rienced is equivalent to a rate of decrease in the acceleration due to gravity of the order of ô€€€10=15 msô€€€2=s = ô€€€2=3 msô€€€3. For our rocket and its payload, the acceleration due to gravity decreases as the rocket gets fur- ther from the launch site - still assuming a ver- tical trajectory. If we take the radius of Earth to be R0 = 6:38 106 m, then the acceleration due to gravity goes like ag = MG (h(t) + R0)2 ; where M and G are constants2 and as before, h(t) is the rocket's height above Earth's surface as a function of time. For an object of constant mass, but changing height, the rate of weight- loss experienced would be given by d(ag)=dt. Using the information gained previously for the height and velocity of the rocket in the `boost phase', nd an expression for the rate of weight- loss experienced by the rocket and it's payload as it continues to climb just at the end of the boost phase. You should use the operating pa- rameters of the Shahab 3 missile employed by Iran (see notes3 4), which has an initial mass of 16000 kg of which 90% can be considered as fuel. Furthermore the Shahab 3 has a boost phase of 95 s by which time it reaches a height of 75 km. How does this rate of weight-loss compare with that experienced in the `vomit comet'?
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