1. ## Rocket Launching

I am having trouble with this problem. Can anyone shed some light?

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 e ects 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 thedesign parameters for the rocket engine is to set aconstant fuel burn rate, then we will have the relationship dm/dt = -b where b is the fuel burn rate constant measured in kg/s. The initial total mass of the rocket can be represented by the constant m0,and we can assume that the fuel is a signi cant 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 =[kb/(m0-bt)]-g where g is the acceleration due to gravity, and kb is the thrust provided by the rocket engine, k being an experimentally determined constant. If we launch the rocket from a stationary position, we can jus-ti ably 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 (km0/b) [(1-X)ln(1-X)+X[1-(gXm0/2kb)]

(b) What is the velocity of the rocket at the end of the 'boost phase'?

(c) Astronauts aboard the 'vomit comet' or its current equivalent, 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 experienced is equivalent to a rate of decrease in the acceleration due to gravity of the order of -10/15ms^-2/s = -2/3ms^-3. For our rocket and its payload, the accleration due to gravity decreases as the rocket gets further from the launch site - still assming a vertical trajectory. If we take the radius of Earth to be R = 6.38x10^6m, then the acceleration due to gravity goes like ab = MG/(h(t)+R)^2, where M and G are constants and as before h(t) is the rocket's height above Earth's surfacae 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(a)/dt.
Using the information gained previously for the height and velocity of the rocket in the 'boost phase', find an expression for the rate of weight-loss experienced by the rocketand it's payload as it continues to climb just at the end of the boost phase. You should use the operating parameters of the Shahab 3 missle employed by Iran, which has an initial mass of 16000kg of which 90% can be considered as fuel. Furthermore the Shahab 3 has a boost phase of 95s by which time it reaches a height of 75km. How does this rate of weight-loss compared with that experienced in the 'vomit comet'?

2. ## very fast

Hi,

You problem is long to read and I do not have much time. But for question a) and b), you have the equation of d2h/d2t, something in 1/t. The second antiderivative is tln(t)-ln(t). So you can easily find h(t) (will be written using this function). After you have h(0)=0 and I think you meant dh(0)/dt=0 also. So you can find h(t) and also dh/t at any time. Now your condition for end of bust is that bt=moX, that gives you the time and you can find the height and the velocity at that moment. If you have problem with question c) (I did not even read it!) let us know, I will try to have a look later ( I used to like these rockets problem, and this one is actually not hard as the equation is already given, what is more interesting is to find the equation of acceleration!)

3. ## little correction

Actually the second antiderivative is tln(t)-t, sorry for the typo!