Problem

Repeat the rocket problem in Example 3.14 in Chapter 3. Include aerodynamic drag, with a r...

Repeat the rocket problem in Example 3.14 in Chapter 3. Include aerodynamic drag, with a rocket diameter of 3 m and drag coefficient of 0.6.

Example 3.14

Work Example 3.13 for the specific values of initial mass Ma = 5000 kg, burn rate me = 35 kg/s, and exit velocity Ve = 2500 m/s relative to the rocket body for a burn duration of t = 100 s. How high will the rocket travel?

SOLUTION: At t = 100 s, the elevation of the rocket is

y( 100 s) = 72,000 m.

The velocity of the rocket at burnout is

After the rocket engine has ceased firing, the rocket will continue to climb under its own inertia for a while. The trajectory after burnout is actually easier to calculate since there is no longer any thrust force acting on the rocket. The only force acting on the rocket is its weight, so Newton’s second law for the rocket is F = ma = −mg, and the acceleration a = −g. So the velocity for the final stage of ascent is V(t) = Vo − gt, and the elevation is y(t) = yo + Vot − gt2. The maximum altitude will be reached when the rocket loses all its inertia, or V = 0. This happens at t = (2030 m/s)/(9.8 m/s2) = 207 s after burnout. The gain in elevation will be (2,030 m/s)(207 s) − (9.8 m/s2)(207 s)2 = 210,000 m, for a total height above the earth of 282,000 m.

The height of the rocket as a function of time is plotted in Figure 3.14, and its velocity as a function of time is plotted in Figure 3.15. Of course, these are only approximations, since aerodynamic drag was not accounted for. Drag will be discussed in Chapter, and the MATLAB program in Appendix F for calculating trajectories does include the effects of drag. It should also be noted that at these high altitudes the value of the gravitational acceleration is less than it is at sea level and is not a constant.

FIGURE 3.14 Height as a function of time for the rocket problem.