Determined by Numerical Analysis of Velocity and Acceleration

Exit Velocity of a Rocket :

Determined by Numerical Analysis of Velocity and Acceleration

Brandon Pruski

Dr. Kalmar-Nagy

Aero 211 Final Project

4/30/2008

“An aggie does not lie, cheat, or steal, nor tolerate those that do”

Brandon Pruski

Table of Contents

Introduction……………………………………………………..……………………………………………………..4

Problem Formulation……………………………………………………..……………………………………….4

Procedure and Safety Protocols……………………………………………………..……………………….4

Analytical Calculations……………………………………………………..…………………………………….5

Experimental Results……………………………………………………..……………………………………….7

Comparisson of Results……………………………………..……………………………………………………7

Discussion……………………………………………………..………………………………………………………..8

Summary……………………………………………………..…………………………………………………………8

Conclusion……………………………………………………..……………………………………………………….8

Works Cited…………………………………………………………………………………………………………….9

Appendix……………………………………………………..……………………………………………………..….10

Introduction

The basis of the project was to create a program that approximated escape velocity of a rocket. Escape velocity has far reaching consequences in the space realms, since it is the minimum velocity required to put an object into orbit.

Problem Formulation

All models have simplifications assumed because the actual conditions cannot be imitated perfectly, or the effects of certain conditions are negligible. The model here utilizes several of these simplifying assumptions. In order to avoid dealing with a non constant mass system, the model assumes a constant velocity at liftoff. Therefore, using numerical analysis, and Newton’s Universal Law of Gravitation, mass can be eliminated from the equation. This is due to the factthat acceleration is then only related to the mass of earth, G (Newton’s gravitational constant), and the square of the radius between the bodies.

This is different than many other models, since they assume a constant thrust, and derive the calculations from the following relation.

Procedure and Safety Protocols

Developing the model requires an understanding of numerical integration and differentiation. Since the model asks for an initial velocity to determine if it is large enough to break through the earth’s atmosphere, both numerical integration and differentiation are required. The following equations form the basis for the entire model.

After manipulation, these two equations allow the model to determine the position, acceleration, and finally the new velocity. This process is then looped in order to find if the velocity is great enough to overcome the attractive force produced by the earth.

Experimental Results

Time (s) / Position (m) / Velocity (m/s) / Acceleration (m/s^2)
0 / 0 / 11000 / -9.79542
1 / 11000 / 10990.2 / -9.76172
1000 / 8.49E+06 / 7053.36 / -1.80294
2000 / 1.48E+07 / 5805.34 / -0.884694
3000 / 2.03E+07 / 5104.44 / -0.560992
4000 / 2.51E+07 / 4631.26 / -0.401424
5000 / 2.96E+07 / 4280.15 / -0.308243
6000 / 3.37E+07 / 4004.08 / -0.247908
7000 / 3.76E+07 / 3778.31 / -0.206015
8000 / 4.13E+07 / 3588.35 / -0.175422
9000 / 4.48E+07 / 3425.05 / -0.152213
10000 / 4.81E+07 / 3282.27 / -0.134069
11000 / 5.14E+07 / 3155.73 / -0.119541
12000 / 5.45E+07 / 3042.31 / -0.107676
13000 / 5.74E+07 / 2939.72 / -0.0978224
14000 / 6.03E+07 / 2846.16 / -0.0895246
15000 / 6.31E+07 / 2760.27 / -0.0824517
16000 / 6.59E+07 / 2680.94 / -0.0763592
17000 / 6.85E+07 / 2607.29 / -0.0710627
18000 / 7.11E+07 / 2538.6 / -0.0664206
19000 / 7.36E+07 / 2474.27 / -0.0623224
19999 / 7.60E+07 / 2413.87 / -0.0586842

Looking at the data gleaned from the model, after 19,999 s of analysis, the velocity of the rocket is still well above zero, and the acceleration is becoming negligible. Therefore, the velocity of 11000 m/s passes the model as high enough to escape the majority of the earth’s gravitational field and enter orbit.

Theoretical vs. Actual Comparisson

When looking at the theoretical calculation in comparison to the actual calculation, it is exactly the same. This is due to the fact that the model is a theoretical model, and the hand calculations included are the governing equations for the results. The main deviation in the calculations comes when different time steps are used to analyze the model. The computational model uses time steps of 1 s, and runs for a total of 19999 s. Included in the theoretical is an approximation for the position, velocity, and acceleration at 1000 s, but the time step used is 999 s, and utilizes the data after 1 s, which brings the net time elapsed to 1000 s. The difference in the parameters is illustrated in the table below.

Time (s) / Position (m) / Velocity
(m/s) / Acceleration (m/s^2) / Time Step
1000 / 8.49E+06 / 7053.36 / -1.08294 / 1
1000 / 1.10E+07 / 9670.56 / -1.32096 / 999

Therefore, even though the theoretical and actual calculations are identical if the time step used is not the same, there is a wide deviation in calculations when large time steps are used.

Discussion

The attritional force experienced by any object leaving earth diminishes as the object moves away from earth, but it never reaches zero. The remaining force is what allows objects to orbit the earth. Therefore, even though there is still a negative acceleration at the end of the 20000 iterations of the program, it is small in comparison to the motion of the rocket because it would then enter a circular orbit, which requires the attractive force to keep it from soaring off into the emptiness of space.

Summary

It was found that 11,000 m/s is a large enough velocity for a rocket to exit the atmosphere of earth, and overcome the attractive gravitational force generated by the planet. This result was also verified by online material that concludes that an exit velocity of 11,000 m/s is required to exit the atmosphere. The online material validates the solutions of our model, and the simplifications that were assumed in order to alleviate complications. There is still room for error, since velocities below the 11,000 m/s threshold also exhibit behavior that showcases them as high enough to overcome the earth’s attractive gravitational field.

Conclusion

Looking at the variation in results, when different time steps are used, showcases the greatest achievement of the model. Computational analysis allows a person to do millions of iterations of the same calculation, with minimal effort expended by the person conducting the analysis. Therefore many errors can be alleviated by decreasing step sizes, which causes the model to take a longer time period to run, but it greatly increases the validity of the results.

Works Cited

Wikipedia Online Encyclopedia