A Mathematical Model for a Mission to mars

1. AMathematical Model for a Mission to mars Glenn Ledder Department of Mathematics University of Nebraska-Lincoln gledder@math.unl.edu

2. Mathematical Models A mathematical model is a mathematical object based on real phenomena and created in the hope that its mathematical behavior resembles the real behavior. Mathematical Modeling the process of creating, analyzing, and interpreting mathematical models

3. Model Structure Math Problem Input Data Output Data Key Question: What is the relationship between input and output data?

4. EXAMPLERanking of Football Teams Mathematical Algorithm Weight Factors Ranking Game Data Game Data: situation dependent Weight Factors: built into mathematical model

5. EXAMPLERanking of Football Teams Mathematical Algorithm Weight Factors Ranking Game Data Modeling Goal: Choose the weights to get the “correct” national championship game.

6. The Lander Design Problem A spaceship goes to Mars and establishes an orbit. Astronauts or a robot go down to the surface in a Mars Lander. They collect samples of rocks and use the landing vehicle to return to the spaceship. What specifications guarantee that the lander is able to escape Mars’ gravity?

7. A Simple Rocket Launch Model“I Shot an Arrow Into the Air” • PlanetData • RadiusR • Gravitational constant g • Design Data • Massm • Initialvelocityv0 (t<0)

8. A Simple Rocket Launch Model“I Shot an Arrow Into the Air” Schematic of the Simple Launch Problem: m, v0 Mathematical Model Flight Data R, g

9. Basic Newtonian Mechanics I Newton’s Second Law of Motion: FΔt=Δ(mv) (“impulse = momentum”) Constantm version: dv dt F m ―=―

10. Basic Newtonian Mechanics II Newton’s Law of Gravitation: R2 F(t)=-mg—— z2(t) Constantm rocket flight equation: dv dt gR2 ―=-—— z2(t)

11. The Height-Velocity Equation dv dt gR2 ―=-—— Gravitational Motion: z2(t) Think of v as a function of z. dv dt dv dz dz dt dv dz —=— —=— v Then dv dz gR2 v ―=-—— Result: z2

12. Escape Velocity dv dz gR2 v ―=-—— Height-Velocity equation: z2 Supposev=0as z→∞ andv=veat z=R. Separate variables and Integrate: ve2=2gR 2vdv= 2gR2z-2dz The Escape Velocity is ve=

13. Nondimensionalization The height-velocity problem dv dz gR2 v ―=-—— v(R)=v0 z2 has 3 parameters. Nondimensionalization: replacing dimensional quantities with dimensionless quantities V=v/veand Z=z/Rare dimensionless

14. z R v0 R v ve Z= ― V0= ― Let V= ― dv dz dv dV dV dZ dZ dz dV dZ ve R ― = ―――=―― Then ve2 R g Z2 dv dz dV dZ gR2 v ―=-—— ―V―=-― z2 V(1)=V0 v(R)=v0

15. The 3-parameter height-velocity problem dv dz gR2 v ―=-—— v(R)=v0 z2 becomes the 1-parameterdimensionless problem dV dZ 1 Z2 2V―=-― V(1)=V0

16. Height-Velocity Curves dV dZ 1 Z2 2V―=-― V(1)=V0 1 Z V2–V02=―–1 The Escape Curve has V0 = 1: ZV2= 1

17. Height-Velocity Curves 1 Z V2–V02=―–1 ZV2>1 ZV2=1 ZV2<1

18. A Two-Phase Launch Model • Phase 1: The vehicle burns fuel at maximum rate. • Phase 2: The vehicle drifts out of Mars’ gravity. Phase 2 Phase 1 (t<0)

19. A Two-Phase Launch Model • Planet Data • RadiusR • Gravitational constantg • DesignData • Vehicle massM • Fuel massP • Burn rateα • Exhaust velocityβ Phase 2 Phase 1 (t<0)

20. A Two-Phase Launch Model We have already solved the Phase 2 problem! Schematic of the Launch Problem: Phase 1 Problem M, P, α, β Success / Failure ZV2 ≥1 /ZV2 <1 R, g

21. Newtonian Mechanics, revisited FΔt=Δ(mv) Newton’s Second Law of Motion: Variablem version, with gravitational force: dv dt dm dt R2 m ― +v –—= F =-mg — z2 RocketFlight equation: dv dt αβ m gR2 ― =—–– —– z2

22. Full Phase 1 Model dv dt αβ m gR2 ― =—–– —– v(0)=0 z2 dz dt ― =v z(0)=R dm dt ― =-α m(0)=M+P 0≤ t≤ P/α

23. Simplification 4 design parameters is too many! dv dt αβ m dv dt gR2 αβ M+P ― =—–– —– ―(0)=—–– g z2 dv dt ―(0) ≥ 0 αβ>g(M+P) αβ g P = —–– M Take maximum fuel!

24. Full Phase 1 Model dv dt αβ m gR2 ― =—–– —– v(0)=0 z2 dz dt ― =v z(0)=R αβ g dm dt ― =-α m(0)=—– β g M α 0≤ t≤ — – —

25. Nondimensionalization gt β v ve z R Z= ― Let V= ― T= ― β ve B= ― Dimensionless exhaust velocity αβ Mg A= ―– Dimensionless acceleration

26. Dimensionless Phase 1 Model dV dT B 1-T B ―=—–– —– V(0)=0 Z2 dZ dT ―= 2BV Z(0)=1 0≤ T≤ 1–A-1 The new model has only 2 parameters, with only 1 in the initial value problem.

27. The Flight Time Function For any given velocity B, let T0 be the time required to reach the escape curve ZV2=1. dV dT B 1-T B ―=—– — V(0)=0 Z2 B T0(B) dZ dT Z(0)=1 ―= 2BV ZV2(T0)=1

28. Success Criterion T0(B) is thetime needed to reach the escape curve in Phase 1. 1–A-1 is the time available beforethe fuel supply is “exhausted.” 1–A-1 ≥T0(B): Success is defined by f(A,B) = A-1+T0(B) ≤ 1

29. The Vehicle Design Curve increasing acceleration increasing exhaust velocity

30. A Successful Launch A=2.5 B=2.0

31. An “Unsuccessful” Launch A=2.0 B=2.5 The vehicle “hovers” at z=4R.Maybe that is ideal!

32. Implications for Mars β ve αβ Mg A= ―– B= ― BODY g (m/sec2) ve (km/sec) Moon 1.62 2.37 Mars 3.72 5.02 αandβneed to be almost double; after 35 years, this is probably OK

33. An Easier Task Why don’t we land on Mars’ smaller moon Deimos instead? The escape velocity is only 7m/sec, which is about 16mph, roughly the speed of the 1600 meter race in this summer’s Olympic Games!