A Mathematical Model for a

Mission to mars

Glenn Ledder

Department of MathematicsUniversity of Nebraska-Lincolngledder@math.unl.edu

Mathematical Models

A mathematical model is a mathematicalobject based on real phenomena andcreated in the hope that its mathematicalbehavior resembles the real behavior.

Mathematical Modeling

the process of creating, analyzing, andinterpreting mathematical models

Model Structure

MathProblem

Input Data Output Data

Key Question:

What is the relationship between input and output data?

EXAMPLERanking of Football Teams

MathematicalAlgorithmGame Data

RankingWeight Factors

Game Data: situation dependent

Weight Factors: built into mathematical model

EXAMPLERanking of Football Teams

MathematicalAlgorithmGame Data

RankingWeight Factors

Modeling Goal: Choose the weights to get the “correct” national championship game.

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 collectsamples of rocks and use the landing vehicleto return to the spaceship.

What specifications guarantee that thelander is able to escape Mars’ gravity?

A Simple Rocket Launch Model“I Shot an Arrow Into the Air”

• Planet Data

– Radius R– Gravitational constant g

• Design Data

– Mass m

– Initial velocity v0

(t<0)

MathematicalModelR, g

Flight Datam, v0

Schematic of the Simple Launch Problem:

A Simple Rocket Launch Model“I Shot an Arrow Into the Air”

Basic Newtonian Mechanics I

Newton’s Second Law of Motion:

F Δt = Δ(mv)(“impulse = momentum”)

Constant m version:

― = ―dvdt

Fm

Basic Newtonian Mechanics II

Newton’s Law of Gravitation:

F (t) = -mg ——

Constant m rocket flight equation:

― = - ——

R2

z2 (t)

z2 (t)

g R2dvdt

The Height-Velocity Equation

― = - ——z2

(t)g R2dv

dtGravitational Motion:

Think of v as a function of z.

dvdt

dvdz

dv dzdz dt

Then — = — — = — v

Result: v ― = - ——z2

g R2dvdz

Escape Velocity

2 gR

0

v e

v ― = - ——z2

g R2dvdz

Height-Velocity equation:

Separate variables and Integrate:

2v dv = 2gR2 z

-2 dz

Suppose v = 0 as z→∞ and v = ve at z = R.

ve2

= 2gRR

The Escape Velocity is ve =

Nondimensionalization

v ― = - ——z2

g R2dvdz

v(R) = v0

The height-velocity problem

has 3 parameters.

Nondimensionalization: replacing dimensional quantities with dimensionless quantities

V = v/ve and Z = z/R are dimensionless

Let V = ― Z = ― V0 = ―v0

R

dvdz

dvdV

dVdZ

dZdz

― = ― ― ― = ― ―ThendVdZ

ve

R

v ― = - ——z2

g R2dvdz

― V ― = - ―ve

2

RdVdZ

gZ2

v(R) = v0 V(1) = V0

vve

zR

v ― = - ——z2

g R2dvdz

v(R) = v0

The 3-parameter height-velocity problem

becomes the 1-parameter dimensionless problem

2V ― = - ―dVdZ

1Z2 V(1) = V0

Height-Velocity Curves

2V ― = - ―dVdZ

1Z

2 V(1) = V0

V 2

– V02 = ― – 1 1

Z

The Escape Curve has V0 = 1 :

ZV 2 = 1

Height-Velocity Curves

V 2

– V02 = ― – 1 1

Z

ZV 2

> 1

ZV 2

= 1

ZV 2

< 1

A Two-Phase Launch Model

• Phase 1:

The vehicle burns fuel at maximum rate.

• Phase 2:

The vehicle drifts out of Mars’ gravity.(t<0)

Phase 1

Phase 2

A Two-Phase Launch Model

(t<0)

• Planet Data– Radius R– Gravitational constant g

• Design Data– Vehicle mass M– Fuel mass P

– Burn rate α– Exhaust velocity β

Phase 1

Phase 2

Phase 1ProblemR, g

Success / Failure ZV 2 ≥ 1 / ZV 2 < 1

M, P, α, β

We have already solved the Phase 2 problem!

Schematic of the Launch Problem:

A Two-Phase Launch Model

Newtonian Mechanics, revisited

Newton’s Second Law of Motion: F Δt = Δ(mv)

Variable m version, with gravitational force:

Rocket Flight equation:

― = —– – —–dvdt

dvdt

dmdt

m ― + v –— = F = -m g —R2

z2

αβ m z2

gR2

Full Phase 1 Model

― = —– – —–dvdt

αβ m z2

gR2

― = vdzdt

― = - αdmdt

v(0) = 0

z(0) = R

m(0) = M + P

0 ≤ t ≤ P/α

Simplification

4 design parameters is too many!

― = —– – —–dvdt

αβ m z2

gR2

―(0) = —– – gdvdt

αβ M+P

―(0) ≥ 0 dvdt

αβ > g (M+P)

Take maximum fuel! P = —– – Mαβ g

Full Phase 1 Model

― = —– – —–dvdt

αβ m z2

gR2

― = vdzdt

― = - αdmdt

v(0) = 0

z(0) = R

m(0) = —–

0 ≤ t ≤ — – —

αβ g

βg

Mα

Nondimensionalization

Let V = ― vve

Z = ― zR

T = ―gt β

B = ― βve

Dimensionless exhaust velocity

A = ―– αβMg

Dimensionless acceleration

Dimensionless Phase 1 Model

― = —– – —–dVdT

B1-T Z2

B

― = 2BVdZdT

V(0) = 0

Z(0) = 1

0 ≤ T ≤ 1 – A-1

The new model has only 2 parameters,with only 1 in the initial value problem.

― = — – —dVdT

B1-T Z2

B

― = 2BVdZdT

V(0) = 0

Z(0) = 1

ZV 2(T0) = 1

The Flight Time Function

For any given velocity B, let T0 be the time required

to reach the escape curve ZV 2

= 1.

B T0(B)

Success Criterion

T0(B) is the time needed to reach the

escape curve in Phase 1.

1 – A-1 is the time available before the fuel supply is “exhausted.”

1 – A-1 ≥ T0(B) :Success is defined by

f (A, B) = A-1 + T0(B) ≤ 1

The Vehicle Design Curve

increasingacceleration

increasing exhaust velocity

A Successful Launch

A = 2.5

B = 2.0

An “Unsuccessful” Launch

The vehicle “hovers” at z = 4R. Maybe that is ideal!

A = 2.0

B = 2.5

Implications for Mars

B = ― βve

A = ―– αβMg

BODY g (m/sec2) ve (km/sec)

Moon 1.62 2.37Mars 3.72 5.02

α and β need to be almost double;after 35 years, this is probably OK

An Easier Task

Why don’t we land on Mars’ smaller moon Deimos instead?

The escape velocity is only 7 m/sec,which is about 16 mph, roughly the speedof the 1600 meter race in this summer’sOlympic Games!