A1256655036nAfKc

1
A Mathematical Model for a Mission to mars
Glenn Ledder Department of Mathematics University
of Nebraska-Lincoln gledder_at_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 ModelI Shot an Arrow
Into the Air

• Planet Data
• Radius R
• Gravitational constant g
• Design Data
• Mass m
• Initial velocity v0

(tlt0)
8
A Simple Rocket Launch ModelI Shot an Arrow
Into the Air
Schematic of the Simple Launch Problem
Mathematical Model
m, v0
Flight Data
R, g
9
Basic Newtonian Mechanics I
Newtons Second Law of Motion
F ?t ?(mv)
(impulse momentum)
Constant m version
dv dt
F m
? ?
10
Basic Newtonian Mechanics II
Newtons Law of Gravitation
R2
F (t) -mg
z2 (t)
Constant m rocket flight equation
dv dt
g R2
? -
z2 (t)
11
The Height-Velocity Equation
dv dt
g R2
? -
Gravitational Motion
z2 (t)
Think of v as a function of z.
dv dt
dv dz
dv dz dz dt
v
Then
dv dz
g R2
v ? -
Result
z2
12
Escape Velocity
dv dz
g R2
v ? -
Height-Velocity equation
z2
Suppose v 0 as z?8
and v ve at z R.
Separate variables and Integrate
ve2 2gR
2v dv 2gR2 z -2 dz
The Escape Velocity is ve
13
Nondimensionalization
The height-velocity problem
dv dz
g R2
v ? -
v(R) v0
z2
has 3 parameters.
Nondimensionalization replacing dimensional
quantities with dimensionless quantities
V v/ve and Z z/R are dimensionless
14
v0 R
z R
v ve
Let V ?
Z ?
V0 ?
dv dz
dv dV
dV dZ
dZ dz
dV dZ
ve R
? ? ? ? ? ?
Then
ve2 R
g Z2
dv dz
dV dZ
g R2
v ? -
? V ? - ?
z2
V(1) V0
v(R) v0
15
The 3-parameter height-velocity problem
dv dz
g R2
v ? -
v(R) v0
z2
becomes the 1-parameter dimensionless problem
dV dZ
1 Z2
2V ? - ?
V(1) V0
16
Height-Velocity Curves
dV dZ
1 Z 2
2V ? - ?
V(1) V0
1 Z
V 2 V02 ? 1
The Escape Curve has V0 1
ZV 2 1
17
Height-Velocity Curves
1 Z
V 2 V02 ? 1
ZV 2 gt 1
ZV 2 1
ZV 2 lt 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
(tlt0)
19
A Two-Phase Launch Model

• Planet Data
• Radius R
• Gravitational constant g
• Design Data
• Vehicle mass M
• Fuel mass P
• Burn rate a
• Exhaust velocity ß

Phase 2
Phase 1
(tlt0)
20
A Two-Phase Launch Model
We have already solved the Phase 2 problem!
Schematic of the Launch Problem
Phase 1 Problem
M, P, a, ß
Success / Failure ZV 2 1 / ZV 2 lt 1
R, g
21
Newtonian Mechanics, revisited
F ?t ?(mv)
Newtons Second Law of Motion
Variable m version, with gravitational force
dv dt
dm dt
R2
m ? v F -m g
z2
Rocket Flight equation
dv dt
aß m
gR2
?
z2
22
Full Phase 1 Model
dv dt
aß m
gR2
?
v(0) 0
z2
dz dt
? v
z(0) R
dm dt
? - a
m(0) M P
0 t P/a
23
Simplification
4 design parameters is too many!
dv dt
aß m
dv dt
gR2
aß MP
?
?(0) g
z2
dv dt
?(0) 0
aß gt g (MP)
aß g
P M
Take maximum fuel!
24
Full Phase 1 Model
dv dt
aß m
gR2
?
v(0) 0
z2
dz dt
? v
z(0) R
aß g
dm dt
? - a
m(0)
ß g
M a
0 t
25
Nondimensionalization
gt ß
v ve
z R
Z ?
Let V ?
T ?
ß ve
B ?
Dimensionless exhaust velocity
aß 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 ZV 2 1.
dV dT
B 1-T
B
?
V(0) 0
Z2
B
T0(B)
dZ dT
Z(0) 1
? 2BV
ZV 2(T0) 1
28
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
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
ß ve
aß Mg
B ?
A ?
BODY g (m/sec2) ve (km/sec) Moon 1.62
2.37 Mars 3.72 5.02
a and ß need to be almost double after 35 years,
this is probably OK
33
An Easier Task
Why dont we land on Mars smaller moon Deimos
instead?
The escape velocity is only 7 m/sec, which is
about 16 mph, roughly the speed of the 1600 meter
race in this summers Olympic Games!