A Mathematical Model for a
Mission to mars
Glenn Ledder
Department of MathematicsUniversity of [email protected]
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!