a mathematical model for a mission to mars glenn ledder department of mathematics university of...
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![Page 1: A Mathematical Model for a Mission to mars Glenn Ledder Department of Mathematics University of Nebraska-Lincoln gledder@math.unl.edu](https://reader034.vdocuments.us/reader034/viewer/2022051619/56649d695503460f94a479c7/html5/thumbnails/1.jpg)
A Mathematical Model for a
Mission to mars
Glenn Ledder
Department of MathematicsUniversity of [email protected]
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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
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Model Structure
MathProblem
Input Data Output Data
Key Question:
What is the relationship between input and output data?
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EXAMPLERanking of Football Teams
MathematicalAlgorithmGame Data
RankingWeight Factors
Game Data: situation dependent
Weight Factors: built into mathematical model
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EXAMPLERanking of Football Teams
MathematicalAlgorithmGame Data
RankingWeight Factors
Modeling Goal: Choose the weights to get the “correct” national championship game.
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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?
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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)
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MathematicalModelR, g
Flight Datam, v0
Schematic of the Simple Launch Problem:
A Simple Rocket Launch Model“I Shot an Arrow Into the Air”
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Basic Newtonian Mechanics I
Newton’s Second Law of Motion:
F Δt = Δ(mv)(“impulse = momentum”)
Constant m version:
― = ―dvdt
Fm
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Basic Newtonian Mechanics II
Newton’s Law of Gravitation:
F (t) = -mg ——
Constant m rocket flight equation:
― = - ——
R2
z2 (t)
z2 (t)
g R2dvdt
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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
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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 =
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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
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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
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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
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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
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Height-Velocity Curves
V 2
– V02 = ― – 1 1
Z
ZV 2
> 1
ZV 2
= 1
ZV 2
< 1
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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
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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
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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
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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
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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
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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
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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
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Nondimensionalization
Let V = ― vve
Z = ― zR
T = ―gt β
B = ― βve
Dimensionless exhaust velocity
A = ―– αβMg
Dimensionless acceleration
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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.
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― = — – —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)
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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
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The Vehicle Design Curve
increasingacceleration
increasing exhaust velocity
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A Successful Launch
A = 2.5
B = 2.0
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An “Unsuccessful” Launch
The vehicle “hovers” at z = 4R. Maybe that is ideal!
A = 2.0
B = 2.5
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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
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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!