Approaching Application Problems Ryan Pietropaolo Ph.D. NCSSM January 2014.
A Trip to Mars Douglas Marks NCSSM. The Problem Find a flight path from the Earth to Mars.
-
Upload
camilla-abigayle-phillips -
Category
Documents
-
view
223 -
download
6
Transcript of A Trip to Mars Douglas Marks NCSSM. The Problem Find a flight path from the Earth to Mars.
Approaches
• Define an Archimedean spiral (Pre cal version)
• Use Kepler’s laws (Physics/Pre cal version)
• Define the forces due to gravity on the rocket (Calculus version)
Differential Equations𝑑2𝒓𝑑𝑡2
=−𝐺⋅𝑚1
𝑟2⋅ 𝒓|𝒓|
𝑟𝑦
𝑥
𝑑2𝑥𝑑𝑡2
=−𝐺⋅𝑚1
𝑟2⋅cos (𝜃)
𝜃 𝑑2 𝑦𝑑𝑡 2
=−𝐺⋅𝑚1
𝑟2⋅sin (𝜃)
𝑑2𝑥𝑑𝑡2
=−𝐺⋅𝑚1
𝑟2⋅ 𝑥𝑟
¿−𝐺⋅𝑚1𝑥
(𝑥2+𝑦 2 )32
𝑑2 𝑦𝑑𝑡 2
=−𝐺⋅𝑚1
𝑟2⋅ 𝑦𝑟
¿−𝐺⋅𝑚1 𝑦
(𝑥2+𝑦 2 )32
Euler’s Method (linear)
𝑣 𝑥𝑛=𝑣𝑥𝑛− 1
+ 𝑑2 𝑥𝑑𝑡 2
(𝑥𝑛− 1 , 𝑦 𝑛−1 ) ⋅Δ 𝑡
𝑦 𝑛=𝑦𝑛−1+𝑣𝑦𝑛 −1⋅Δ𝑡𝑥𝑛=𝑥𝑛−1+𝑣 𝑥𝑛 −1
⋅Δ𝑡
𝑣 𝑦𝑛=𝑣𝑦𝑛 −1
+𝑑2 𝑦𝑑 𝑡2
(𝑥𝑛−1 , 𝑦𝑛−1 ) ⋅Δ𝑡
𝑥0=1 𝑣 𝑥0=0
𝑑2𝑥𝑑𝑡2
(𝑥 , 𝑦 )=−𝐺⋅𝑚1𝑥
(𝑥2+ 𝑦2 )32
𝑑2 𝑦𝑑𝑡 2
(𝑥 , 𝑦 )=−𝐺⋅𝑚1 𝑦
(𝑥2+ 𝑦2 )32
𝑣 𝑦0=0.0191
Euler’s Method (quadratic)
𝑣 𝑥𝑛=𝑣𝑥𝑛− 1
+ 𝑑2 𝑥𝑑𝑡 2
(𝑥𝑛− 1 , 𝑦 𝑛−1 ) ⋅Δ 𝑡
𝑦 𝑛=𝑦𝑛−1+𝑣𝑦𝑛 −1⋅Δt+ 1
2𝑑2 𝑦𝑑 𝑡2
(𝑥𝑛− 1 , 𝑦𝑛−1 ) ⋅Δ𝑡 2
𝑣 𝑦𝑛=𝑣𝑦𝑛 −1
+𝑑2 𝑦𝑑 𝑡2
(𝑥𝑛−1 , 𝑦𝑛−1 ) ⋅Δ𝑡
𝑥0=1 𝑣 𝑥0=0
𝑑2𝑥𝑑𝑡2
(𝑥 , 𝑦 )=−𝐺⋅𝑚1𝑥
(𝑥2+𝑦2 )32
𝑑2 𝑦𝑑𝑡 2
(𝑥 , 𝑦 )=−𝐺⋅𝑚1 𝑦
(𝑥2+𝑦2 )32
𝑣 𝑦0=0.0191
Modeling the Planets’ Orbit• Model the orbits as circles.
• Earth’s orbit has a radius of 1 AU and a period of 365 days.
• Mar’s orbit has a radius of 1.52 AU and a period of 687 days.
𝑥𝑒(𝑡)=cos ( 2𝜋365 ⋅𝑡)𝑦 𝑒(𝑡)=sin ( 2𝜋365 ⋅ 𝑡)
𝑥𝑚 (𝑡 )=1.52 ⋅cos ( 2𝜋687 ⋅𝑡)𝑦𝑚(𝑡)=1.52⋅sin ( 2𝜋687 ⋅𝑡)
• How long does it take to get to Mars?
181.5 days
• What is the space ships location when it intersects the Mars Orbit?
(-1.1317, 1.0145)
Where should Mars be at launch?
Solving Equations
𝑥𝑚 (𝑡 )=1.52 ⋅cos ( 2𝜋687 ⋅𝑡)=−1.1317𝑦𝑚 (𝑡 )=1.52 ⋅sin ( 2𝜋687 ⋅𝑡)=1.0145
𝑡=263.578⇒
263.578−181.5=82.078
𝑥𝑚 (𝑡 )=1.52 ⋅cos ( 2𝜋687 ⋅(𝑡+82.078))𝑦𝑚 (𝑡 )=1.52 ⋅sin ( 2𝜋687 ⋅(𝑡+82.078))
Getting Home (Euler’s Method)
𝑣 𝑥𝑛=𝑣𝑥𝑛− 1
+ 𝑑2 𝑥𝑑𝑡 2
(𝑥𝑛− 1 , 𝑦 𝑛−1 ) ⋅Δ 𝑡
𝑦 𝑛=𝑦𝑛−1+𝑣𝑦𝑛⋅Δ t+ 1
2𝑑2 𝑦𝑑𝑡2
(𝑥𝑛− 1 , 𝑦 𝑛−1 ) ⋅Δ𝑡 2
𝑣 𝑦𝑛=𝑣𝑦𝑛 −1
+𝑑2 𝑦𝑑 𝑡2
(𝑥𝑛−1 , 𝑦𝑛−1 ) ⋅Δ𝑡
𝑥0=𝑥𝑚(𝑡𝑙) 𝑣 𝑥0=0.012
𝑦01.52
𝑑2𝑥𝑑𝑡2
(𝑥 , 𝑦 )=−𝐺⋅𝑚1𝑥
(𝑥2+𝑦2 )32
𝑑2 𝑦𝑑𝑡 2
(𝑥 , 𝑦 )=−𝐺⋅𝑚1 𝑦
(𝑥2+𝑦2 )32
𝑣 𝑦0=0.012
𝑥01.52
• How long is the return trip?
183.25 days
• Solve the equations.
• Leave Mars after 715 days after launch
When to launch?
Trip Length Breakdown
Outward Journey 181.5 days
Time On Mars 533.5 days
Return Trip 183.25 days
Neil Degrasse Tyson talking about a trip to Mars.