a trip to mars
DESCRIPTION
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). Gravitational Force. - PowerPoint PPT PresentationTRANSCRIPT
A Trip to Mars
Douglas MarksNCSSM
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)
Gravitational Force
where
𝑚1 𝑚2
𝑟
Acceleration of Mass 2 due to the Gravitational Force
where
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)
What Happens?
• 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))
Second Attempt
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=0.012𝑦01.52
𝑑2𝑥𝑑𝑡2
(𝑥 , 𝑦 )=−𝐺⋅𝑚1𝑥
(𝑥2+ 𝑦2 )32
𝑑2 𝑦𝑑𝑡 2
(𝑥 , 𝑦 )=−𝐺⋅𝑚1 𝑦
(𝑥2+ 𝑦2 )32
𝑣 𝑦0=0.012𝑥01.52
Getting Home (Short Stay)
• How long is the return trip?
183.25 days
• Solve the equations.
• Leave Mars after 715 days after launch
When to launch?
There and Back Again
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.
Other Questions
• What is the space ship’s relative velocity to Mars when they meet?
• How does including the Gravitational Force from the Earth and Mars affect the path?
• How much shorter can you make the trip with better rockets?