A Trip to Mars Calculus

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Transcript A Trip to Mars Calculus

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
𝑚1
𝑚2
𝑟
𝑚1 ⋅ 𝑚2
𝐹=𝐺⋅
𝑟2
where 𝐺 = 6.67 ⋅
10−11
𝑚3
𝑘𝑔⋅𝑠 2
Acceleration of Mass 2 due to the
Gravitational Force
𝑚1 ⋅ 𝑚2
𝐹=𝐺⋅
𝑟2
where 𝐺 = 6.67 ⋅
10−11
𝐹
|𝑎2 | =
𝑚2
𝑚1
|𝑎2 | = 𝐺 2
𝑟
𝑚3
𝑘𝑔⋅𝑠 2
Differential Equations
𝑑2 𝒓
𝑚1 𝒓
= −𝐺 ⋅ 2 ⋅
𝑑𝑡 2
𝑟
𝒓
𝑟
𝑦
𝑑2𝑦
𝑚1
= −𝐺 ⋅ 2 ⋅ sin(𝜃)
𝑑𝑡 2
𝑟
𝜃
𝑥
𝑑2𝑥
𝑚1 𝑥
=
−𝐺
⋅
⋅
𝑑𝑡 2
𝑟2 𝑟
𝑚1 𝑥
= −𝐺 ⋅
𝑥2 + 𝑦2
𝑑2𝑥
𝑚1
= −𝐺 ⋅ 2 ⋅ cos(𝜃)
𝑑𝑡 2
𝑟
3
2
𝑑2𝑦
𝑚1 𝑦
=
−𝐺
⋅
⋅
𝑑𝑡 2
𝑟2 𝑟
𝑚1 𝑦
= −𝐺 ⋅
𝑥2 + 𝑦2
3
2
Euler’s Method (linear)
𝑥0 = 1
𝑦0 = 0
𝑑2𝑥
(𝑥, 𝑦) = −𝐺 ⋅
𝑑𝑡 2
𝑣𝑥𝑛
𝑣𝑥0 = 0
𝑚1 𝑥
𝑥2
+
3
2
𝑦 2
𝑑2𝑥
= 𝑣𝑥𝑛−1 + 2 𝑥𝑛−1 , 𝑦𝑛−1 ⋅ Δ𝑡
𝑑𝑡
𝑥𝑛 = 𝑥𝑛−1 + 𝑣𝑥𝑛−1 ⋅ Δ𝑡
𝑣𝑦0 = 0.0191
𝑑2𝑦
(𝑥, 𝑦) = −𝐺 ⋅
𝑑𝑡 2
𝑣𝑦𝑛
𝑚1 𝑦
𝑥2
+
3
2
𝑦 2
𝑑2𝑦
= 𝑣𝑦𝑛−1 + 2 𝑥𝑛−1 , 𝑦𝑛−1 ⋅ Δ𝑡
𝑑𝑡
𝑦𝑛 = 𝑦𝑛−1 + 𝑣𝑦𝑛−1 ⋅ Δ𝑡
Euler’s Method (quadratic)
𝑥0 = 1
𝑦0 = 0
𝑑2 𝑥
(𝑥, 𝑦) = −𝐺 ⋅
𝑑𝑡 2
𝑣𝑥𝑛
𝑣𝑥0 = 0
𝑚1 𝑥
𝑥2
+
3
𝑦2 2
𝑑2 𝑥
= 𝑣𝑥𝑛−1 + 2 𝑥𝑛−1 , 𝑦𝑛−1 ⋅ Δ𝑡
𝑑𝑡
1 𝑑2 𝑥
𝑥𝑛 = 𝑥𝑛−1 + 𝑣𝑥𝑛−1 ⋅ Δ𝑡 +
𝑥𝑛−1 , 𝑦𝑛−1 ⋅ Δ𝑡 2
2
2 𝑑𝑡
𝑣𝑦0 = 0.0191
𝑑2 𝑦
(𝑥, 𝑦) = −𝐺 ⋅
𝑑𝑡 2
𝑣𝑦𝑛
𝑚1 𝑦
𝑥2
+
3
2
𝑦 2
𝑑2 𝑦
= 𝑣𝑦𝑛−1 + 2 𝑥𝑛−1 , 𝑦𝑛−1 ⋅ Δ𝑡
𝑑𝑡
1 𝑑2 𝑦
𝑦𝑛 = 𝑦𝑛−1 + 𝑣𝑦𝑛−1 ⋅ Δt +
𝑥𝑛−1 , 𝑦𝑛−1 ⋅ Δ𝑡 2
2
2 𝑑𝑡
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.
2𝜋
𝑥𝑒 (𝑡) = cos
⋅𝑡
365
𝑥𝑚
2𝜋
𝑦𝑒 (𝑡) = sin
⋅𝑡
365
𝑦𝑚 (𝑡) = 1.52 ⋅ sin
2𝜋
𝑡 = 1.52 ⋅ cos
⋅𝑡
687
2𝜋
⋅𝑡
687
What Happens?
Where should Mars be at launch?
• 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)
Solving Equations
𝑥𝑚
2𝜋
𝑡 = 1.52 ⋅ cos
⋅ 𝑡 = −1.1317
687
𝑦𝑚 𝑡 = 1.52 ⋅ sin
2𝜋
⋅ 𝑡 = 1.0145
687
⇒
263.578 − 181.5 = 82.078
𝑥𝑚 𝑡 = 1.52 ⋅ cos
2𝜋
⋅ (𝑡 + 82.078)
687
𝑦𝑚 𝑡 = 1.52 ⋅ sin
2𝜋
⋅ (𝑡 + 82.078)
687
𝑡 = 263.578
Second Attempt
Getting Home (Euler’s Method)
𝑥0 = 𝑥𝑚 (𝑡𝑙 )
𝑦0 = 𝑦𝑚 (𝑡𝑙 )
𝑑2 𝑥
(𝑥, 𝑦) = −𝐺 ⋅
𝑑𝑡 2
𝑣𝑥𝑛
𝑚1 𝑥
𝑥2
+
3
2
𝑦 2
𝑑2 𝑥
= 𝑣𝑥𝑛−1 + 2 𝑥𝑛−1 , 𝑦𝑛−1 ⋅ Δ𝑡
𝑑𝑡
1 𝑑2 𝑥
𝑥𝑛 = 𝑥𝑛−1 + 𝑣𝑥𝑛 ⋅ Δ𝑡 +
𝑥𝑛−1 , 𝑦𝑛−1 ⋅ Δ𝑡 2
2
2 𝑑𝑡
𝑣𝑥0 = 0.012
𝑦0
1.52
𝑣𝑦0 = 0.012
𝑑2 𝑦
(𝑥, 𝑦) = −𝐺 ⋅
𝑑𝑡 2
𝑣𝑦𝑛
𝑥0
1.52
𝑚1 𝑦
𝑥2 + 𝑦2
3
2
𝑑2 𝑦
= 𝑣𝑦𝑛−1 + 2 𝑥𝑛−1 , 𝑦𝑛−1 ⋅ Δ𝑡
𝑑𝑡
1 𝑑2 𝑦
𝑦𝑛 = 𝑦𝑛−1 + 𝑣𝑦𝑛 ⋅ Δt +
𝑥𝑛−1 , 𝑦𝑛−1 ⋅ Δ𝑡 2
2
2 𝑑𝑡
Getting Home (Short Stay)
When to launch?
• How long is the return trip?
183.25 days
• Solve the equations.
• Leave Mars after 715 days after 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?