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A Trip to Mars Douglas Marks NCSSM. The Problem Find a flight path from the Earth to Mars.

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A Trip to Mars Douglas Marks NCSSM
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A Trip to Mars

Douglas MarksNCSSM

The Problem

β€’ Find a flight path from the Earth to Mars.

Approaches

β€’ Define an Archimedean spiral (Pre cal version)

β€’ Use Kepler’s laws (Physics/Pre cal version)

β€’ Define the forces due to gravity on the rocket (Calculus version)

Gravitational Force

where

π‘š1 π‘š2

π‘Ÿ

Acceleration of Mass 2 due to the Gravitational Force

where

Differential Equations𝑑2𝒓𝑑𝑑2

=βˆ’πΊβ‹…π‘š1

π‘Ÿ2β‹… 𝒓|𝒓|

π‘Ÿπ‘¦

π‘₯

𝑑2π‘₯𝑑𝑑2

=βˆ’πΊβ‹…π‘š1

π‘Ÿ2β‹…cos (πœƒ)

πœƒ 𝑑2 𝑦𝑑𝑑 2

=βˆ’πΊβ‹…π‘š1

π‘Ÿ2β‹…sin (πœƒ)

𝑑2π‘₯𝑑𝑑2

=βˆ’πΊβ‹…π‘š1

π‘Ÿ2β‹… π‘₯π‘Ÿ

ΒΏβˆ’πΊβ‹…π‘š1π‘₯

(π‘₯2+𝑦 2 )32

𝑑2 𝑦𝑑𝑑 2

=βˆ’πΊβ‹…π‘š1

π‘Ÿ2β‹… π‘¦π‘Ÿ

ΒΏβˆ’πΊβ‹…π‘š1 𝑦

(π‘₯2+𝑦 2 )32

Euler’s Method (linear)

𝑣 π‘₯𝑛=𝑣π‘₯π‘›βˆ’ 1

+ 𝑑2 π‘₯𝑑𝑑 2

(π‘₯π‘›βˆ’ 1 , 𝑦 π‘›βˆ’1 ) β‹…Ξ” 𝑑

𝑦 𝑛=π‘¦π‘›βˆ’1+𝑣𝑦𝑛 βˆ’1⋅Δ𝑑π‘₯𝑛=π‘₯π‘›βˆ’1+𝑣 π‘₯𝑛 βˆ’1

⋅Δ𝑑

𝑣 𝑦𝑛=𝑣𝑦𝑛 βˆ’1

+𝑑2 𝑦𝑑 𝑑2

(π‘₯π‘›βˆ’1 , π‘¦π‘›βˆ’1 ) ⋅Δ𝑑

π‘₯0=1 𝑣 π‘₯0=0

𝑑2π‘₯𝑑𝑑2

(π‘₯ , 𝑦 )=βˆ’πΊβ‹…π‘š1π‘₯

(π‘₯2+ 𝑦2 )32

𝑑2 𝑦𝑑𝑑 2

(π‘₯ , 𝑦 )=βˆ’πΊβ‹…π‘š1 𝑦

(π‘₯2+ 𝑦2 )32

𝑣 𝑦0=0.0191

Euler’s Method (quadratic)

𝑣 π‘₯𝑛=𝑣π‘₯π‘›βˆ’ 1

+ 𝑑2 π‘₯𝑑𝑑 2

(π‘₯π‘›βˆ’ 1 , 𝑦 π‘›βˆ’1 ) β‹…Ξ” 𝑑

𝑦 𝑛=π‘¦π‘›βˆ’1+𝑣𝑦𝑛 βˆ’1β‹…Ξ”t+ 1

2𝑑2 𝑦𝑑 𝑑2

(π‘₯π‘›βˆ’ 1 , π‘¦π‘›βˆ’1 ) ⋅Δ𝑑 2

𝑣 𝑦𝑛=𝑣𝑦𝑛 βˆ’1

+𝑑2 𝑦𝑑 𝑑2

(π‘₯π‘›βˆ’1 , π‘¦π‘›βˆ’1 ) ⋅Δ𝑑

π‘₯0=1 𝑣 π‘₯0=0

𝑑2π‘₯𝑑𝑑2

(π‘₯ , 𝑦 )=βˆ’πΊβ‹…π‘š1π‘₯

(π‘₯2+𝑦2 )32

𝑑2 𝑦𝑑𝑑 2

(π‘₯ , 𝑦 )=βˆ’πΊβ‹…π‘š1 𝑦

(π‘₯2+𝑦2 )32

𝑣 𝑦0=0.0191

Modeling the Planets’ Orbitβ€’ Model the orbits as circles.

β€’ Earth’s orbit has a radius of 1 AU and a period of 365 days.

β€’ Mar’s orbit has a radius of 1.52 AU and a period of 687 days.

π‘₯𝑒(𝑑)=cos ( 2πœ‹365 ⋅𝑑)𝑦 𝑒(𝑑)=sin ( 2πœ‹365 β‹… 𝑑)

π‘₯π‘š (𝑑 )=1.52 β‹…cos ( 2πœ‹687 ⋅𝑑)π‘¦π‘š(𝑑)=1.52β‹…sin ( 2πœ‹687 ⋅𝑑)

What Happens?

β€’ How long does it take to get to Mars?

181.5 days

β€’ What is the space ships location when it intersects the Mars Orbit?

(-1.1317, 1.0145)

Where should Mars be at launch?

Solving Equations

π‘₯π‘š (𝑑 )=1.52 β‹…cos ( 2πœ‹687 ⋅𝑑)=βˆ’1.1317π‘¦π‘š (𝑑 )=1.52 β‹…sin ( 2πœ‹687 ⋅𝑑)=1.0145

𝑑=263.578β‡’

263.578βˆ’181.5=82.078

π‘₯π‘š (𝑑 )=1.52 β‹…cos ( 2πœ‹687 β‹…(𝑑+82.078))π‘¦π‘š (𝑑 )=1.52 β‹…sin ( 2πœ‹687 β‹…(𝑑+82.078))

Second Attempt

Getting Home (Euler’s Method)

𝑣 π‘₯𝑛=𝑣π‘₯π‘›βˆ’ 1

+ 𝑑2 π‘₯𝑑𝑑 2

(π‘₯π‘›βˆ’ 1 , 𝑦 π‘›βˆ’1 ) β‹…Ξ” 𝑑

𝑦 𝑛=π‘¦π‘›βˆ’1+𝑣𝑦𝑛⋅Δ t+ 1

2𝑑2 𝑦𝑑𝑑2

(π‘₯π‘›βˆ’ 1 , 𝑦 π‘›βˆ’1 ) ⋅Δ𝑑 2

𝑣 𝑦𝑛=𝑣𝑦𝑛 βˆ’1

+𝑑2 𝑦𝑑 𝑑2

(π‘₯π‘›βˆ’1 , π‘¦π‘›βˆ’1 ) ⋅Δ𝑑

π‘₯0=π‘₯π‘š(𝑑𝑙) 𝑣 π‘₯0=0.012

𝑦01.52

𝑑2π‘₯𝑑𝑑2

(π‘₯ , 𝑦 )=βˆ’πΊβ‹…π‘š1π‘₯

(π‘₯2+𝑦2 )32

𝑑2 𝑦𝑑𝑑 2

(π‘₯ , 𝑦 )=βˆ’πΊβ‹…π‘š1 𝑦

(π‘₯2+𝑦2 )32

𝑣 𝑦0=0.012

π‘₯01.52

Getting Home (Short Stay)

β€’ How long is the return trip?

183.25 days

β€’ Solve the equations.

β€’ 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

β€’ What is the space ship’s relative velocity to Mars when they meet?

β€’ How does including the Gravitational Force from the Earth and Mars affect the path?

β€’ How much shorter can you make the trip with better rockets?


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