1. ME 340 Final Project Bryan Johnson, Devin LeBaron
Rocket Man
ME 340 Final Project
Bryan Johnson, Devin LeBaron

2. Introduction

3. The Problem
How close could a spaceship fly to the sun before melting?

4. The Sun
Temp = 5800 K Radius = 6.95e8 m Assume blackbody (ε=1)

5. The Space Ship
Assume Aluminum body Melting Temp = 933 K To approximate the shape… We will use a sphere

6. Energy Balance Gs = 4εσT^4/α
In space, convection and conduction don’t apply
αGs*Ac
Ac = πD^2/4
As = πD^2
E*As = αGs*Ac
εσT^4*As = αGs*Ac
εσT^4 = αGs/4
Gs = 4εσT^4/α
E*As
Space Ship

7. Emissivity and Absorptivity
At Tm = 933 K:
F1 = ; F2 =
ε(Tm) = 0.2(0.4301) ( ) (1 – )
ε(Tm) = 0.58
At Ts = 5800 K:
F1 = ; F2 =
α(Ts) = 0.2(0.9904) +
0.9( ) +
0.35( )
α(Ts) =
Chart based on graph from Fig from the book

8. Radiation of the Sun Ab
qs = Es*As qs = σT^4*As qs = Gs*Ab Gs*Ab = σT^4*As Gs = σT^4*As/Ab Gs = σT^4*Rs^2/(D+Rs)^2
Ship D As

9. Combine the formulas D = 7.32e9 m
Gs = σTs^4*Rs^2/(D+Rs)^2 Gs = 4εσTm^4/α 4εσTm^4/α = σTs^4*Rs^2/(D+Rs)^2 Solving for D: D = [(αTs^4*Rs^2)/(4εTm^4)]^.5 – Rs
ε = 0.58
α =
Tm = 933 K
Ts = 5800 K
Rs = 6.95e8 m
D = 7.32e9 m

10. The Answer
The final distance to melting your aluminum space ship would be about:
7.32e9 m
= 4,550,000 miles
Mercury is about 5.8e10 m from the sun. Which means you’d be pretty darn close.

11. Conclusions and Recommendations
What are you doing flying to the sun? Just don’t. But if you must, please use a better material with a higher melting point, better heat resistance and lower absorptivity.

12. References
Incropera, Frank P., and Frank P. Incropera. Fundamentals of Heat and Mass Transfer. Hoboken, NJ: John Wiley, Print. Rocketman. Perf. Harland Williams, Jessica Lundy, and William Sadler. Disney, 1997.