#format jsmath
= Stadyshell Falling =

Without light pressure or horizontal velocity, how long would it take for a stadyshell to fall into the sun?

Falling into the sun is an orbit with an aphelion of the starting height $r$ = 50 AU, and a perhelion of 0.  The semimajor axis a = $r/2$, and the eccentricity is 1.0.  The falling time is half the orbital period for this orbit,  $ T = \pi \sqrt{ r^3 / 8 \mu } $.   

We could look up $ \mu $ - but we can calculate it pretty accurately - without a calculator!  The earth is in an orbit with a semimajor axis of 1 AU, and the period is a year.  so  $ ( 1 year ) = 2 \pi \sqrt{ (1 AU)^3 / \mu } $.  Solving, $ \mu = 4 {\pi}^2 {AU}^3/{year}^2 $.  So the falling time is:

$ T = \pi \sqrt{ 125000 {AU}^3 / ( 32 * {\pi}^2 {AU}^3/{year}^2 ) }  = \sqrt{ 125000 / 32 }$ years $ = \sqrt{ ( 125/2 )^2 } $ years.
So '''T = 62.5 years'''.  Which would be exact if the earth had a 1 AU semimajor axis, which it does not, see below.

With a calculator, we have a handy way to estimate solar $ \mu $ to 4 decimal places. 

Remember that there are (100 + 3) leap years per 400 years ;  1 year = 365.2575 days = 31558248 seconds, 1 AU = 149.6e9 meters, estimated $\mu$ = 1.327176698e20 m^3^/s^2^.  The book value is 1.32712440018e20 m^3^/s^2^ (estimate is 39 ppm high), a year is 365.256363004 days (estimate is 3 ppm high), and earth's semimajor axis is 1.49598023e9 meters (estimate is 13 ppm high).  Which leaves a 6 ppm discrepancy, but pretty good for semi-POMA numbers.

How long does it take to fall half way down?  

We can project the zero-width elliptical orbit on a unit circle, and compute the area swept halfway through the circle (Kepler's second law) and ratio that to a half circle.  The pie shaped area swept is half the total area, and the triangle to that halfway point is $ ( 1/2 ) / ( \pi / 2 ) = (1/\pi ) $, so the time is 62.5 years * $ 1/2 + 1/pi $ = 51.14 years .  Other times can be solved similarly, with trigonometry.  

We can approximate with constant gravity for the first few years of drop.  At 50 AU, g = $ ( \pi^2 / 625 ) {AU}/{year}^2 ~ \approx ~ 0.0158 {AU}/{year}^2 $, so 1 year of falling is 0.008 AU, and 5 years of falling is 0.2 AU.  Not much, and a 0.2% increase in temperature at black body equilibrium (except we are sideways for this to happen, and the thinsat is not getting any sunlight, and closer to 2.7 K ).   The constant gravity approximation