# 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