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| With a calculator, we have a handy way to estimate $ \mu $, if we remember 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. | With a calculator, we have a handy way to estimate $ \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. |
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 
2 
r38
We could look up (1AU)3 =42AU3year2
125000AU3(32
2AU3year2)=12500032=(1252)2
With a calculator, we have a handy way to estimate \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 m3/s2. The book value is 1.32712440018e20 m3/s2 (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 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 = $ ( \pi2 / 625 ) {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 ).