#Solve sec(pi/2-theta) | Explore Tumblr posts and blogs

vash3r · 1 year ago

Text

my answer

first estimate, something on the order of a few light-seconds. some quick points of comparison from numbers i have memorized:

1 light-second: ~399,000,000 m/s * 1s = ~400,000 km

sun-earth distance: 1 AU (?? km i forgor), a few light-minutes (4~8?)

diameter of earth: ~40,000 km (radius ~10,000 km)

distance to "earth orbit": ~100km? (atmosphere: ~10km)

moon radius approx 1/4 that of earth? def at least 1/10, moon is a very big moon (=> because volume is cube, only 1/64th the mass: seems an appropriate order of magnitude)

from "Eclipse" (vaguely remembered): moon-sun distance ratio (~=radius ratio, due to eclipses) is about 1:400 ?

refined guess after thinking a little: 1-4 light-seconds.

now it's time to get nerdsniped.

recall equation for gravity: F= (G*m1m2/d^2). i forgot G but we can skip it because we know g=9.8 at surface of earth (1e4 km); at moon orbit (estimated 4e5 km) we need the following to hold: that the moon orbits the earth approx 1x every 29 days (sidereal but let's ignore that for now) (29*86400=30*_-_ =2592000-86400s = 2,508,000s).

let's introduce some variables:

Rmo[radius of moon orbit] = (unknown, estimated 4e8m)

Re[radius of earth] = 1e7 (m)

Cmo[circumference of moon orbit] = Rmo*2pi (m)

Tmo[time of moon to orbit once] = approx 2.5 e6 (sec)

Vmo[velocity of moon in orbit] = Cmo/Tmo = 2pi*Rmo/Tmo (m/s)

Ame[acceleration of moon due to Earth gravity] = 9.8 m/s^2 * (Re/Rmo)^2

we want to solve an equation for circular motion for radius given the centrifugal/centripetal acceleration and orbital period. i forgot the equation though and sadly i can't just do trig on a tiny triangle (not only because i don't have arctan tables memorized).

calculus to the rescue? let's solve the integral of acceleration along the circumference for a quarter-turn (because we know the resulting change in velocity): starting at the standard spot (+Rmo, 0) with velocity [0, +v] (in the positive-y direction) and going counterclockwise: integrate S[0 to pi/2] (a [-cos theta, -sin theta] dtheta) = [-v, -v]. except instead of just integrating along angle, we want to integrate along time; so

S{0 to Tmo/4} (a [-cos (2pi*t/Tmo), -sin (2pi*t/Tmo)] dt) = [-v,-v]

a * (Tmo/2pi) * |0 to Tmo/4| [-sin(2pi*t/Tmo), cos(2pi*t/Tmo)] = [-v,-v]

a * (Tmo/2pi) * [-1,-1] = v*[-1,-1]

a * Tmo/2pi = v (in units of seconds). we've successfully rederived the circular motion equation! now substituting in v=Cmo/Tmo and a=Ame, we get

Ame = 2pi * Cmo/Tmo^2

9.8 *(Re/Rmo)^2 = (2pi)^2 * Rmo/Tmo^2

Rmo = cbrt(Tmo^2 * Re^2 * (2pi)^2 * 9.8)

simplifying to order of magnitude:

Rmo = cbrt(2.5e6^2 * 1e7^2 * 2^2 * pi^2 * 1e1)

Rmo = cbrt((5^2)* 1e12 * 1e14 * 1e1 * 1e1)

Rmo = cbrt(25e28) ~ 2.9e9.3 m, or approx 600,000 km, about 1.5 lightseconds. without using books, internet, or a calculator.

EDIT: turns out while my estimates in terms of lightseconds were relatively accurate, many of my other numbers were super wrong: lightspeed (which is 299,xxx,xxx m/s instead of 399,xxx,xxx), saying "earth diameter" when i meant circumference, using earth diameter in place of radius, multiplying by 2pi^2 instead of dividing, and forgetting an order of magnitude in the answer lmao, so the formula should be

Rmo = cbrt(Tmo^2 * Re^2 * 9.8 / (2pi)^2) = cbrt(1.4e12 * 36e12) = cbrt(54e24) and from 54 between 27 and 64, we get approx 3.8e8 which is practically spot on

Guesstimation game: no googling!

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