Musician/ stay at home dad

I own an independent record label.

Freight conductor/ locomotive engineer.

Day job - Cybersecurity Analyst.
Side job - Adjunct Professor at local University.

Can we please get some more "warehouse" or other blue collar jobs? Im getting really depressed here! Lol

Seriously though I hate you all

I've been a sensor data analyst for the USAF, mostly working with satellite data. All very classified, you know. I've been a celestial mechanic (and I still play with orbits as a hobby) and a jackleg signal processor working with radar/IR signals, ambiguity diagrams, probability of detection or of false alarm given a signal-to-noise ratio. Now I'm retired, but I still pitch in to help physics students cheat on their homework at Yahoo Answers.

Typical question.

Two masses, M₁ and M₂, are at time t₀ separated by a distance, d, at rest with respect to each other, and acted upon by no forces other than their mutual gravitational attraction. How much time elapses, beginning at the moment their separation is d/2 and ending at the moment their separation is d/3?

Step 1. This is a plunge orbit problem. First set up the Vis Viva equation, solved for the speed in orbit.

M = M₁ + M₂

v = √[GM(2/r − 1/a)]

Since gravity is a central force (classically anyway) all of the future motion of the masses with respect to each other will be radial, meaning we can change v to dr/dt and create an ordinary, non-linear differential equation, solved for dt

dt = dr / √[GM(2/r − 1/a)]

Since the semimajor axis of the orbit will be half the apoapsis (initial) distance,

a = d/2

∫ dt = ∫ dr / √[2GM(1/r − 1/d)]

Integrating both sides gives:

t−t₀ = √[d/(2GM)] { √(rd−r²) + d Arctan[√(d/r−1)] }

By the way, we note that when r«d then r ≈ 0, and

t−t₀ ≈ π √[d³/(8GM)]

Anyway, when r=d/2

t₁−t₀ = √[d/(2GM)] { √(d²/2−d²/4) + d Arctan[√(2d/d−1)] }
t₁−t₀ = √[d/(2GM)] { d/2 + πd/4 }
t₁−t₀ = { 1/2 + π/4 } √[d³/(2GM)]

When r=d/3

t₂−t₀ = √[d/(2GM)] { √(d²/3−d²/9) + d Arctan[√(3d/d−1)] }
t₂−t₀ = {√(2)/3 + Arctan[√(2)] } √[d³/(2GM)]

Therefore,

t₂−t₁ = {√(2)/3 + Arctan[√(2)] − 1/2 − π/4 } √[d³/(2GM)]
t₂−t₁ ≈ 0.1413229755 √[d³/(2GM)]

I'm really bored since the update left me without ESO access. Hopefully, my new computer will arrive Very Soon!

Train conductor on the central trains in Denmark
But have alot of year behind me with 3D and Photoshop. And still doing it freelance.

I 'm the current vice president of Mars, but all I really do is eat pie.

Lysette wrote: »

kongkim wrote: »

Train conductor on the central trains in Denmark
But have alot of year behind me with 3D and Photoshop. And still doing it freelance.

A wonder, that this job was not rephrased yet like that of cabin crews in commercial airlines - steward/stewardess, cabin crew, flight attendants and whatever they may be called in future. Well, just going 100 years back, a computer was a human being not a machine - time change a lot - but train conductors are still train conductors, very traditional.

Yah but in Denmark the train system works so we need to check peoples tickets so they can't change it other then if the change the whole system and that will not happen in many years
And its well paied

Waffennacht wrote: »

Can we please get some more "warehouse" or other blue collar jobs? Im getting really depressed here! Lol

Seriously though I hate you all

Coal miner and taxidermist here.

911 Dispatcher/Operator