Blog post Number 15 Written:05-30-2021 Uploaded 04-29-2022
I know I am bad at math, but bear with me, I have actually done some of the astrodynamic calculations for my books, but really only the second book with… we’ll you’ll see. Anyway, I was trying to figure out orbital velocity, because I needed to know how long it took a ship to make an orbit. Problem was, I didn’t know the ship’s weight…
By astrodynamics, I mean orbital mechanics. Let me dig out my notes and see what I’ve got here, since it’s been a couple years since I wrote the second book. *spends an hour rummaging through papers and notes*
Wow, these notes are dated 05-07-2018. I hadn’t realised it had been so long. I guess that means I need to finish writing book 5 this year to stay up to par of a book a year, which is a tall order since I haven’t really started on it… have I? *opens other word document* 14,394 words, I guess it’s started, at least 2k of those are just random notes to myself though… anyway math… (As of this posting it sits just under 40,000 words. It’s been a rough year.)
I think for simplicity’s sake I am going to write down word for word my old handwritten notes and then look at the math again and see how wrong I was.
“Is this orbit possible question
Orbital distance =1.4 AU (130,138,139.22 miles)
Orbital velocity = 0.23 LS (154,241,824.67 mph)
Orbital period = 5.30 hours or 5 hours 18 minutes
Kepler 452 mass =1.037x(2×10^30 kg) or 10609425264 pounds
Orbital circumfrance = 817,681,987.70 miles
Kepler 452 diameter =959,679.36 miles
Kepler 452 gravitational force = 1.109 of the sun or 11.33 mile/s/s
Mass of Tsaritsyn battle cruiser = ?
Will decrease velocity and increase orbital period to fit the timeline the Uradain ship will simply reverse orbit and enter planetary orbit rather then solar orbit.
2.64487272E^-8 .004302 * .0000067874*.9058
=4.302*10^-3*6.7874*10^-6*1.104^-1 (km/s)
If this formula is roughly equal to desired velocity then the orbit is possible
248,228,155 KMH = SQ RT (2×10^30kg)*(6.673*10^-11 N*M^2/kg^2)/209,437.019KM
798,268.214 KmH = SQRT 637,233,140,894.82
This means my orbital velocity is 311 times too fast for that orbit
This gives an orbital period of 68.8 days
496,020,872mph = .00074LS or .07 Light speed.
Yamamoto’s velocity 1 mil kph =0.0009 LS
So 10 mil is .009 LS (.9%)
So 11 mil is 1.01 LS
G=4.302×10^-3 pc M^-1 (km/s)^2 =SQRT (G*M)/R
Kepler 303.87 m/s/s =679 mph
Suns gravity of 274 m/s/
1 solar mass = 2X10^30 KG 1 solar radius= 432,288 miles
1 AU = 149,597,870,700 meters =92,955,807.3 miles
LS =299,792,458 m/s =670,616,629 mph = 1,079,252,848.8 kph”
That’s a typed version of my page of handwritten notes there is another side to that page, but those involve spoilers the world ain’t ready for yet. Yes, my math formulas count as spoilers.
Anyway, my simple bad at math brain picked an arbitrary speed for my ship, to complete a solar orbit in an allotted amount of time, but my chosen speed was 311 times faster than was possible. But All I did was calculate the circumference of the orbit based on the distance from the star and then figure out how long it would take to travel that distance at a given speed. I know that’s a very rudimentary approach and doesn’t properly account for the star’s gravitational pull, but remember, I don’t know the mass of my ship in question (The battlecruiser Tsaristyn) and I’m not good at math. I found out, to be at a more reasonable speed it would still take almost 70 days to complete a solar orbit, so that’s why I put the scene in question over the planet Kepler 452B and not the star Kepler 452. I mean they were going to end up planeside either way, but I needed a specific time interval in order for some of the narrative to work and timeline to line up, more like five-ish hours, not 70 days, so again, planetary orbit not solar orbit. It’s not precise, or even right, but it was enough to tell me which ballpark to play in.
Chace Randolph 