Droo

1 Droo Day - around 14 hrs
Rotational Velocity at equator-158 m/s
Mass= 2.383561897×10^23kg.
Radius-1274200 m
Area-2.04x 10^13 m2
Escape Velocity-4997.4 m/s
Revolution period=58 days apprx

   Luna

Radius: 354,000 m
Mass: 3.00E+21 kg
Orbital Velocity at 30 km: 722.1 m/s
Orbital Height above Droo (Avg.): 37,200,500 m
Revolution period-4d6hrs apprx
Rotation period-4d 6 hr apprx(tidal locking)

   Cylero

Radius: 678,000 m
Mass: 2.554717E+22 kg
Orbital Velocity at 150 km: 1435 m/s
Orbital Height above the Sun: 22,655 Mm
Escape Velocity: 2242.68 m/s

Feel free to tell more !
Let's bring a science revolution to SR2!


22 Comments

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    1,660 tsampoy

    If only science was added to sr2

    7 months ago
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    6 Maal

    @MAHADI Cylero does have same orbital speed as Kerbin in KSP dho

    7 months ago
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    154 MAHADI

    let's go to Mars!

    7 months ago
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    @Aerospacer thanks for finding error,I'll fix it

    8 months ago
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    @PriyanshuRoy I am look to wiki.
    Must be sqrt(2GM/r)

    8 months ago
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    148 GR00G0

    Sorry, never saw your post! I was hoping for some kind of community thread for everything about the game. If I get any mor new data I will post a comment here!

    8 months ago
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    101 Rexy

    @jacksawild Your delta V may be slightly more accurate. I solved deltaV by using the rocket equation but alas I utilized the vacuum exhaust velocity for both atmospheric and orbital flight which will contribute some error. My method was manual reading of rocket masses at each stage using 8 different rockets all entering into a 115x115 then summing a total dV.

    I'll try to put together a much more thorough experimental run at some point over the next few days.

    The average atmospheric density between surface and 100km is about 5.21 PSI

    8 months ago
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    @Aerospacer I used formula sqrt (GM/r)

    8 months ago
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    @Rexy "I've been doing various test runs and delta V required to achieve LDO appears to be approximately 5.57 km/s"

    Which way are you launching? Going east I can get 100km * 100km orbit with about 4.3km/s. Even less with an efficient design.

    According to my totally scientific sample size of 1, the edge of Droo SOI is 95775.5km

    8 months ago
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    P.S. Recalculate of escape velosity.
    It's about 4997.4 m/s.
    Zero-level orbital speed = 3533.7 m/s.

    8 months ago
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    +1 interesting fact about Droo -
    His average density ~27.5 t/m^3.
    Sadly, the same mistake as in KSP - it's unrealistic...
    (Osmium density ~22.6 t/m^3)

    +1 8 months ago
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    101 Rexy

    I've been doing various test runs and delta V required to achieve LDO appears to be approximately 5.57 km/s

    Atmosphere air pressure can also be roughly approximated by the function:
    p(h)=14.7exp(-(x/40000)^2)
    units being PSI,

    that was calculated experimentally taking engine thrust measurements every kilometer to look at dropoff, best fitting a gaussian packet, determining engine chamber pressure in vacuum, then solving for external air pressure as a function of altitude.

    +1 8 months ago
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    Also did some SCIENCE with Cylero and came up with this:
    Radius: 678,000 m
    Mass: 2.554717E+22 kg
    Orbital Velocity at 150 km: 1435 m/s
    Orbital Height above the Sun: 22,655 Mm
    Escape Velocity: 2242.68 m/s

    +1 8 months ago
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    I got some info on Luna. Not sure how accurate it is, so I will probably update it soon, but here are some rough estimates:

    Radius: 354,000 m
    Mass: 3.00E+21 kg
    Orbital Velocity at 30 km: 722.1 m/s
    Orbital Height above Droo (Avg.): 37,200,500 m

    +1 8 months ago
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    @jacksawild You need more upvotes for your info

    +1 8 months ago
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    @DrCoconut1245
    Earth mass is: 5.972 × 10^24 kg
    radius is: 6.3781×10^6 m

    So earth is about 5 times wider and 25 times more massive.

    +1 8 months ago
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    @jackawild Can we get comparisons to earth? The only thing I recognize is that the surface gravity is almost earths.

    8 months ago
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    @PriyanshuRoy

    M=ar^2/G

    a = surface gravity = 9.798 m/s^2
    r^2 = radius squared: 1274200^2 = 1623585640000 meters
    G = gravity constant = 6.6726 x 10-11N-m^2/kg^2

    So a bit more precise: 2.38406199992806×10^23kg, which alters droostationary height to 8803730 meters.

    +3 8 months ago
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    @jacksawild Thanks! But I didn't need to place a geostationary satellite.I just want to bring together the information.
    PS how did you find mass of droo?

    8 months ago
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    27 eonn44

    @PriyanshuRoy thanks

    8 months ago
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    Good info. So to calculate geostationary altitude:

    G = 6.67381x10^-11
    M = 2.383561897×10^23 (mass of Droo)
    t = 2540160000 (14 hours * 60 * 60)

    cuberoot((6.67381x10^-11 * 2.383561897×10^23 * 2540160000) / 4* pi^2) = 10,077,834

    Subtract radius of Droo (1274200 ) = 8803634.6 metres

    About as close as I could get without RCS: https://youtu.be/GY3HpcblSpM

    It has a period of 14hours and it seems to be sitting over the same continent. So I'd say it's pretty damn close.

    +4 8 months ago
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    172 TR2002

    Cool!

    8 months ago

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