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HawaiianHippie
Lv 4
HawaiianHippie asked in Science & MathematicsPhysics · 1 decade ago

How long would it take a spaceship capable of 3-g acceleration to travel 1 light-month?

I'm writing a sci-fi story, and need a ballpark answer.

The ship needs to accelerate to the midpoint, then decelerate to a stop, covering a distance of 1 light-month (roughly 32.4 billion km). It can maintain a continuous acceleration / deceleration of roughly 3-g (roughly 30 km/sec/sec).

How long does this trip take?

Thanks!

Update:

Sorry, my math was wrong, a light-month is about 778 billion km

Update 2:

DG, I need that ship to arrive with 0 velocity, so it accelerates halfway, and decelerates halfway. Can I just double that final figure? No, that would be too simple, wouldn't it?

Update 3:

Um, it isn't hard SF, so I'm not so worried about accurately portraying the effects of relativity. The ship's FTL engines failed a light-month short of the destination. Their conventional engines are capable of sustained 3-g thrust. How long does the crew have, from their perspective, until they reach home?

4 Answers

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  • D g
    Lv 7
    1 decade ago
    Favorite Answer

    c =3*10^8 m/s

    3 g's = 3*9.8 m/s^2 = 29.4 m/s^2

    1 min = 60 seconds

    1 hour = 60 minutes

    1 day = 24 hours

    1 month = 30 days

    1 month = 30 * 24 * 3600 sec

    ........... =2.592* 10^6 s

    light would travel in one month

    x = ct

    = 3*10^8 *2.592*10^6 s

    = 7.776 * 10^14 meters

    So light travels 7.776 * 10^14 meters in 30 days.

    a ship accelerating the whole trip

    x = 1/2at^2

    solve for t

    t^2 = 2x/a

    t = root(2x/a)

    t = 7.273 * 10 ^6 seconds

    divide that by number of seconds in a month

    1 month = 30 * 24 * 3600 sec

    ........... =2.592* 10^6 s

    t = 2.81 months to reach the spot

    ONLY IF you keep on accelerating the whole trip.

    the velocity at that point would be

    v = 3*9.8*7.273 * 10 ^6

    = 2.138 * 10^8 m/s

    about 71 percent of c

  • billrussell42
    Lv 7
    1 decade ago

    For an observer on the ship or one stationary?

    for the stationary observer, about 1 month, as the ship will be traveling close to the speed of light most of the time. For an observer on the ship, a lot less time.

    Exact calculations involve relativity equations. And I don't have a lot of experience in that area, but I do know the calculations can get complicated.

    .

  • Steve
    Lv 7
    1 decade ago

    The halfway point is at 3.888E14 m

    Round trip time t is twice the time to the midpoint:

    t = 2*√(2x/a) = 32.199E6 sec = 372.67 days

    Vmax = a*t/2 = 3*32.199E6/2 = 48.3E3 km/sec (about .16c, so relativity is not a consequential factor)....

  • guru
    Lv 6
    1 decade ago

    it depends on the ship max speed parameters.

    I would not exceed the speed of light.

    It’s probably dangerous to do so.

    Hope this answers your question and helps you

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