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Posers and Puzzles

Posers and Puzzles

  1. Subscriber sonhouse
    Fast and Curious
    02 Apr '10 20:20 / 1 edit
    So this genius kid figures out a few what now is sci fi, a small anti-matter power source, and a kind of propulsion unit that would accelerate everything at the same rate so you can do accel at a million G's without getting squashed.
    So his maiden voyage is to the moon, we call it 384,000 Km or 240,000 miles away take your pick.
    The question is, how long does it take for him to land on the moon at 1 million G's. What is his average velocity and what is the peak velocity?

    We assume he is already in orbit so you don't have to deal in atmospheric drag and his orbit is equidistant from the moon so all we worry about is the 384,000 km. Just take off from a couple hundred miles high and land on the moon. How much time?
  2. Standard member forkedknight
    Defend the Universe
    02 Apr '10 20:39 / 3 edits
    Originally posted by sonhouse
    So this genius kid figures out a few what now is sci fi, a small anti-matter power source, and a kind of propulsion unit that would accelerate everything at the same rate so you can do accel at a million G's without getting squashed.
    So his maiden voyage is to the moon, we call it 384,000 Km or 240,000 miles away take your pick.
    The question is, how long ...[text shortened]... land on the moon at 1 million G's. What is his average velocity and what is the peak velocity?
    Accel = A
    Velocity = Ax
    Dist = 1/2*Ax^2

    A = 1 000 000 * G
    G = 9.8 m/s^s

    Assuming positive acceleration for the first half of the journey, and negative acceleration for the second half.

    Dist = (Dist to moon)/2 = 192 000 000 m
    192 000 000 = 1/2 * A * x2
    x ~= 6.26s
    Max Velocity = A*x = 9 800 000 m/s^2 * 6.26s = 61 344 926 m/s

    Total time = 2x = 12.52s
    Ave velocity = 384 000 000m / 12.52s = 30 672 563 m/s

    *edit*
    I haven't looked at or thought about basic calculus in quite some time. Sorry if I solved your problem before others had a chance.
  3. Standard member forkedknight
    Defend the Universe
    02 Apr '10 20:48
    To the moon and back in less time than it takes me to run 200m; that's pretty impressive.
  4. Subscriber sonhouse
    Fast and Curious
    03 Apr '10 17:01
    Originally posted by forkedknight
    To the moon and back in less time than it takes me to run 200m; that's pretty impressive.
    Like I said, a bat out of hell
    Can you derive a simple formula for this? Where you accelerate halfway, turn the engine around and decel the rest of the way. In one formula?
    That is to say, given the accel rate and distance, what is the time of flight from start of accel run to end of decel run.
  5. 03 Apr '10 18:06 / 4 edits
    Originally posted by sonhouse
    Like I said, a bat out of hell
    Can you derive a simple formula for this? Where you accelerate halfway, turn the engine around and decel the rest of the way. In one formula?
    That is to say, given the accel rate and distance, what is the time of flight from start of accel run to end of decel run.
    Obviously twice the time to get half way, because one half is a mirror image of the other.

    So (ignoring relativity)

    D/2 = 1/2*A(t/2)^2
    D/2 = At^2/8
    d = At^2/4
    sqrt(d/A) = t/2
    t = 2.sqrt(d/A)


    For the example given:
    t = 2.sqrt( 384000000/9800000) = 12.52 secs or so, as forked knight said

    If we include relativity, I think it would seem like 12.52 seconds to the person accelerating, but would seem somewhat longer than that to an observer.
  6. Subscriber sonhouse
    Fast and Curious
    03 Apr '10 19:52
    Originally posted by iamatiger
    Obviously twice the time to get half way, because one half is a mirror image of the other.

    So (ignoring relativity)

    D/2 = 1/2*A(t/2)^2
    D/2 = At^2/8
    d = At^2/4
    sqrt(d/A) = t/2
    t = 2.sqrt(d/A)


    For the example given:
    t = 2.sqrt( 384000000/9800000) = 12.52 secs or so, as forked knight said

    If we include relativity, I think it would seem like ...[text shortened]... .52 seconds to the person accelerating, but would seem somewhat longer than that to an observer.
    I did it this way: from the old standby S=(AT^2)/2 I got T=2*(S/A)^1/2. Pretty simple but all in one formula. Comes in at 12.585 seconds, using 240,000 miles and 32 ft/sec/second *1E 6.
  7. 04 Apr '10 06:12 / 2 edits
    Originally posted by sonhouse
    I did it this way: from the old standby S=(AT^2)/2 I got T=2*(S/A)^1/2. Pretty simple but all in one formula. Comes in at 12.585 seconds, using 240,000 miles and 32 ft/sec/second *1E 6.
    That's the same as mine, if your S is distance.
  8. Subscriber sonhouse
    Fast and Curious
    04 Apr '10 12:34
    Originally posted by iamatiger
    That's the same as mine, if your S is distance.
    I think the S came from German, it seems to be the standard way of denoting distance in accel formulae. At 1 million G it takes about 30 seconds to get close to c.