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UPDATE: 5/12: I'm making progress on this project and I can now ask a few, more specific questions. However, to expedite things, would anyone be willing to allow me to contact them directly? It would speed things up considerably.

Until then, (and please excuse my inability to calculate variations of the formulas offered below) my current question is:

Traveling vertically straight up at an acceleration of 3Gs, how long would it take to reach an altitude of 250 miles? And how fast would you be traveling upon arrival (no need to decelerate)?

ORIGINAL QUESTIONS:

I'm a screenwriter developing a screenplay for a possible movie and I need some help calculating acceleration so the facts in the film are accurate. I also need solutions with different variables (described below) so I can build the story properly (I tried doing the math myself, but my algebra skills have atrophied).

  1. Accelerating vertically at a constant 1G, how long would it take to reach an altitude of 250 miles? (Please note, this is not into orbit, just altitude)

  2. How about to an altitude of 22,236 miles?

  3. How fast would you be traveling when you reached 250 miles and 22,236 miles respectively?

  4. I also need answers for acceleration at 3Gs, 6Gs and 9Gs.

  5. If I only had 30 minutes to reach 250 miles altitude, how many Gs would I experience and how fast would I be traveling upon arrival?

  6. What if I only had 20 minutes?

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    $\begingroup$ Try playing with this calculator at calculator soup, for distance, and this one, for speed. $\endgroup$
    – kim holder
    Commented Apr 7, 2016 at 17:56
  • $\begingroup$ Are you sure you want "straight up"? You might want to watch the recording of the Space-X SES-9 live stream. (youtube.com/watch?v=muDPSyO7-A0) This might help you understand how long it takes to get into orbit. The commentators do a very good job of explaining in lay-terms what is needed to get into orbit. To reach orbit, you need to go sideways really fast, not really far up. Also see: XKCD: what-if.xkcd.com/58 $\endgroup$
    – MattD
    Commented Apr 8, 2016 at 16:13
  • $\begingroup$ Putting 1+1 together, you are talking about going up a space elevator. I believe the passenger would also experience horizontal acceleration. The numbers in the answers below are only vertical acceleration. In addition to going up 22,236 miles, you will also need to increase your horizontal velocity to the speed at which the elevator is moving horizontally at that altitude. It will be moving about 5.6 times as fast - about 5600 mph. $\endgroup$
    – MattD
    Commented Apr 8, 2016 at 19:01
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    $\begingroup$ @MattD I have a feeling NeilWiser does't need to achieve orbit. ("Please note, this is not into orbit, just altitude") If one just want to get up there but don't need to stay up there, then there is no need for an orbit. The goal might be to pass by a satellite and interact with it briefly or maybe hitch a ride on a Vogon Constructor Ship. $\endgroup$
    – uhoh
    Commented Apr 9, 2016 at 0:59
  • $\begingroup$ Yes, this is for a Space Elevator (I forgot to mention that), so orbital elements aren't relevant. However, I didn't consider the interesting implications of horizontal acceleration/velocity. Unfortunately, I may have to fudge on those affects. We're talking movies here, not actual rocket science. 😉 $\endgroup$
    – Neal Wiser
    Commented Apr 9, 2016 at 19:06

1 Answer 1

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Distance travelled ($h$) after a given time ($t$) at a given acceleration ($a$), is calculated by:

$$h=\frac{t^2a}{2}$$

Re-arranging that into:

$$t=\sqrt{\frac{2h}{a}}$$

Should answer your first and second question, giving $286.5s$ for 250 miles, and $2702s$ for 22,236 miles.

That velocity is just: $v=ta$, is what you need for your third question, $2,809 m/s$ and $26,493 m/s$ respectively. Insert any other acceleration into the previous formula to get the new time, and use that together with the acceleration value in this formula to get the required numbers for your fourth question, namely:

$$v=\sqrt{2ha}$$

Finally, for number five and six, the required acceleration given altitude and time is:

$$a=\frac{2h}{t^2}$$

That gives you $0.02533 g$ and $0.05698 g$, for 30 minutes and 20 minutes respectively.

This is all of course assuming you mean change in velocity with your acceleration values, not the felt acceleration. (e.g. you feel 1g by just standing).

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  • $\begingroup$ Thanks for the answers, but let me make sure I understand what you're saying;<p> 1) It would only take 286.5 seconds to reach an altitude of 250 miles accelerating at 1G?<p> 2) And 2,702 seconds to reach an altitude of 22,236 miles?<p> 3) And the speeds would be:<br> - 2,809m/s at 250 miles?<br> - 26,493m/s at 22,236 miles?<p> What do those speeds translate as when converted to MPH?<p> 4) For 30 minutes to reach 250 miles altitude only requires an acceleration of 0.02533g<br> - And 0.05698g for 20 minutes? $\endgroup$
    – Neal Wiser
    Commented Apr 7, 2016 at 21:56
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    $\begingroup$ @NealWiser 1) yes. 2) yes. 3) yes. yes. if converting between mph and m/s is difficult, maybe start with something else than space travel? 4) yes. yes. $\endgroup$
    – SE - stop firing the good guys
    Commented Apr 7, 2016 at 22:05
  • $\begingroup$ The rotation of the earth adds a sideways velocity of 463 m/s -times the cosine of the latitude of the launch. That may affect the arrival time, and definitely affects the arrival location! $\endgroup$
    – uhoh
    Commented Apr 8, 2016 at 2:32
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    $\begingroup$ @NealWiser amazingly, if you type "2,809 m/s in mph" directly into google, it says "6283.554 miles per hour!" Google can "understand" and calculate many things. You can even plot functions! $\endgroup$
    – uhoh
    Commented Apr 8, 2016 at 2:49
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    $\begingroup$ Since the question is looking for acceleration in a space elevator, one would need to accelerate and then decelerate. Assuming you want to speed up at the same rate you slow down, accelerate half of the way up in half the time, and decelerate similarly. So, solve the same using half the time and half the distance. This yields: 0.50 g's for 30 minutes and 1.12 g's for 20 minutes. $\endgroup$
    – MattD
    Commented Apr 12, 2016 at 14:52

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