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I have been researching a lot and can't make any sense of it all, so I am trying my luck here.

I want to program a simple formula that gives me the travel time in space.

I have a constant thrust of let's call it 'a'. The spaceship starts to decelerate at the midpoint, 'm'.

the total distance is 'd', so m=d/2

How long does the complete voyage take and what is the speed at the midpoint and at any given time, please?

No factoring of lightspeed/relativity stuff, please. ;)

Help would be much appreciated. Thanks.

Sir Slarti
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    do you want constant acceleration, or constant thrust? These give very different results. – Hobbes Jan 17 '19 at 17:01
  • Related How fast will 1g get you there? – James Jenkins Jan 17 '19 at 17:32
  • well, first thank you all for replying. Ok, to clarify, I have constant mass in the ship and an engine that produces a constant thrust.

    Is the acceleration instant or would it build up? What would be the graph like for such a thing?

     – Sir Slarti Jan 18 '19 at 14:38
  • @SirSlarti The acceleration will be low at first, and then increase as your ship weighs less due to consuming propellant. – ikrase Mar 15 '20 at 01:01
  • For really high speeds you need to use the relativistic formulas: https://en.wikipedia.org/wiki/Space_travel_using_constant_acceleration x( t)= {c^2/a} ( sqrt(1+(at/c)^2 ) -1 ); v(t)= {at} / { sqrt(1+(at/c)^2) }; – HJC Jan 17 '21 at 16:40

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for a straight line, the speed at the midpoint is $a\frac{t}{2}$. The average speed is half that $v_{avg}=\frac{a\frac{t}{2}}2 =a\frac{t}{4}$ and the total time is $t=\frac{d}{v_{avg}}$ and so $$\begin{align} t=&\frac{d}{a\frac{t}{4}}\\ t=&\frac{4.d}{a.t}\\ t^2=&\frac{4d}{a}\\ t=&2\sqrt{\frac{d}{a}} \end{align}$$

In the real world it'll actually be orbital mechanics, and as Hobbes points out, you'll actually be dealing with constant thrust and a reducing mass giving an increasing acceleration.

The speed at an instant in time is $v_i=a\left(\frac{t}2-ABS{\left(\frac{t}2-i\right)}\right)$

  • Thanks for this. From what I have understood here is that this is not taking into account the deceleration midpoint. Right? – Sir Slarti Jan 18 '19 at 15:07
  • no, @SirSlarti, that takes it into account -- as should be clear from the maths: average speed with constant acceleration throughout the trip would be $v=\frac{a.t}2$ –  Jan 18 '19 at 18:39