May somebody guide me what exactly determines the length of a launch window? When doing the (basic) celestrial mechanics caclulations for e.g. a transit from Earth to Venus, one usually assumes point-like planets, which would give more exact times when to start from Earth to reach Venus via a Hohmann transfer orbit, but what about the allowed variance around that time?


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    $\begingroup$ A good starting point are "porkchop plots" - essentially the window is defined by how much excess performance your launch vehicle has $\endgroup$ – asdfex Jan 2 at 10:24
  • $\begingroup$ @asdfex I get the point with additional fuel. I am however also wondering if the finite (non-zero) size of Venus matters somehow? I mean, I don't want to hit a point on the orbit of Venus, I want to enter an orbit around Venus which is much larger and probably easier (in terms of possibly widening the launch window), isn't it? $\endgroup$ – B--rian Jan 2 at 11:17
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    $\begingroup$ @B--rian The orbital circumference of Venus is 56000 times its diameter. In terms of determining the practical launch window, the effect of the size of the target is way down in the rounding error. $\endgroup$ – Christopher James Huff Jan 2 at 17:02
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    $\begingroup$ @B--rian when you consider Venus, you do want to hit a point in the orbit (a precise one at that). When you consider the "How do I get there?" question, it all boils down to what is the optimal launch time to hit the point in the sky that will get you to whatever transfer to Venus is being used. On either side of that optimal time will be a number of seconds where you can launch, and still have enough fuel to get there. From a planning standpoint there will be a narrow range of acceptable obits around Venus, so you do need to hit a fairly particular point to enter an acceptable orbit. $\endgroup$ – David C. Rankin Jan 3 at 18:17

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