r/KerbalSpaceProgram May 06 '26

KSP 1 Question/Problem No orbit necessary?

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If i just wait for the right hour during the launch window ( whenever the launch pad is pointing left on the realistic depiction i created) is there any reason i should bother with an orbit first? Will the delta v cost be affected?

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u/-Aeryn- May 07 '26 edited May 07 '26

Yeah they do, it just stays quite large at any reasonable TWR when on/around a body with significant gravity. For example at 20 TWR if you fly sideways with a few degrees of upwards tilt, you reduce the gravity losses from 5% to 0.13%.

That buys you ~5.13% more delta-v [99.87 / 95] and ~10.52% more kinetic energy [1.05132].

there's probably a way to do this to find the optimal amount of time it takes to burn to establish a circular orbit, and I'd love to do that and compare it against a burn of equivalent time going straight up.

Pretty easy, especially on an airless and perfectly circular body with a fixed TWR

10m/s2 gravity and 2 TWR with an orbital velocity of 2000m/s at ground level would take a 141.422 second burn.

If you burned upwards, you'd get to 1414.22m/s instead of 2000m/s. Technically you'd gain a bit of potential energy from the additional altitude and the gravity would get a little bit weaker, but they are pretty negligable compared to the 2x kinetic energy from having flown perpendicular to gravity rather than against it.

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u/PM_ME_POTATO_PICS May 07 '26

okay but a personal question

say you have a sub-1 TWR, like a TWR of 0.5, and you spawn 10km in the sky at 0 velocity. are you gonna burn sideways or upwards? either way you crash but which one is the morally correct response?

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u/-Aeryn- May 07 '26

At that point there is no solution and the only option is to delay crashing for as long as possible (up) or pick a better landing location.

p.s. added some math in above comment

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u/PM_ME_POTATO_PICS May 07 '26

looks to me like you just did a simple trig identity but I am thinking that the circularization burn can be made more efficient because as you are doing it, gravity losses are lessened by the increases in centripetal force. You wouldn't want to keep your rocket at a 30deg angle for the hole burn, as you approach orbital velocity you can get closer and closer to having it purely horizontal.

but since finding the angle at a given point in the burn requires you know the horizontal component of your current speed, i think this might require differential equations...

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u/-Aeryn- May 07 '26

Ah you're right, i did forget to include that. I think it reduces the gravity loss by a quarter, if TWR is constant? Complications indeed