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## Main Question or Discussion Point

I've seen much about jerk, and how it's generally nearly instantaneous, and for general acceleration, that's fine. However, if I lift at a constant acceleration upward slightly stronger than gravity is pulling me downward, the gravitation pull of the Earth will offset part of my force, so that my acceleration upward is lower. Then, as my vessel rises higher, Earth's gravity will be reduced, and will offset less of my force.

For example, if I lift at 1.05G, at the surface, Earth's gravity will reduce my force by 9.80 m/s

Any help you can give with the equations would be very useful - otherwise, I'll have to resort to fair approximations, which I'd rather not do. And yes, I had calculus, but I haven't used it for several decades ... Thanks ... :-)

For example, if I lift at 1.05G, at the surface, Earth's gravity will reduce my force by 9.80 m/s

^{2}, and my resultant force will be 0.05G (0.49 m/s^{2}). However, at an altitude of 160km (low Earth orbit), Earth's gravity will be only 9.33 m/s^{2}, so my resultant force will be 0.96 m/s^{2}. At 400 km altitude (Int'l Space Station), Earth's gravity is reduced further, to only 8.68 m/s^{2}, making my resultant force 1.61 m/s^{2}. If I theoretically lift to geosynchronous orbit, 35,800 km, Earth's pull reduces to 0.22 m/s^{2}. So the difficulty I'm having is a), what is my instantaneous acceleration at specific altitudes (such as LEO or ISS altitudes, or every 1000km up to geo - and I'm pretty sure I have this equation, from sqrt(current acceleration*distance travelled)), and b) how long will it take? I can figure this for consistent acceleration, but as you see, this isn't the case here.Any help you can give with the equations would be very useful - otherwise, I'll have to resort to fair approximations, which I'd rather not do. And yes, I had calculus, but I haven't used it for several decades ... Thanks ... :-)