3 ms·

> The sum of this infinite series is finite, but it is still large. Out of curiosity, if g was much greater, could it be infinite? What order of magnitude woul

• Avshalom 8y ago
  The ability to launch from the surface is proof of it's finiteness. As a corralary it should be finite as long as you're not inside a blackhole. It should be asymptotic up to that point. Mathematically infinite at the horizon and then completely unbounded inside it (in the sense that it converges to infinity versus not converging at all ( see a "flat" universe versus hyperbolic)).
• a1369209993 8y ago
  Yes; if the acceleration from gravity exceeds the rocket's thrust-to-mass ratio, the rocket cannot make any upward progress against gravity. The best engine listed on the Wikipedia article was the Merlin 1D with 180.1 gravities, so g ≥ 1767 meters per square second would suffice to keep it on the ground, less if you account for the fuel tanks and such.
  • darkmighty 8y ago
    You can always make a mass driver system (which is not limited by the rocket equation), but given how insanely hard it is at our own gravity, it would be more insanely difficult at a few times great g. So there's definitely a quite low maximum gravity allowing practical space access.