Gokul43201
Staff Emeritus
Science Advisor
Gold Member
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I'm not forgetting that. That's why I said, "without doing the integral" - that is to say that I'm eyeballing the solution to the differential equation. But even if you assume the final propellant mass is zero, and do the integral, will you be able to cover that order of magnitude shortfall? I doubt it. Include the extra energy needed to overcome drag and the extra mass from the rocket, and it looks even worse than that.signerror said:This is invalid: you're forgetting that propellant is removed throughout the burn time of a rocket - the mass reaching LEO is much smaller than what is launched.
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