# What ratio of fuel is needed for launching a mass up to low earth orbit compared to the Lagrange point? [closed]

I've heard an argument that moon landing was at best improbable as it would have required too much more fuel than simply achieving LEO. It seems to me that it would be relatively little more. I'm just looking for a percentage of fuel burned after clearing the atmosphere by a craft headed for the Lagrange point.

To further simplify the problem, we can assume that the launch is straight up, the Moon is stationary with respect to the launch point and the final velocity upon reaching the Lagrange point is 0.

For practical purposes, I think fuel weight would have to be included in the ratio, but that stage separation can be neglected as it tremendously complicates things and the entire launch mass, minus all fuel, reaches the Lagrange point.

Bonus: To what degree does it matter whether the moon is new or full?

## closed as unclear what you're asking by Russell Borogove, Nathan Tuggy, Mark Omo, JCRM, uhohDec 14 '17 at 6:49

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• The "lunar module"?? – Organic Marble Dec 12 '17 at 21:26
• LOL. No? How about "the mass"? – Tom Russell Dec 13 '17 at 4:37
• I think comparing Saturn V with a rocket capable of bringing the CSM+LM to Earth orbit would be a good start. – Hobbes Dec 24 '17 at 17:54
• "Relative distance" is still a misconception for understanding the cost of reaching the moon, but I can take a stab at the question in this edited form. – Russell Borogove Dec 24 '17 at 18:25
• 1) Distance is pretty much irrelevant, it's the velocity change that matters. 2) Fuel use is exponential on velocity change. Thus it matters what fuel you're using. That being said, you're looking at 9.3km/sec minimum to orbit and 6.4 from there to the lunar surface (and some for safety). – Loren Pechtel Dec 24 '17 at 21:17