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Comment by umeshunni | Hacker News Reader

That's 14 kilometers for those wondering.

Thanks, I initially saw the article title as 'Apollo 10 Stopped just 47 feet from the Moon' and I thought, gee that's rather strange, why would you do that. Also, how would you do that. Thanks for saving me from embarrassment. [edit] I always read the comments before the article.

No atmosphere on the moon, technically you could orbit it at 47 feet. Just hope you miss any mountains and large boulders.

Even assuming no obstacles, gravitational perturbations due to irregularities in the distribution of the moon's mass would upset the orbit and cause a crash.

Nevertheless, there's a short story from the SF golden age exactly featuring a satellite orbiting a few feet off the ground. Including perfectly drilled tunnels in the elevations ...

Ah, here we go - "The Holes Around Mars" by Jerome Bixby: https://scifi.stackexchange.com/questions/143541/short-story...

You couldn't get into a circular orbit that was always 47 feet from the surface. That's impossible anyway, because a great circle route across the moon's surface is not a perfect circle, it dips down into craters and up over mountains. But could you achieve a stable elliptical orbit whose lowest approach to the surface was 47 feet? It sounds plausible to me, if not particularly safe.

Actually it's hard to get a stable orbit around the moon at all. See https://en.wikipedia.org/wiki/Lunar_orbit#Perturbation_effec...

I’m wondering if Apollo mission was actually using feet or meters for calculations.

https://github.com/chrislgarry/Apollo-11/blob/master/Luminar...

Code seems to use SI, but imperial in comments.

Wikipedia says: "Although data was stored internally in metric units, they were displayed as United States customary units."

https://en.wikipedia.org/wiki/Apollo_Guidance_Computer#DSKY_...

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