• dalekcaan@lemm.ee
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    1 day ago

    Why are you standing two meters away from her, you weirdo? She have COVID or something?

    • felsiq@lemmy.zip
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      1 day ago

      If we set $F_moon/F_me = 1$ so the attraction is equal, her mass and the gravitational constant cancel out so we end up with $$ (M_moon / r^2_moon) / (M_me / r^2_me) = 1 $$ which we can rearrange to $$ r_me = r_moon sqrt(M_me/M_moon) $$ which is 0.011m, or 1.1cm
      (My lemmy client doesn’t show latex markdown but hopefully at least some do so this looks less dogshit)

      Edited to add: this also treats human beings like point masses, which is a less than useful approximation for close distances. The answer isn’t that you’d need to be nearly touching, it’s that your centers of mass would need to be nearly touching, ie your bodies would have to physically overlap