Distance from Earth where gravity becomes negligible This is a basic little problem I thought up when trying to remember some physics, and I wanted to see if it's at all correct: trying to figure out the distance from earth where gravity becomes negligible...
Solving gravity for distance is trivial: $r = \sqrt{\frac{GMm}{F_g}}$
For the definition of "negligible", I just assumed that $F_g<1\mathrm{N}$ should be fine, so with that assumption, assuming also that $m=100$kg for an average sized human
$$r = \sqrt{\frac{GMm}{F_g}} > \sqrt{GMm}\approx 200,000\,\mathrm{km}$$
For the mass of a Mars-like planet, $m=6.4\mathrm e\, 23$, I get an obviously much further distance of about $1.6\mathrm e\,16$ km.
I'm not physically inclined, so do these answers make any sort of sense?
 A: Classical gravity falls off with $\frac{1}{r^2}$, so there isn't really a point where it becomes "negligible" unless you choose an arbitrary cutoff point beyond which something is "negligible". I'm not sure what you mean by $F_g < 1$, (what are the units on this?) and a better definition would be in terms of the mass-independent acceleration. For example, if you wanted to define "negligible" as exerting an acceleration of $\epsilon \frac{m}{s^2}$, then:
$$ r_\text{negligible} = \sqrt{\frac{GM}{\epsilon}}. $$
If we pick $\epsilon=0.001\frac{m}{s^2}$, then for Earth, a "negligible" radius is about $6.31\times10^8$ meters, or about $1.6$ times the orbital radius of the moon, and for Mars, this result is $2.07\times10^8$ meters. I suspect the ridiculously large result of $1.6\times10^{16}$ meters you were getting was due to a unit error, since you didn't specify the units $F_g$.
However, it doesn't really make sense to set a hard cutoff boundary for "negligible", since we could have just as easily defined $\epsilon = 0.002\frac{m}{s^2}$ or $\epsilon = 0.1\frac{m}{s^2}$. It all depends on the specific scenario you're dealing with.
