# 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?

• I think negligible is very ambiguous. Given enough time even a small force will accelerate an object to a non-negligible speed. I think a better way for you to define negligible in this context in in terms of time (i.e doesn't accelerate the object to some given speed), or compared to another force (e.g. the gravitational force keeping the human on whatever object they are on), or another well-defined number. – Robbie Feb 25 '16 at 5:45
• Why the downvotes, btw? – galois Feb 25 '16 at 6:47
• This doesn't make any sense, whatsoever, since the effects of gravity are independent of the mass of a (small) test object (for non-small objects we have to use the reduced mass). The "splat" time for something that starts falling from 200,000km is roughly 60 hours... so gravity is certainly not negligible. – CuriousOne Feb 25 '16 at 6:47
• Wouldn't that be more helpful to explain in an answer, than downvoting and leaving it be? I'm not quite sure what level of physics expertise you guys are expecting here – galois Feb 25 '16 at 6:51
• Down votes are not personal and I certainly try not to make them personal. They are a tool to indicate to other users of the site that a question is poorly posed or that the concepts in the body of the question should not be adopted. – CuriousOne Feb 25 '16 at 7:31

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.