# Range of Earth's gravitational field

We know that the acceleration due to gravity acting on a body situated h meter away from the surface of the earth is given by,

$$g' = (1 - 2h/r)g,$$ where $$\,r$$ = Radius of the earth ($$R$$) + $$h$$. Now we can find the range of the earth's gravitational field, i.e, where does the value of g' become $$0$$.

Now, $$g'$$ becomes $$0\,$$ if $$\,h = R/ 2$$. Radius of the earth = $$6.37\cdot10^6$$ m, so $$R/ 2 = 3.18\cdot10^6$$ m. Therefore, at a distance $$3.18\cdot10^6$$ m distance away from the surface of the earth, the gravitational field of the earth becomes zero or the value of $$g$$ becomes zero. Is my math correct? If not, why?

Range of the gravitational field of any body, not just the earth, extends upto infinity. The formula you are using is an approximation that only applies when $$h< which is definitely not the case when $$h=R/2$$.
The more general formula for $$g$$ is $$g = GM_e/r^2$$ where $$G$$ is the universal gravitational constant, $$M_e$$ is the mass of the earth and $$r$$ is the distance of the object from the centre of the earth. As you can see, the value of $$g$$ gets smaller and smaller but never zero as $$r$$ gets larger and larger. If we now, substitute $$r = R+h$$ where $$r$$ is the radius of the earth, and $$h$$ is the distance from the surface of the earth, then apply the first order approximation for $$h< then we get the formula you used.
Think about how you got the formula. We had, $$g$$=$$GM\over R^2$$ and $$g\prime$$=$$GM\over (R+h)^2$$ . Now from them we get, $$g\prime$$=$$R^2\over (R+h)^2\times g$$ $$\implies$$ $$g\prime$$=$$g\over(1+h/R)^2$$ From here, by an approximation, $$h< we get your formula.
But if you want to get $$g\prime$$=0, the denominator must go to infinity, which implies $$h$$ goes to infinity. Which is a basic assumption of the gravitational field.