Range of Earth's gravitational field

Question

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?

Answer 1

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 $h<<R$ 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 $h<<R$ then we get the formula you used.

Answer 2

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, $h<<R$ 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.