Calculate percentage change in gravity

Question

The acceleration due to gravity $g$ is given by: $g=GM/r^2$, where $M$ is the mass of the Earth, $r$ is the distance from the center of the earth, and $G$ is a constant. Show that if we change $r$ by a small amount $Δr$, the change $Δg$ in the acceleration is approximately equal to $-2gΔr/r$. Use this approximation to compute the percentage in $g$ when you move from sea level to the top of Pike's peak, which is 4.315 km above sea level. (You'll need to know that the radius of the Earth is $6400$ km.)

I have been trying to do this for awhile but I just can't get to the $-2gΔr/r$ but I just can't. I know I am supposed to use the tangent line approximation, but something goes wrong. Any help?

Answer 1

$$g = \frac{GM}{r^2}$$

$$\frac{dg}{dr} = -2\frac{GM}{r^3}$$

$$\Delta g \approx -2 \frac{GM}{r^3} \Delta r$$

Divide by the first equation:

$$\frac{\Delta g}{g} \approx -2 \frac{r^2}{r^3} \Delta r$$

$$\therefore \Delta g \approx -2g \frac{\Delta r}{r} $$

Answer 2

Assuming you cannot use derivatives. $$ \begin{split} \Delta g &= g(r + \Delta r) - g \\ &= \frac{GM}{(r+\Delta r)^2} - \frac{GM}{r^2} \\ &= \frac{GM}{r^2} \left[ \frac{r^2}{(r+\Delta r)^2} - 1\right] \\ &= g \left[ \frac{r^2}{r^2+2r\Delta r + (\Delta r)^2} - \frac{r^2+2r\Delta r + (\Delta r)^2} {r^2+2r\Delta r + (\Delta r)^2}\right] \\ &= g \left[ \frac{-2r\Delta r - (\Delta r)^2} {r^2+2r\Delta r + (\Delta r)^2} \right] \\ &= -g \Delta r \left[ \frac{2r - \Delta r} {r^2+2r\Delta r + (\Delta r)^2} \right] \\ \end{split} $$ and since $\Delta r \to 0$, the numerator of the fraction looks like $2r$ while the denominator looks like $r^2$, so the whole thing is $$ \Delta g \approx -g \Delta r \frac{2r}{r^2} = -2g\frac{\Delta r}{r}, $$ as desired.