Gravitational field at height from the surface of a planet

Question:
Find the magnitude of the gravitational acceleration near the surface of a planet of radius $R$ at height, $h$ to the second order. Let $g_0$ be the gravitational acceleration at $h=0$.

Answer:
Use the universal law of gravitation.
\[
mg = \frac{GMm}{r^2}
\]
So
\[
g = \frac{GM}{r^2}
\]
The distance $r$ is the radius of the planet and the height, $r=R+h$
\[
g = \frac{GM}{(R+h)^2}
\]
We can arrange it as follows:
\begin{eqnarray}
& & g  = \frac{GM}{R^2}\frac{1}{\left(1+\frac{h}{R} \right)^2}  \\
& & g = g_0 \frac{1}{\left(1+\frac{h}{R} \right)^2}
\end{eqnarray}
Since $h \ll R$, we can use expansion, $\sum^{\infty}_{n=0}\frac{f^{(n)}(a)}{n!}(x-a)^n$. The second order of the approximation is
\[
g=g_0 \left[ 1 - 2\frac{h}{R} + 3\left(\frac{h}{R}\right)^2 \right]
\]

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