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:

\[  g  = \frac{GM}{R^2}\frac{1}{\left(1+\frac{h}{R} \right)^2}  \]

\[  g = g_0 \frac{1}{\left(1+\frac{h}{R} \right)^2}   \]

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

\[  g=g_0 \left[ 1 – 2\frac{h}{R} + 3\left(\frac{h}{R}\right)^2 \right]  \]