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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Gravitational force: Moon causing a tidal force on the Earth's ocean

Question:
The gravitational force between the moon and Earth creates a tidal force. From the figure, $a$ is the distance between the moon and the Earth. $M$ and $m$ are the masses of Earth and moon, respectively. $r$ denotes the radius of Earth. Find the differential tidal acceleration.

Answer:
The tidal force is obtained by the difference of gravitational fields between C (center of mass) and S (place to get tidal force). This can be associated with the differential tidal acceleration. Let us write down each gravitational acceleration.
\begin{eqnarray}
g_C &=& \frac{Gm}{a^2}  \\
g_S &=& \frac{Gm}{(a+r)^2}
\end{eqnarray}
The difference of them is the tidal acceleration.
\begin{eqnarray}
g_C -g_S &=& \frac{Gm}{a^2} - \frac{Gm}{(a+r)^2}  \\
   &=& \frac{Gm}{a^2}\left(1-\frac{a^2}{(a+r)^2}\right)  \\
   &=& \frac{Gm}{a^2}\left(1-\frac{1}{(1+\frac{r}{a})^2}\right)  \\
   &\sim& \frac{Gm}{a^2}\left(1-\left\{1-2\frac{r}{a}\right\}\right)
\end{eqnarray}
The above uses approximation. Hence, we have
\[
g_{\mathrm{tidal}} = \frac{2Gmr}{a^3}
\]

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Gravitational field of a hollow sphere

Question:
A hollow sphere has region $a<r<b$ filled with mass of uniform density $\rho$. Find the magnitude of the gravitational field between $a$ and $b$.

Answer:
Utilize Gauss's  law for gravitational fields.
\[
\int_{S} g d\alpha = -4\pi GM
\]
As we know, if there is no mass in a sphere, no gravitational field is detected. Thus, when $r<a$, $g=0$. We can also find the field when $r>b$.
\[
\int_{S} g d\alpha = -4\pi GM  \\
\rightarrow 4 \pi r^2 g = -4 \pi GM  \\
\rightarrow g = \frac{GM}{r^2}
\]
The integral of left hand side gives surface area of a sphere. For $a<r<b$, the mass, $M$, depends on the volume.
\[
M_{a-b} = \int \rho dV = \rho \int^{r}_{a}r^2 \int^{\pi}_{0}\sin \theta d\theta \int^{2\pi}_{0}d\phi \\
=\rho\frac{4\pi}{3}(r^3-a^3)
\]
From Gauss's law,
\[
-4\pi r^2 g = -4\pi G \rho\frac{4\pi}{3}(r^3-a^3)
\]
Therefore, we have the gravitational field in $a<r<b$.
\[
g = \frac{4\pi}{3}G\rho\left(r-\frac{a^3}{r^2}\right)
\]

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