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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A binary star system and Kepler's law

Question:
Consider a binary star system to find the mass of the stars. The distance between them is found out to be $a$, and the period of revolution $T$. Assume that the masses of two stars are equal. Find the mass from the conditions.

Answer:
This method allows us to find the total mass of the system in laboratory by knowing the distance between stars and the period. Kepler's third law indicates the relationship between the period and distance.
\[
\frac{T^2}{a^3} = \frac{4\pi^2}{G(m_1+m_2)}
\]
where $G$ is the gravitational constant. Since $m_1+m_2 = 2m$, we can solve for the mass.
\[
m = \frac{2\pi^2 a^3}{GT^2}
\]
This gives the mass of one star.

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Galilean transformation

Question:
Physics laws cannot be altered by the observer's frames of reference. The transformation from a lab reference  frame to a moving reference frame is known as Galilean transformation. Show that Newton's laws are invariant under Galilean transformation.

Answer:
Galilean transformation subtracts the amount of displacement regarding velocity of the moving frame of reference to make the equivalent observation from the lab frame. Namely,
\[
x' = x - vt, \quad t' = t
\]
We assume that the time elapses equally for both frames; and the relative velocity is constant. Thus, the velocity becomes
\[
v' = \frac{dx'}{dt'}=\frac{d}{dt}(x-vt)=\frac{dx}{dt}-v
\]
The acceleration becomes
\[
a' = \frac{d^2x'}{dt'^2}=\frac{dv'}{dt'}=\frac{d}{dt}\left(\frac{dx}{dt}-v \right)=\frac{d^2x}{dt^2}=a
\]
Again, note that $v$ is a constant. Therefore, the Newton's equation of motion (laws) must be invariant under Galilean transformation.
\[
F'=m\frac{d^2x'}{dt'^2}=ma=F
\]
In other words, the physics law is the same from any observers.

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Gravity: Force and potential energy

Question:
Newtonian theory of gravity can be modified at short range. The potential energy between two objects is given as
\[
U(r) = -\frac{Gmm'}{r}(1-ae^{-\frac{r}{\lambda}})
\]
What is the force between $m$ and $m'$ for short distances ($r \ll \lambda$)?

Answer:
The force is conservative and calculated by the derivative of the potential energy with respect to the distance.
\begin{eqnarray}
F &=& -\frac{dU}{dr}  \\
 &=& -\frac{Gmm'}{r^2}(1-ae^{-\frac{r}{\lambda}})+\frac{Gmm'}{r}\frac{a}{\lambda}(1-ae^{-\frac{r}{\lambda}}) \\
 &=& -\frac{Gmm'}{r^2}\left(1-ae^{-\frac{r}{\lambda}}\left(1+\frac{r}{\lambda}\right)\right)
\end{eqnarray}
Since $r \ll \lambda$, $\frac{r}{\lambda}\approx 0$. Therefore, the force for short ranges is
\[
F = -\frac{Gmm'}{r^2}(1-a)
\]

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