A planet orbits a star in an elliptical orbit: Kepler's second law

Question:
A planet orbits a star in an elliptical orbit. The distance at aphelion is $2a$ and the distance at perihelion is $a$. Find the ratio of the planets speed at perihelion to that at aphelion.

Answer:
According to Kepler's second law, the area swept out per unit time by a radius from the star to a planet is constant. The area can be expressed by
\[
dA = \frac{1}{2}r rd\theta
\]
Take the derivative in terms of time.
\[
\frac{dA}{dt}=\frac{1}{2}r^2 \frac{d\theta}{dt}
\]
$\frac{d\theta}{dt}$ is the angular velocity, $\omega$. We know angular momentum $L=mr^2\omega$. Thus,
\begin{eqnarray}
\frac{dA}{dt} &=& \frac{1}{2}r^2\omega  \\
  &=& \frac{L}{2m}
\end{eqnarray}
This shows that the equal area per unit time indicates constant angular momentum, $L$. We get back to the original expression of the angular momentum, $L=mr^2\omega$. We also know that $v=r\omega$ and $v$ is the tangential speed, so $L=mvr$. Compare the angular momenta at perihelion and aphelion.
\[
L = mv_{\mathrm{p}}a = mv_{\mathrm{a}}2a
\]
Therefore the ratio of the speeds is
\[
v_{\mathrm{p}}:v_{\mathrm{a}}=2:1
\]
We can state that Kepler's second law is essentially equal to the conservation of angular momentum. The more radius, the slower the planet has. The less radius, the faster the planet gets.

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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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Conservation of energy and spring and pendulum bob

Question:
A pendulum bob is released from the height of $h$ to hit a spring that creates the force $F=-kx-bx^3$ in terms of the displacement. If the pendulum has mass $m$, find the compression displacement of the spring.

Answer:
Find the potential energy of spring. Since it is a conservative force, we integrate it in terms of displacement.
\[
U = -\int F dx = \int kx+bx^3 dx = \frac{1}{2}kx^2+\frac{1}{4}bx^4
\]
The potential energy of the bob is $mgh$. This can be transferred into the spring energy, so
\[
mgh =  \frac{1}{2}kx^2+\frac{1}{4}bx^4
\]
Rearrange it to solve for $x$:
\begin{eqnarray}
& & x^4 +\frac{2k}{b}x^2 = \frac{4mgh}{b}  \\
& & \left(x^2+\frac{k}{b}\right)^2 - \frac{k^2}{b^2} = \frac{4mgh}{b}  \\
& & x^2 + \frac{k}{b} = \sqrt{\frac{4mgh}{b}+\frac{k^2}{b^2}}  \\
& & x  = \sqrt{\sqrt{\frac{4mgh}{b}+\frac{k^2}{b^2}} -\frac{k}{b}}
\end{eqnarray}

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Equation for a linear line: Knowing the coordinates of two points

Question:
A line goes through points (-3, 9) and (2, 4). Find the equation of the linear line.

Answer:
There are two methods. You can use a line equation with two unknowns:
\[
y = ax + b
\]
where $a$ is the slope and $b$ is the intercept. Plug in the coordinates to solve for simultaneous equations.
\begin{eqnarray}
9 &=& -3a + b \\
4 &=& 2a + b
\end{eqnarray}
(1)-(2) gives $5 = -5a$. Thus, $a=-1$. Plug back in either equation and we get $b=6$. The line equation is
\[
y = -x + 6
\]
The other method is to utilize following formula:
\[
(y-y_1)=\frac{y_2-y_1}{x_2-x_1}(x-x_1)
\]
The factor, $\frac{y_2-y_1}{x_2-x_1}$, corresponds to the slope. After plugging in the coordinates, we can obtain the same result.

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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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Radio active decay: Half life

Question: 
Find the half life for the process of radioactive decay. Assume that $\lambda$ is the decay constant.

Answer:
The radioactive decay is understood as the comparison of the numbers of nucleus at time $t$ and time $t+\Delta t$. The difference of the numbers can be expressed as $N(t)-N(t+\Delta t)$, and this is proportional to a rate, $\lambda$.
\[
N(t)-N(t+\Delta t) = -\lambda N(t) \Delta t
\]
The right hand side indicates how the number decreases for $\Delta t$. We can rearrange it as
\[
\frac{N(t)-N(t+\Delta t)}{\Delta t}=\frac{dN}{dt}=-\lambda N(t)
\]
This is a first order of differential equation with respect to time. Again, rearrange and integrate it as follows:
\[
\int^N_{N_0} \frac{dN}{N}=\int^{t}_{0} -\lambda dt
\]
This gives
\[
\ln \frac{N}{N_0} = -\lambda t \quad \rightarrow \quad N=N_0e^{-\lambda t}
\]
In order to find the half life, let $N=\frac{N_0}{2}$. Then derive the time, $t_{0.5}$.
\begin{eqnarray*}
& & \frac{N_0}{2}=N_0e^{-\lambda t_{0.5}}  \\
\rightarrow & & \frac{1}{2} = e^{-\lambda t_{0.5}}  \\
\rightarrow & & 2 = e^{\lambda t_{0.5}}  \\
\rightarrow & & \lambda t_{0.5} = \ln 2  \\
\rightarrow & & t_{0.5} = \frac{\ln 2}{\lambda}
\end{eqnarray*}
Thus, the half life is given by $\frac{0.693}{\lambda}$.

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