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\begin{document}
\title{A2 Physics, Thermal Physics and Gravity, 3-4}
\author[1]{Noah Phipps}%
\affil[1]{Affiliation not available}%
\vspace{-1em}
\date{\today}
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\newpage
\section{Thermal Physics}
\subsection{Internal Energy}
\textbf{Internal energy is the sum of the randomly distributed kinetic energy and the potential energy of all particles in a body.}
\\Kinetic energy can be:
\begin{itemize}
\item Translational
\item Rotational
\item Vibrational
\end{itemize}
The energy of a system can be increased by doing work on it or heating it.
\begin{itemize}
\item Doing work is the energy transfer due to a force
\item Heating is a thermal energy transfer
\end{itemize}
A system can do work against an external force in some circumstances:
\begin{itemize}
\item $CO_2$ will expand rapidly if released from a high pressure container
\item Does work against the atmosphere, rapidly losing energy
\item Cools enough to solidify to dry ice
\end{itemize}
\subsubsection{The First Law of Thermodynamics}
\textit{The change in the internal energy of a system is equal to the sum of the energy transferred from or to the system by heating, and the energy transferred from or to the system as a result of work being done against or by an external force.}
\begin{itemize}
\item Kinetic energy is directly related to temperature
\end{itemize}
\subsection{Specific Heat Capacity}
\textbf{\textit{Specific heat capacity, $c$, is the energy required to raise the temperature of 1kg of a material by 1K without any change of state, measured in $Jkg^{-1}K^{-1}$}}
\begin{equation*}
Q=mc\Delta\theta
\end{equation*}
$c_{\text{water}}=4190Jkg^{-1}K^{-1}$; a fairly large value. Hence, a large body of water will take alot of energy to heat.
\subsubsection{Determining $c$ Electrically}
\begin{itemize}
\item Immersion heater in a block of a metal
\item Time and temperature change recorded
\item $Q$ calculated as the energy transferred by the heater, using its current, voltage and power
\end{itemize}
\subsubsection{Determining $c$ by the Method of Mixtures}
\begin{itemize}
\item Water boiled to $100^\circ C$ with object
\item Object placed in cool water
\item Temperature changes equated
\end{itemize}
\subsubsection{Increasing Interal Energy by doing Work}
If an object is dropped, and hits the ground without rebounding, all $GPE$ will have been transferred to $KE$, and then to internal energy. This leads on to:
\subsubsection{Determining $c$ by considering Work}
\begin{itemize}
\item Lead shot in a tube
\item Length of tube and temperature of lead recorded
\item Tube repeatedly turned over, moving lead through length
\item Total $GPE$ changes calculated, and equated to new temperature of lead
\item Gives a very rough estimate
\end{itemize}
\subsubsection{Flow Calculations}
Water is good for storing energy because of its high $c$, so it is used as a heat transfer liquid. With flow calculations, it is often used to use:
\begin{gather*}
\text{rate}=\frac{\Delta v}{\Delta t}=\frac{\mathrm{d} v}{\mathrm{d} t}\\
E=pt\\
\rho=\frac{m}{v}\\
\end{gather*}
\subsection{Specific Latent Heat}
When a substance changes state, work has to be done to break the intermolecular bonds. Whilst this is happening, potential energy will increase whilst kinetic energy remaincs constant. Work often has to be done against the surroundings if an object is expanding.
\subsubsection{Specific Latent Heat of Vaporisation}
\textbf{The energy required to change 1kg of a liquid into 1kg of a gas with no change in temperature.}
\subsubsection{Specific Latent Heat of Fusion}
\textbf{The energy required to change 1kg of a liquid into 1kg of a solid with no change in temperature.}
For a substance of mass $m$, and specific latent heat of fusion $l$, the energy transferred is given by
\begin{align*}
Q=ml,
\therefore l=\frac{Q}{m}.
\end{align*}
\subsubsection{Heating}
\begin{itemize}
\item Energy is supplied at a constant rate
\item Temperature will rise till melting point
\item Substance will melt at constant temperature
\item Temperature of liquid will rise once all melted
\item Liquid will boil at constant temperature
\item Temperature of gas will rise once all melted
\item Process can be recorded with a data logger
\end{itemize}
\subsubsection{Cooling}
Cooling will not take place at a constant rate, as it is dependent on the temperature of the surroundings. Thermal energy will be \textit{dissipated} from a substance that is cooling.
\subsection{Gas Laws}
\subsubsection{Boyle's Law}
\textbf{The volume of a fixed mass of gas at constant temperature is inversely proportional to its pressure.}
\begin{gather*}
pV=\text{constant}\\
p_1V_1=p_2V_2
\end{gather*}
\subsubsection{Charles's Law}
\textbf{The volume of a fixed mass of gas at constant pressure is directly proportional to its kelvin temperature.}
\begin{gather*}
\frac{V}{T}=\text{constant}\\
\frac{V_1}{T_1}=\frac{V_2}{T_2}
\end{gather*}
\begin{itemize}
\item Gives rise to isotherms on a pressure volume gas, each representing a different temperature
\end{itemize}
\subsubsection{The Pressure Law}
\textbf{For a fixed mass of gas at constant volume, the pressure is directly proportional to the kelvin temperature.}
\begin{gather*}
\frac{p}{T}=\text{constant}\\
\frac{p_1}{T_1}=\frac{p_2}{T_2}
\end{gather*}
\subsubsection{Gas Laws}
\textbf{Avogradro's law} states that in constant conditions, $V\propto N$, where $N$ is the number of molecules.
\begin{itemize}
\item All of above are empirical laws
\end{itemize}
\newpage
\subsection{The Ideal Gas}
Makes several assumptions:
\begin{itemize}
\item Gases have a large number of identical molecules
\item Molecules themselves have a negligible volume
\item Molecules possess no internal potential energy
\item All molecules collide perfectly elastically
\item Molecules move randomly, and in straight lines
\item Molecules exert no forces on each other except in collisions
\item Molecules spend significant more time travelling than colliding
\item Newtons laws of motions can be applied to the molecules
\item These are reasonable because:
\begin{itemize}
\item Brownian motion provides evidence for random motion
\item Gases can be compressed, showing the molecules are far apart
\item Gas molecules are not observed to lose temperature or slow, so collisions must be elastic
\end{itemize}
\end{itemize}
Combining the empirical gas laws,
\begin{equation*}
pV=NkT
\end{equation*}
where $k=1.38\times 10^{-23}JK^{-1}$ and is the \textbf{Boltzmann constant}. Hence,
\begin{equation*}
pV=nRT
\end{equation*}
where $R=N_Ak=8.31JK^{-1}mol^{-1}$, and is the \textbf{molar gas constant}. $n$ is the number of moles, and\\
\textbf{1 mole is defined as the number of atoms in 12 gramps of carbon-12, equal to $6.02\times 10^{23}$}\\
Hence,
\begin{equation*}
\frac{p_1V_1}{T_1}=\frac{p_2V_2}{T_2}
\end{equation*}
\subsubsection{Work Done by an Expanding Gas}
A gas will do work whilst expanding, unless in a vacuum.
\begin{equation*}
W=p\Delta V
\end{equation*}
\subsection{The Molecular Kinetic Theory Model}
\textbf{Brownian motion} is the rapid, random movement of small particles.
The kinetic molecular theory model states that \textbf{an increase in temperature results in an increase in the average speed of the gas molecules}.
\subsubsection{Deriving Molecular Kinetic Energy}
Considering $N$ molecules and their kinetic energies,
\begin{gather*}
KE_{total}=\frac{1}{2}mc_1^2+\frac{1}{2}mc_2^2+\frac{1}{2}mc_N^2...\\
\therefore KE_{molecule}=\frac{\frac{1}{2}mc_1^2+\frac{1}{2}mc_2^2+\frac{1}{2}mc_N^2...}{N}\\
\therefore =\frac{\frac{1}{2}m(c_1^2+c_2^2+c_N^2...)}{N}\\
\text{The mean square speed, $(c_{rms})^2$, is defined as } (c_{rms})=\frac{(c_1^2+c_2^2+c_N^2...)}{N}\\
\therefore KE_{molecule}=\frac{1}{2}m(c_{rms})^2
\end{gather*}
For a cube of sides $x$, $y$ and $z$, length $L$, each molecule will have 3 components of its velocity. Hence,
\begin{gather*}
c_1^2=u_1^2+v_1^2+w_1^2\\
\text{For a molecule hitting the right hand wall, $\Delta\text{momentum}=mu_1-(-mu_1)=2mu_1$}\\
\text{The average time between these collisions is $\frac{2L}{u_1}$, giving $\frac{u_1}{2L}$ collisions per second}\\
\text{Using N2L, }F=\frac{mu_1^2}{L}\\
\text{Summing forces, }F_{total}=\frac{m}{L}(u_1^2+u_2^2+U_N^2...)\\
\text{Using $u_{rms}^2$, }F_{total}=\frac{Nm(u_{rms})^2}{L}\\
\text{Since pressure $=\frac{F}{A}$ and the area is $L^2$, with the volume being $L^3$, }p=\frac{Nm(u_{rms})^2}{V}\\
(c_{rms})^2=(u_{rms})^2+(v_{rms})^2+(w_{rms})^2\\
\text{Assuming random motion and distribution, } (u_{rms})^2=(v_{rms})^2=(w_{rms})^2\\
\therefore (u_{rms})^2=\frac{1}{3}(c_{rms})^2\\
\end{gather*}
Finally,
\begin{equation*}
pV=\frac{1}{3}Nm(c_{rms})^2 \tag{Usual form of the kinetic theory equation}
\end{equation*}
\subsubsection{Dependence of Molecular Kinetic Energy on Temperature}
\begin{itemize}
\item Temperature rise increases average speed of molecules
\item Equating the ideal gas and kinetic theory equation gives:
\end{itemize}
\begin{gather*}
\frac{1}{3}Nm(c_{rms})^2=NkT\\
m(c_{rms})^2=3NkT\\
\text{Hence, the average kinetic energy is given by:}\\
\frac{1}{2}m(c_{rms})^2=\frac{3NkT}{2}\\
\end{gather*}
This shows that the average molecular kinetic energy is directly proportional to the kelvin temperature.\\
Since $k=\frac{R}{N_A}$,
\begin{equation*}
\frac{1}{2}m(c_{rms})^2=\frac{3RT}{2N_A}
\end{equation*}
\newpage
\section{Gravity}
\subsection{Newton's Law of Gravity}
\begin{equation*}
\text{Gravitational Force}\propto \frac{1}{\text{Distance}^2}
\end{equation*}
\begin{itemize}
\item Attractive force between all masses
\item Follows an inverse square law, as determined by Newton
\item \textbf{Newton's Law of Universal Gravitation -} \textit{Any two point masses attract each other with a force, $F$, this is directly proportionate to the product of their masses, $m_1 m_2$, and inversely proportionate to the square of their separation, $r^2$.}
\item Point mass refers to a spherically symmetrical object, with a uniform density
\end{itemize}
\begin{equation*}
F=\frac{GMm}{r^2},
\end{equation*}
where $G=6.67\times 10^{-11}\text{Nm}^2\text{kg}^{-2}$, with $G$ being the gravitational constant.
\subsubsection{Calculation of G}
\begin{itemize}
\item Small lead spheres attached to long rod, suspended from a fibre
\item Much larger spheres brought close to each smaller one
\item Suspended rod oscillates due to attraction
\item Force between spheres then calculable
\end{itemize}
\subsection{Gravitational Fields}
Fields are used to describe non-contact interactions. A \textit{source} creates a field in space. Other objects then experience a force, based on the strength of this field, which is represented by a vector.\\\\The mass of an object creates a \textit{gravitational field} around itself. The \textit{gravitational field strength} is the gravitational field strength per unit mass at a given point;
\begin{equation*}
g=\frac{F}{m}=\frac{GM}{r^2}
\end{equation*}
A gravitational field can be represented by vector lines. The field around a spherical mass is \textit{radial}, although the field close to the surface of the earth is taken to be uniform. It is not dependent upon the mass of the object experiencing a force.
\\\\
Between any two objects, there is a neutral point where the gravitational attraction of both will cancel out - such as between the Earth and Moon. This is given by
\begin{equation*}
\frac{GM_1}{r_1^2}=\frac{GM_2}{r_2^2}.
\end{equation*}
\subsection{Gravitational Potential}
GPE is the energy of an object due to its position in a gravitational field. Over a large distance, $g$ is not constant. The work done to move a mass from a distance $r$ to an infinite distance is given by
\begin{equation*}
W=\frac{GMm}{r}.
\end{equation*}
Therefore, the gravitational potential $v$ at a point is defined as \textit{the work done per unit mass to move a mass from infinity to that point.} As the mass is being moved into the field, this is given by
\begin{equation*}V=-\frac{GM}{r}.
\end{equation*}
As $r \to \infty$, $V \to 0$.\\
As $r \to 0$, $V \to -\infty$.\\
The work done to move a mass from a given point to another is given by
\begin{equation*}
\Delta W = m \Delta V
\end{equation*}
\subsubsection{Equipotentials}
At specific distances from the earth, the gravitational potential is equal. Therefore, surfaces of equal potential can represent a gravitational field - equipotentials with equal intervals can be plotted on a diagram. The work required to move from one line to another is equal, regardless of which pair of adjacent lines are used. When plotting a graph of distance against field strength, \textit{the gradient is equal to $-g$.($g=-\frac{\Delta V}{\Delta r}$} Furthermore, the area under a graph of $g$ against $r$ is equal to $\Delta V$.
\subsection{Orbots of Planets, Moons and Satellites}
The force of gravity on a stellar object can be equated to the centripetal force giving:
\begin{equation*}
F=mr\omega^2 =\frac{mv^2}{r},
\end{equation*}
where $\omega= \frac{2\pi}{T}$, the angular velocity. This can be equated to Netwon's law of gravity to find the mass of the mass of the central object.
\subsubsection{Kepler's Laws}
\begin{enumerate}
\item \textit{The Law of Orbits -} All planets move in elliptical orbits, with the sun at one focus
\item \textit{The Law of Areas -} A line that connects a planet to the sun will sweep out equal areas in equal times
\item \textit{The Law of Periods -} The square of the period of any planet is proportional to the cube of the semimajor axis of its orbit (The semimajor axis is taken to be the radius, $r$
\end{enumerate}
The law of periods gives the relationship
\begin{equation*}
T^2 \propto r^3.
\end{equation*}
By equation Netwon's law of gravitation and the equation for centripetal force, the above relationship can be verified;
\begin{equation*}
T^2 = \frac{4\pi^2}{Gm}r^3.
\end{equation*}
\subsubsection{Satellites}
The orbital speed of a satellite is given by the equation
\begin{equation*}
v = \sqrt{\frac{GM}{r}},
\end{equation*}
showing that lower orbit satellites require higher orbital velocities.\\\\
\textbf{Synchronous Orbits}
\begin{itemize}
\item Time period equal to rotational period of body being orbited
\item Geosynchronous if around Earth
\item Geostationary if in same direction as Earths rotation
\item Useful for communication satellites, GPS
\end{itemize}
\textbf{Polar Orbits}
\begin{itemize}
\item Orbit passes above both poles in one period
\item Plane of orbit perpendicular to equatorial plane
\item Typically lower orbit than geostationary satellites
\item Useful for surveillance, monitoring climate, oil spills etc.
\end{itemize}
\subsubsection{Energy of an Orbiting Satellite}
The total energy of a satellite is the sum of its kinetic and graviational potential energies; $E_t = E_p + E_k$.
\begin{gather*}
E_p = mV = -\frac{GMm}{r} \\
\frac{mv^2}{r} = \frac{GMm}{r^2} \\
\therefore E_k = \frac{1}{2} mv^2 = \frac{GMm}{2r}\\
\therefore E_t = -\frac{GMm}{2r} \\
\end{gather*}
The negative energy is due to the value of zero for gravitational potential being defined as infinity.
\begin{itemize}
\item The Earth's atmosphere can affect satellites
\item Air resistance causes satellite to fall into lower orbit
\item $E_k$ increases while $E_t$ decreases
\item Velocity increases
\item Heat increases, satellite burns up
\end{itemize}
\newpage
\subsubsection{Escape Velocity}
\begin{itemize}
\item Velocity at which an object must be travelling to escape the gravitational field of an astronomical body
\item Equal to the work done to move the object from a radius $R$ to infinity
\end{itemize}
\begin{gather*}
\Delta W = m \Delta V = m\frac{GM}{R} \\
\text{If this is equal to the kinetic energy, then;}\\
\frac{1}{2} mv^2 = m\frac{GM}{R}\\
v=\sqrt{\frac{2GM}{R}}
\end{gather*}
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