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\begin{document}


\begin{center}
{\Large{\bf Astrophysical Fluid Dynamics -- Problem Set 1}}

%{\large{\bf Due in class Monday Jan 29}}

\end{center}

\vspace{0.1in}
\ni
Readings: Shu pages 3--6; Tritton 5.1--5.5; White 4.2, 4.3, 4.5; Pringle \& King Sections 1 through 1.4, and 1.7

\vspace{0.1in}
\ni
{\large{\bf Problem 1.}}
Stupid Human Tricks

\ni In grade school\footnote{And perhaps beyond} my friends and
I\footnote{Or maybe it was just me}
used to suck soda up a straw, cap the top end of the straw with
a finger, and watch with amazement as the soda stayed in the straw,
even when we removed the straw from the glass.

\ni Explain this phenomenon. What exactly is pushing against the bottom of the
straw to keep the soda from falling? What is the tallest straw with which
you can perform this trick?

\vspace{0.1in}
\ni
{\large{\bf Problem 2.}}
Tip of the Iceberg

\ni An iceberg is sighted with volume $V$ above the water's surface.
What volume of ice must reside below the water's surface?
Take the density of berg ice, which is composed of frozen freshwater and
full of bubbles, to be
$\rho_{\rm ice} = 0.9 \gm \cm^{-3}$, and the density
of seawater to be $\rho_{\rm sea} = 1.03 \gm \cm^{-3}$.

Try not to merely invoke Archimedes' principle, but instead try to derive the answer from hydrostatic equilibrium.


\vspace{0.1in}
\ni
{\large{\bf Problem 3.}}
Scaling the Heights

\ni (a) Write down an expression for the variation of gas density
$\rho$ (units of mass per volume)
with height $z$ above the Earth's surface, assuming the
gas is at constant temperature $T$ and in hydrostatic equilibrium.
Take the gas to be made of a single
kind of molecule of weight $\mu$ (units of mass). Express
in terms of the density at ground level $\rho_0 \equiv \rho(z=0)$
and the {\it density scale height} $h \equiv kT/(\mu g)$, where
$g$ is the gravitational acceleration. Neglect variations of $g$ with $z$.

\ni (b) Utterly neglecting collisions with other molecules, what
is the maximum height a molecule would attain if launched from
$z=0$ with a typical thermal velocity? Is this close to $h$?

\ni This calculation is 
misleading insofar as it ignores
collisions between molecules, which are crucial for describing air
as a continuum fluid. Nevertheless, it provides a mnemonic for
remembering the scale height, and it illustrates the sometimes surprisingly
close connection between fluid mechanics and particle mechanics
(kinetic theory).

\ni (c) At what height $z$ would you expect the formula in (a),
which depends on the continuum hypothesis, to fail? This height
marks the location of the {\it exobase} in planetary atmospheres.
Just consider how intermolecular collisions validate the continuum approximation. 

\ni (d) Write down an expression for the hydrostatic variation of gas density
$\rho$ with height $z$ above the midplane of a circumstellar disk
at radius $r$. As in $(a)$, assume constant $T$ and $\mu$.
Take the gravitational field to be that from the star alone
(ignore the self-gravity of the disk). Work in the limit that
$z \ll r$. Express in terms of the density at the midplane
$\rho_0$ and the density scale height $h \equiv (kT/\mu)^{1/2} \Omega^{-1}$,
where $\Omega$ is the Keplerian orbital angular frequency.

\ni The height $h$ is often written $h = c_s / \Omega$, where
$c_s = (kT/\mu)^{1/2}$ is the speed of sound waves
in gas that behaves isothermally [by behaving isothermally,
we mean that the gas keeps the same temperature regardless of
how it is compressed or expanded by the sound wave. This
problem does not concern a sound wave {\it per se}, but
people talk about the sound speed anyway because
sound waves (pressure disturbances) are the means by which a medium
establishes hydrostatic equilibrium.]

\ni (e) As in (b), calculate the maximum height that a gas molecule
would attain if launched upwards from the midplane with a typical thermal
speed, ignoring collisions. Compare to $h$.

\ni (f) For the disk to be geometrically ``thin'' ($h/r \ll 1$),
what must be true about $c_s/v_{\rm K}$? Here $v_{\rm K}$
is the Keplerian orbital velocity.

\vspace{0.1in}
\ni
{\large{\bf OPTIONAL Problem 4.}}
To $d/dt$ or Not to $d/dt$

\ni Give a physical explanation for why the left-hand side of the momentum 
equation reads $\rho d u / dt$ and not $d(\rho u)/dt$, where $u$ is 
velocity, $\rho$ is density, and $d/dt$ is the convective derivative.
What is being assumed about the fluid parcel in the Lagrangian view?


\end{document}




