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Chemical equilibrium imposes ${\mu }_e+{\mu }_p={\mu }_H+{\mu }_{\gamma }$. For blackbody photons, ${\mu }_{\gamma }=0$. Define the positive Hydrogen binding energy by $I=(m_e+m_p-m_H)c^2$, approximately $13.6\mathrm{eV}$. Multiplying the two Maxwell-Boltzmann distribution number densities and dividing by the neutral-hydrogen density givesUse $g_e{g}_p/g_H=1$ and $m_p/m_H\simeq 1$ to obtain the Saha ionization equationThe assumptions are nonrelativistic, dilute, nondegenerate species at a common temperature, ground-state hydrogen with the stated degeneracy factors, negligible mass correction in the prefactor, and sufficiently rapid reactions to maintain chemical equilibrium.

Apply part (a) first to the position $\mathbf{r}$:Applying it again to both terms on the right giveswhere $(\dot{\mathbf{\omega }}{)}_S=(\dot{\mathbf{\omega }}{)}_{S^{\prime }}$ because $\mathbf{\omega }\times \mathbf{\omega }=0$. Substituting the first formula into the last term produces

In the Earth-fixed rotating reference frame, the sideways apparent force is the Coriolis acceleration multiplied by the ski-doo's mass:At the South Pole, $\mathbf{\omega }$ is vertical and antiparallel to the local outward normal, while $\mathbf{v}$ is horizontal, soThe force points to the left of the direction of travel, which is westward when the ski-doo is moving north after crossing the pole.

An isolated collapse emits transient matter and gravitational waves. The exterior is expected to settle to an asymptotically flat stationary spacetime. Because the star is uncharged, the black-hole uniqueness theorem identifies the regular final state with a Kerr black hole. Its higher multipole moments are fixed by its conserved charges: the black-hole no-hair theorem leaves only the total mass $M$ and angular momentum $J$. Thus the late-time spacetime is characterized by $M$ and $J$.

Let $M(t)=m_0-\alpha t$ be the trolley's remaining mass. During $dt$, the expelled mass $-dM$ has ground-frame velocity $v+u$. Conservation of momentum gives, to first order,and hence $Mdv=udM$. Integration yields the rocket equationAt the stopping time, $M=m_0{e}^{-v_0/u}$. Therefore the time spent ejecting gas is

Assume $\rho$ is mass per unit equilibrium length, the tension has constant magnitude $T$, the string has negligible bending stiffness, and there is no damping or external transverse force. Assume small slopes $|y_x|\ll 1$ and purely transverse motion. Then a short segment has mass $\rho \Delta x$ to first order. If $\alpha$ is the angle of its tangent, the small-slope approximation gives $\sin \alpha \simeq \tan \alpha =y_x$.

The vertical components of tension at the two ends give net transverse force $T[y_x(x+\Delta x,t)-y_x(x,t)]$. Newton's second law therefore givesto first order. Dividing by $\Delta x$ and taking its limit to zero yieldsThus $c$ is the propagation speed of small transverse waves.

For a volume source of constant mass density $\rho$, use $dm=\rho d{V}^{\prime }$. Its Newtonian gravitational potential is the volume integralThe source and observation position vectors are ${\mathbf{r}}^{\prime }$ and $\mathbf{r}$, respectively; $G$ is the gravitational constant. With this sign convention positive source mass gives negative Newtonian gravitational potential.

For the surface, a source element has mass $dm=\sigma d{S}^{\prime }$. Superposing its Newtonian gravitational potential gives the surface integralHere $d{S}^{\prime }$ is the source surface area element, ${\mathbf{r}}^{\prime }$ the source position vector, and $G$ the gravitational constant. The same zero-at-infinity convention applies.

Use the convention that the Newtonian gravitational potential tends to zero at infinity and the gravitational field is $-\nabla \varphi$. A source element of mass $dm$ at position vector ${\mathbf{r}}^{\prime }$ contributes $-Gdm/|\mathbf{r}-{\mathbf{r}}^{\prime }|$, where $G$ is the gravitational constant. For the curve, $dm=\gamma d{s}^{\prime }$, soHere $d{s}^{\prime }$ is arc length along the source curve; the expression is finite away from it. This normalization fixes the otherwise arbitrary additive constant.

Use source spherical coordinates with radius $s$, polar angle $\alpha$ measured from the positive $z$-axis, and azimuth $\theta$. The lower half of the spherical shell has $a\le s\le 2a$, $\pi /2\le \alpha \le \pi$, and $0\le \theta \le 2\pi$. Its volume element is $s^2\sin \alpha dsd\alpha d\theta$. At the observation point $(0,0,h)$, $h>0$, the separation is $\sqrt{s^2+h^2-2sh\cos \alpha }$, soThe angular integral is elementary: put $q=\cos \alpha$ to obtainWe used $\sqrt{(s+h{)}^2}=s+h$ since both $s,h$ are positive. HenceThus the axial potential of a uniform hemispherical shell, at every point of the positive $z$-axis, isThere is no need to split into $h<a$, $a<h<2a$, and $h>2a$: the observation point is outside the lower hemisphere in all three cases. As checks, $\varphi (h)\to -3\pi G\rho a^2$ as $h\downarrow 0$, and at large $h$,The last factor is exactly the total mass of the lower half of the spherical shell.

Use source polar coordinates $x^{\prime }=s\cos \theta$, $y^{\prime }=s\sin \theta$ on the semicircular disk, with $0\le s\le a$ and $0\le \theta \le \pi$. Its surface area element is $sdsd\theta$, and the separation from $(0,0,h)$ is $\sqrt{s^2+h^2}$. ThusThis Newtonian gravitational potential is half that of the full disk because the separation is independent of the polar angle. At large $h$ it approaches $-G(\pi a^2\sigma /2)/h$, with the correct total mass.

Write the observation point as $(0,0,h)$ with $h>0$, reserving source coordinates for the integral. For the straight segment, $d{s}^{\prime }=d{x}^{\prime }$ and the separation is $\sqrt{x^{\prime 2}+h^2}$. Therefore its Newtonian gravitational potential isEquivalently $\varphi (h)=-2G\gamma \mathrm{arsinh}(a/h)$. At large $h$ it approaches $-G(2a\gamma )/h$, the Newtonian potential of a point mass with the segment's total mass.

Write $\mu =GM$. In polar coordinates, Newton's second law per unit mass gives the radial and transverse equations of motionThe transverse equation impliesThis is conservation of angular momentum per unit mass. For the nonradial trajectory in the next part, choose orientation so $h>0$. With $u=1/r$ and primes denoting derivatives with respect to $\theta$, the chain rule givesSubstituting into the radial equation yields $-h^2{u}^2(u^{\prime \prime }+u)=-\mu u^2$, henceThe change of independent variable requires $h\ne 0$; a purely radial trajectory instead has constant $\theta$.

Far from the planet, the magnitude of angular momentum per unit mass is the magnitude of the cross product of position vector and velocity. The perpendicular distance from the incoming asymptote is the impact parameter $d$, soChoose $\theta =0$ at closest approach. The general solution of the differential equation in a, with $u^{\prime }(0)=0$, isThe specific orbital energy isAt infinity $\epsilon =v^2/2$, giving $e=\sqrt{1+h^2{v}^2/{\mu }^2}>1$. Thus the hyperbolic orbit hasA different zero of polar angle simply replaces $\theta$ by $\theta -{\theta }_0$.

The incoming and outgoing asymptotes have radial directions $-{\theta }_{\infty }$ and ${\theta }_{\infty }$, where $\cos {\theta }_{\infty }=-1/e$ and $\pi /2<{\theta }_{\infty }<\pi$. The incoming velocity points opposite its radial direction, so the deflection angle between incoming and outgoing velocities is $\delta =2{\theta }_{\infty }-\pi$. HencePutting $\delta =\pi /2$ gives

By symmetry, the centre of mass is $G=(0,0,\ell /2)$. Choose horizontal axes through $G$ and the vertical symmetry axis as the third axis, writing $Z=z-\ell /2$. With uniform mass density $\rho$, the inertia tensor has entriesReflection in each coordinate plane makes all the mixed integrals $\int xydm$, $\int xZdm$, and $\int yZdm$ vanish. Rotational symmetry about the vertical axis gives $I_{11}=I_{22}=A$. Thus these are principal axes andThe axial moment of inertia of a hollow cylinder isIts mass is $M=\pi \rho \ell (b^2-a^2)$. Dividing and factoring the difference of squares gives

Choose positive $x$ downhill and positive $\theta$ in the corresponding rolling direction, with $x=\theta =0$ initially. For rolling without slipping, the contact point has zero velocity, so $\dot{x}-b\dot{\theta }=0$. Integrating the rolling constraint givesLet $F$ be the uphill static friction. Along the inclined plane, Newton's second law for the centre of mass givesThe weight passes through the centre of mass, and the normal reaction has its line of action through it. Thus only static friction supplies torque about $G$. The angular velocity is along the cylinder's symmetry axis, so the inertia tensor gives angular momentum $C\dot{\theta }$ in that direction. The torque equation is thereforeEliminating static friction and inserting $C$ givesThe extra term represents the rotational inertia that must be accelerated along with the translational mass.

For a uniform solid cylinder, set $a=0$ in the moment of inertia formula, givingFor the hollow cylinder, $a>0$ makes the denominator larger:on a downward incline with $\sin \phi >0$. Thus the solid cylinder accelerates faster, assuming rolling without slipping for both. More of the hollow cylinder's mass lies far from the rotation axis, increasing its moment of inertia relative to $M{b}^2$.

The specific heat capacity $c$ is the energy needed per unit mass per unit increase in temperature. With $c$ constant, a small element of volume $dV$ has mass $\rho dV$ and thermal energy $\rho cTdV$, measured relative to the reference at which $T=0$. Summing over $\Omega$ givesChoosing a different reference temperature only adds a time-independent constant to this energy, so it does not affect the heat balance with volumetric heat generation.

For any closed surface enclosing a region $V$, Gauss's law for gravity states that the outward flux of the gravitational field isHere $G$ is the gravitational constant and $M_V$ is the enclosed mass; the negative sign expresses attractive gravity.

Use the gravitational potential sign conventions specified by this paper: $\mathbf{g}=\nabla \phi$. Thus $\phi$ is the negative of the usual Newtonian gravitational potential $\Phi$ defined by $\mathbf{g}=-\nabla \Phi$. For example, setting the value at infinity to zero gives $\phi =GM/r$ for a point mass. The divergence theorem now givesSince $V$ is arbitrary, continuity of the integrands implies the local Poisson's equationConversely, integrating this Poisson's equation over any $V$ and applying the divergence theorem recovers Gauss's law for gravity. At a discontinuity of the mass density, the local equation is interpreted piecewise, or globally in the sense of distributions.

On a sphere of radius $r$, the radial gravitational field has constant magnitude. Thus Gauss's law for gravity becomeswhere $M(r)$ is the enclosed mass. In the hollow core $M(r)=0$. In the material shell, put $u=r/r_0$ and use the spherical coordinates volume element:Outside the shell the enclosed mass is the totalConsequently,At $r=0$ the displayed value refers to the radial derivative of the constant core value. The expressions match at both shell boundaries. There is no surface sheet of mass, so a jump in the volume mass density does not produce a jump in the gravitational field. The core has zero gravitational field, and the exterior gravitational field equals that of a central point mass $M_{\ast }$, as in the shell theorem.

On the surface of revolution, a normal vector is proportional to $-H^{\prime }(r){\mathbf{e}}_r+{\mathbf{e}}_z$. Thus the normal force and gravity have no ${\mathbf{e}}_{\theta }$ component. The ${\mathbf{e}}_{\theta }$ component of Newton's second law, using the given linear drag force, isMultiplication by $r$ therefore givesThe printed target equation omits the factor $1/m$. The matching original PDF, page 5, gives both ${\mathbf{F}}_D=-\alpha \dot{\mathbf{r}}$ and $\dot{h}=-\alpha h$, so this is a source inconsistency. If its $\alpha$ in the target equation is intended to mean the drag coefficient per unit mass, put $\lambda =\alpha /m$ and write $\dot{h}=-\lambda h$. Equivalently, if the intended force was ${\mathbf{F}}_D=-m\alpha \dot{\mathbf{r}}$, the requested convention givesThese formulas determine $\dot{\theta }$ once $r(t)$ is known; they do not assume a constant radius.