**See the attached pages at the back of the submission for hand-written work.**

*A certain planet has density $\rho(r)$ that decreases linearly going from the core to the surface, so that $\rho(r)=\rho_0(1-2r/3R)$, where $r$ is the radial coordinate and $R$ is the radius of the planet.*

**(a)** derive an expression for the mass $M_r$ contained within radius $r$.

The density function describes the density of a particular thin shell with radius $x$. The infinitesimal volume of the thin shell is:

\[\mathop{dV}=4\pi x^2 \mathop{dx}\]The mass for these shell is then:

\[\begin{aligned} m(x)&=\rho(x)\mathop{dV}\\ &=\rho_0\left(1-\frac{2x}{3R}\right)\cdot 4\pi x^2 \mathop{dx} \end{aligned}\]Then we can integrate from x = 0 → r to get the mass inside the radius $r$.

\[\begin{aligned} M_r&=\frac{4\pi\rho_0}{3R}\int_0^r(1-2x)x^2\mathop{dx}\\ &=\frac{4\pi\rho_0}{3R}\int_0^r x^2-2x^3\mathop{dx}\\ &=\frac{4\pi\rho_0}{3R} \left[ \frac{1}{3}x^3-\frac{1}{2}x^4 \right]^r_0\\ &=\frac{4\pi\rho_0}{3R} \left( \frac{1}{3}r^3-\frac{1}{2}r^4 \right)\\ \end{aligned}\] \[\begin{aligned} M_r&=4\pi\rho_0\int_0^r\left(1-\frac{2x}{3R}\right)x^2\mathop{dx}\\ &=4\pi\rho_0\int_0^rx^2-\frac{2x^3}{3R}\mathop{dx}\\ &=4\pi\rho_0\left[\frac{x^3}{3}-\frac{x^4}{6R}\right]^r_0\\ &=4\pi\rho_0\left(\frac{r^3}{3}-\frac{r^4}{6R}\right) \end{aligned}\]**(b)** Derive dP/dr.

Start by considering an imaginary cylinder with mass $dm$ and height of $dr$ radially from the center of the sphere. Then start from Newton’s law of force:

\[F=ma\]Where forces are the forces from top pressure and bottom pressure of the cylinder acting on the cylindrical mass. And acceleration is just double-derivative of the radial position: $\ddot r$. Formalizing the equation:

\[(dm)(\ddot r) = - F_g + F_{P_B} - F_{P_T}\]Where

\[F_g = \frac{G M_r dm}{r^2} \simeq \frac{G M_r}{r^2}\]And since we’re considering hydrostatic equilibrium, the acceleration should be 0: $\ddot r = 0$.

Putting it all together, and cancelling $F_{P_B} - F_{P_T}$ into $A dP$, we get:

\[\rho \ddot r = \frac{G M_r \rho}{r^2}-\frac{dP}{dr},\quad\ddot r = 0\]Isolate for dP/dr, and substitute $M_r$ for the expression from part (a), and substitute the expression for $\rho=\rho(r)$:

\[\frac{dP}{dr}=\frac{G}{r^2}\times 4\pi \rho\left(\frac{r^3}{3}-\frac{r^4}{6R}\right)\times \rho_0 \left(1-\frac{2r}{3R}\right)\]Simplifying the above equation and we get a three-term polynomial:

\[\frac{dP}{dr}=4\pi\rho_0 ^2G\left(\frac{r}{3}-\frac{7 r^2}{18 R} + \frac{2r^3}{18 R^2}\right)\]* (c) To get pressure, we integrate*.

To find C, we set $r=R$, and assume that the pressure at surface $P(R)=0$:

\[\begin{aligned} P(R)=0&=\left(\frac{1}{6}r^2-\frac{7}{54R}r^3+\frac{1}{36R^2}r^4+C\right)\\ 0&=\frac{7R^2}{108}+C\\ C&=-\frac{7R^2}{108} \end{aligned}\]Putting it all together:

\[P(R)=4\pi\rho_0^2 G \left(\frac{1}{6}r^2-\frac{7}{54R}r^3+\frac{1}{36R^2}r^4-\frac{7R^2}{108}\right)\]**(a)** Find degeneracy pressure in number density.

Assuming that the star is full of H_{2} gas with uniform density. The mass of the hydrogen atom is 1.67×10^{-27} kg. So that means 1kg of Hydrogen has 5.988×10^{26} atoms. This is our number density for the electrons since a hydrogen atom only has one electron.

Then for a given mass density $\rho$ in the units kg/m^{3}, we have $\rho\times5.988\times10^{26}$ number density in the units e^{-}/m^{3}.

Then we plug into the equation given from class:

\[P=\frac{1}{20}\left(\frac{3}{\pi}\right)^{2/3}\frac{h^2}{m_e}n_e^{5/3}\\ P=\frac{1}{20}\left(\frac{3}{\pi}\right)^{2/3}\frac{h^2}{m_e}(\rho\times 5.988\times 10^{26})^{5/3}\]**(b)** Equation of boundary between non-degeneracy and degeneracy.

Starting from

\[P_{termal}=\frac{\rho}{m_p}k_BT\]Then we plug in the expressions from part (a), then isolate for T:

\[\begin{aligned} T&= \frac{P_{thermal} m_p}{\rho k_B}\\ &= \frac{\left(\frac{1}{20}\left(\frac{3}{\pi}\right)^{2/3}\frac{h^2}{m_e}(\rho\times 5.988\times 10^{26})^{5/3}\right) m_p}{\rho k_B}\\ &=\left(\frac{3\rho}{\pi}\right)^{2/3}\frac{h^2(5.988\times10^{26})^{5/3}m_p}{20m_ek_B}\\ \end{aligned}\]The entire rightmost side can be evaluated down to a constant:

\[\frac{h^2(5.988\times10^{26})^{5/3}m_p}{20m_ek_B}=1242\]So therefore:

\[T=1242\left(\frac{3\rho}{\pi}\right)^{2/3}\]**(c)** Plot

Here is a log-log plot of the above function:

Where the y-axis is log_{10}(T) and x-axis is log_{10}(ρ). The region of degeneracy is **above** the graph.

**(d)** Saturn

For a mass density of 3000 kg/m^{3}, we plug it into the equation we derived in (b):

The actual temperature is $\sim 40,000K$, so there is no degeneracy. So it lies in part of the graph that is below the boundary curve.

**(a)** Find maximum orbit frequency

We can assume that surface mass is attached to the neutron star if it’s not *orbiting*. Then we can start by calculating the required circular orbit speed at the surface:

Plugging in the constants, we get a orbit speed of $v_c=1.36\times 10^8$ m/s which is very fast.

The equatorial circumference of the neutron star is $2\pi(10\text{km})$, and therefore the period is distance divided by speed:

\[P=\frac{2\pi 10,000}{1.36\times 10^8}=0.4609\times 10^{-3} \text{s}\]**(b)** equatorial bulge

The relationship between the polar and equatorial radius is given by:

\[\frac{E-P}{R}=\frac{5\Omega^2R^3}{4GM}\]So we plug everything in where $\Omega$ is the angular frequency defined by:

\[\Omega = 2\pi P^{-1}\]Then

\[E-P = \frac{5\Omega^2 R^4}{4G (1.4 M_\odot)}=1.374 \text{km}\]To solve for E and P, we can also assume that the volume of the ellipsoid and the sphere don’t change:

\[V_{sphere}=V_{ellipsoid}\\ r^3=E^2 P, \quad r=10\text{km}, \quad P=E-1.374\text{km}\]We have two equations and two unknowns, so we solve for both:

\[E = 10.48\text{km},\quad P=9.106\text{km}\]