$$\require{cancel}$$

# Exercise 11.1.1

K.E. is $$\frac{1}{2}mv^2$$, where $$v = \dot a \ell$$, so the test particle's kinetic energy is

\begin{equation*} \begin{aligned} \frac{1}{2}m\dot a^2 \ell ^2 \end{aligned} \end{equation*}

# Exercise 11.1.2

P.E. is $$-\frac{GM(<d)m}{d}$$, where by $$M(<d)$$ we mean the mass contained in the sphere of radius $$d$$, so $$M(<d) = \frac{4}{3}\pi d^3 \rho.$$

Therefore the test particle's potential energy is

\begin{equation*} \begin{aligned} -G \frac{4}{3} \pi \rho d^2 m \end{aligned} \end{equation*}

# Exercise 11.1.3

\begin{equation*} \begin{aligned} \frac{1}{2}m\dot a^2 \ell ^2 - G \frac{4}{3} \pi \rho d^2 m = \kappa \end{aligned} \end{equation*}

# Exercise 11.1.4

Recall that $$d = a \ell \; \Longrightarrow \; \ell = \frac{d}{a}$$, substituting this in and rearranging our equation we get
\begin{equation*} \begin{aligned} \frac{1}{2}md^2 \frac{\dot a^2}{a^2} - \frac{1}{2}md^2 \frac{8\pi G\rho}{3} = \kappa \end{aligned} \end{equation*}
dividing through by $$\frac{1}{2}md^2$$ gives
\begin{equation*} \begin{aligned} \frac{\dot a^2}{a^2} - \frac{8\pi G\rho}{3} = \frac{2\kappa}{m}\frac{1}{d^2} \end{aligned} \end{equation*}
Then we substitute back in $$d = a \ell$$ and solve for $$\big(\frac{\dot a}{a}\big)^2$$:
\begin{equation*} \begin{aligned} \left(\frac{\dot a}{a}\right)^2 = \frac{8\pi G \rho}{3} + \frac{2\kappa}{ml^2} \times \frac{1}{a^2} \end{aligned} \end{equation*}