= Solution
Let $R=|\mathbf x|$, $\mathbf n=\mathbf x/R$ and $t_R=t-R/c_0$. Convolving the source terms in the <Lighthill acoustic analogy> with the <retarded acoustic Green function>, and moving their spatial derivatives outside the integral, gives
$$
\rho'=\partial_{x_i}\partial_{x_j}\int\frac{T_{ij}(\mathbf y,t-|\mathbf x-\mathbf y|/c_0)}{4\pi c_0^2|\mathbf x-\mathbf y|}\,d^3y
-\partial_{x_i}\int\frac{f_i(\mathbf y,t-|\mathbf x-\mathbf y|/c_0)}{4\pi c_0^2|\mathbf x-\mathbf y|}\,d^3y.
$$
The integrals have compact spatial support. For an acoustically compact source, its size $l$ satisfies $\omega l/c_0\ll1$, so the source-dependent retardation is negligible at leading order. Also take $R\gg l$. Define $S_{ij}=\int T_{ij}\,d^3y$ and $F_i=\int f_i\,d^3y$. In the radiation terms the spatial derivatives act on $t_R$, with $\partial_{x_i}t_R=-n_i/c_0$. Derivatives of $1/R$ produce smaller near-field terms. Thus the <far-field acoustic force and stress moments> give
$$
\boxed{\rho'(\mathbf x,t)=\frac{n_in_j\ddot S_{ij}(t_R)}{4\pi c_0^4R}
+\frac{n_i\dot F_i(t_R)}{4\pi c_0^3R}.}
$$
Using $n_i=x_i/R$ gives the equivalent powers of $R$ requested. The force term is an <acoustic dipole>; the <Lighthill stress tensor> produces an <acoustic quadrupole>.
For a small body in low-speed flow, put $M=u/c_0\ll1$. A typical unsteady force has size $F=O(\rho_0u^2l^2)$ and time scale $l/u$, whereas $S=O(\rho_0u^2l^3)$. Therefore
$$
\boxed{\frac{\rho'_D}{\rho_0}=O\!\left(\frac lR M^3\right),
\qquad \frac{\rho'_Q}{\rho_0}=O\!\left(\frac lR M^4\right).}
$$
These are the <acoustic dipole Mach-number scaling> and <compact acoustic quadrupole Mach-number scaling>. Since $p'=c_0^2\rho'$ and outgoing acoustic intensity is of order $p'^2/(\rho_0c_0)$, their sound powers scale as
$$
P_D=O(\rho_0u^6l^2/c_0^3),\qquad P_Q=O(\rho_0u^8l^2/c_0^5).
$$
Thus an unsteady force on the body can produce the dominant low-<Mach number> sound. If the integrated force or its time derivative vanishes, that dipole contribution is absent. Force on the body and force on the fluid have opposite signs.
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