= Solution
Let the free surface be at $r=R_1$ and take the <Newtonian gravitational potential> to vanish at infinity. <Mass conservation> fixes $R_1=(3M_1/(4\pi\rho_1))^{1/3}$. The enclosed mass is $m(r)=4\pi\rho_1r^3/3$, and <hydrostatic equilibrium> requires $dp_0/dr=-\rho_1Gm(r)/r^2$. With vacuum outside, $p_0(R_1)=0$, giving
$$
\boxed{p_0(r)=\frac{2\pi G\rho_1^2}{3}(R_1^2-r^2),\qquad
\Phi_0(r)=
\begin{cases}
-\dfrac{GM_1}{2R_1}\left(3-\dfrac{r^2}{R_1^2}\right),&r\leq R_1,\\
-\dfrac{GM_1}{r},&r\geq R_1.
\end{cases}}
$$
The <pressure> is zero outside. A prescribed constant external <pressure> would simply add that constant to $p_0$.
The bulk <Eulerian and Lagrangian fluid perturbations> obey $\delta\rho=0$ because the <mass density> is uniform and the <Lagrangian displacement (fluid mechanics)> has zero <divergence>. The linearized <Euler momentum equation> in the interior is
$$
-\omega^2\boldsymbol\xi=-\nabla\left(\frac{\delta p}{\rho_1}+\delta\Phi\right).
$$
Taking the <curl> gives $\omega^2\nabla\times\boldsymbol\xi=0$. Thus every nonzero-frequency <normal mode> has an <irrotational flow> displacement; the <simply connected space> formed by the interior admits $\boldsymbol\xi=\nabla U$ and <incompressibility> gives the <Laplace equation> $\nabla^2U=0$. For a regular center, one <spherical harmonic> component is $U=A_l r^lY_l^m$, with $|m|\leq l$.
It is essential to retain the <surface gravity perturbation of a uniform-density sphere>. Although the bulk <mass density> perturbation is zero, the displaced density discontinuity gives
$$
\delta\rho=\rho_1\xi_r(R_1)\delta(r-R_1),\qquad
\xi_r(R_1)=lA_lR_1^{l-1}Y_l^m.
$$
The <Poisson equation> implies that $\delta\Phi$ is continuous and its outward radial derivative has the jump
$$
\left[\partial_r\delta\Phi\right]_{\mathrm{out}-\mathrm{in}}=4\pi G\rho_1\xi_r(R_1).
$$
Using regularity at the center and decay at infinity, write
$$
\delta\Phi_{\mathrm{in}}=C_l r^lY_l^m,\qquad
\delta\Phi_{\mathrm{out}}=C_l R_1^{2l+1}r^{-l-1}Y_l^m.
$$
The derivative jump is $-(2l+1)C_lR_1^{l-1}Y_l^m$, so $C_l=-4\pi G\rho_1lA_l/(2l+1)$.
Integrating the bulk <Euler momentum equation> gives $\delta p/\rho_1=\omega^2U-\delta\Phi$ for $l\geq1$. At the free surface the <Lagrangian pressure perturbation> vanishes:
$$
0=\delta p+\xi_r p_0'(R_1),\qquad
\frac{\delta p}{\rho_1}=\frac{GM_1}{R_1^2}\xi_r.
$$
Combining these relations yields the <Kelvin stellar mode> frequency
$$
\omega_l^2=l\frac{GM_1}{R_1^3}-\frac{4\pi G\rho_1l}{2l+1}
=\boxed{\frac{2l(l-1)}{2l+1}\frac{GM_1}{R_1^3}},\qquad l\geq1.
$$
The $2l+1$ choices of $m$ are degenerate because of <spherical symmetry>. For $l=1$, $rY_1^m$ is a <linear combination> of coordinates in a <Cartesian coordinate system>, so $\nabla U$ is a constant <displacement>: \b[the zero frequency is rigid translation of the whole isolated body]. Such a translation has no restoring force. The constant $l=0$ potential generates no displacement; a radial breathing motion is excluded by <incompressibility> and a regular center.
The printed irrotational assertion needs the nonzero-frequency qualification, or restriction to the potential <incompressible stellar surface modes>. It is false for all neutral displacements: $\boldsymbol\xi=\boldsymbol\Omega\times\mathbf r$ has zero <divergence>, zero normal displacement at the surface, and $\delta p=\delta\Phi=0$, so it is a zero-frequency displacement, but $\nabla\times\boldsymbol\xi=2\boldsymbol\Omega\ne0$. This does not change the <Kelvin stellar mode> spectrum just derived.
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