= Solution
\b[Local equivalence and tides.] In <Newtonian mechanics>, <inertial mass> $m_{\rm i}$ measures the response to a <force>, whereas passive <gravitational mass> $m_{\rm g}$ couples to an imposed <Newtonian gravitational field>. A freely falling body obeys
$$
m_{\rm i}\mathbf a=m_{\rm g}\mathbf g,\qquad
\mathbf a=\frac{m_{\rm g}}{m_{\rm i}}\mathbf g.
$$
The observed composition independence of <free fall> says that this ratio is universal. With a common choice of units it is one. This is the Newtonian content of the <weak equivalence principle> for <test particles>; it does not by itself specify all nongravitational physics.
The <Einstein equivalence principle> further says that local nongravitational experiments in a sufficiently small freely falling laboratory have the form predicted by <special relativity>, independently of the laboratory's location and motion. In such a <local inertial frame>, the <metric tensor> is locally the <Minkowski metric> and the <Christoffel symbols> vanish at the selected event. The <strong equivalence principle> extends the universality claim to local gravitational experiments and bodies with significant gravitational self-energy. None of these statements removes measurable <tidal forces> across a finite apparatus.
In an ideal small freely falling lift, a released ball initially floats relative to the cabin, because both undergo the same <free fall>. Two separated balls, however, need not retain their separation: they sample different parts of the <gravitational field>. For a point source, linearizing $\mathbf g=-GM\mathbf r/r^3$ across a small separation $\boldsymbol\xi$ gives the <tidal tensor> prediction
$$
\boxed{\frac{d^2\boldsymbol\xi}{dt^2}
=\frac{GM}{r^3}\big[3\widehat{\mathbf r}(\widehat{\mathbf r}\cdot\boldsymbol\xi)-\boldsymbol\xi\big].}
$$
A radial separation stretches with coefficient $2GM/r^3$, whereas a transverse separation is compressed with coefficient $-GM/r^3$. Initially parallel <velocities> therefore develop relative motion even though the individual bodies experience zero <proper acceleration>.
The relativistic statement is <geodesic deviation>. With the <Riemann curvature tensor> convention
$$
R^\mu{}_{\nu\rho\sigma}
=\partial_\rho\Gamma^\mu{}_{\sigma\nu}-\partial_\sigma\Gamma^\mu{}_{\rho\nu}
+\Gamma^\mu{}_{\rho\alpha}\Gamma^\alpha{}_{\sigma\nu}
-\Gamma^\mu{}_{\sigma\alpha}\Gamma^\alpha{}_{\rho\nu},
$$
a connecting vector between neighboring freely falling <worldlines> satisfies
$$
\boxed{\frac{D^2\xi^\mu}{d\tau^2}
=-R^\mu{}_{\nu\rho\sigma}U^\nu\xi^\rho U^\sigma.}
$$
Here $D/d\tau$ is the <covariant derivative> along the reference <geodesic>, and $U^\mu$ its <four-velocity>. <Curvature> can be nonzero where the <Christoffel symbols> have been made zero. In the weak, static, slow-motion limit this equation reduces to $\ddot\xi^i=-\partial_i\partial_j\Phi\,\xi^j$, recovering the <tidal tensor> above. A finite falling lift can detect <gravity> through these relative <accelerations>; local equivalence does not imply a globally uniform field.
\b[Weak-field connection and potential.] Take $x^0=ct$ and the $+---$ <metric signature>. There is an index error in the printed formula for the <Christoffel symbols>: its first differentiated metric component must be $g_{\nu0}$, not $g_{\mu0}$. The correctly contracted expression is
$$
\Gamma^\mu{}_{00}=\frac12g^{\mu\nu}
\left(2\partial_0g_{\nu0}-\partial_\nu g_{00}\right).
$$
The offending term vanishes under the stipulated time independence, so this correction does not change the intended calculation. Write $g_{\mu\nu}=\eta_{\mu\nu}+h_{\mu\nu}$ and treat both $h$ and its <derivatives> as first order. To that order $g^{00}=1-h_{00}$, $g^{0i}=O(h)$, and $g^{ij}=-\delta^{ij}+O(h)$. A stationary <metric tensor> gives
$$
\Gamma^\mu{}_{00}=-\frac12g^{\mu j}\partial_jh_{00}.
$$
For the temporal component the inverse-metric coefficient is already first order, so
$$
\boxed{\Gamma^0{}_{00}=O(h\,\partial h)\approx0.}
$$
For a spatial component the zeroth-order inverse metric contributes:
$$
\boxed{\Gamma^i{}_{00}\approx\frac12\delta^{ij}\partial_jh_{00}
=\frac12\frac{\partial h_{00}}{\partial x^i}.}
$$
The time component of the approximate <geodesic equation> says that $dt/d\tau$ is constant to the retained order. In its spatial component, convert the <proper time> derivatives to coordinate-time derivatives and cancel $(dt/d\tau)^2$:
$$
\frac{d^2x^i}{dt^2}\approx-c^2\Gamma^i{}_{00}
\approx-\frac{c^2}{2}\partial_i h_{00}.
$$
Comparison with the Newtonian equation $d^2\mathbf x/dt^2=-\boldsymbol\nabla\Phi$ identifies $h_{00}=2\Phi/c^2+C$. For an isolated source, choose the <gravitational potential> to tend to zero at infinity and normalize the time coordinate there, giving $C=0$. Thus
$$
\boxed{g_{00}\approx1+\frac{2\Phi}{c^2}.}
$$
The quantity $\Phi$ is the Newtonian <gravitational potential>, or <gravitational potential energy> per unit <mass>, in $\mathrm{J\,kg^{-1}}=\mathrm{m^2\,s^{-2}}$. For a point source, $\Phi=-GM/r$. A static clock has $d\tau=\sqrt{g_{00}}\,dt\approx(1+\Phi/c^2)dt$, connecting the <Newtonian limit> with <gravitational time dilation>.
\b[Frequency of the orbiting source.] Treat the gas blob as a <test particle> on a <timelike circular Schwarzschild geodesic>, with pressure and other nongravitational forces negligible. Treat our very distant location as infinity, so the observer's own <gravitational potential> is negligible. These are the assumptions that make a unique answer in terms of $R_S/r$ possible.
Put $f=1-R_S/r$ and use a local static observer at the emission point as an intermediate measuring frame. The remote observer lies on the symmetry axis of the orbit. A ray arriving on that axis has zero conserved axial <angular momentum>, so its local direction is perpendicular to the emitter's azimuthal <velocity>. Bending within a meridional plane does not alter this orthogonality. The transverse <relativistic Doppler effect> therefore gives
$$
\nu_{\rm stat}=\frac{\nu_{\rm em}}{\gamma_v},\qquad
\gamma_v=\frac1{\sqrt{1-v^2/c^2}},
$$
where $v$ is the orbital <speed> measured by that local static observer. The <gravitational redshift between static Schwarzschild observers> then gives
$$
\frac{\nu_{\rm ob}}{\nu_{\rm em}}=\frac{\sqrt{1-R_S/r}}{\gamma_v}.
$$
A gravitational <circular orbit> has $v^2=GM/r+O((GM/r)^2/c^2)$ in the weak-field limit, so $v^2/c^2=R_S/(2r)+O((R_S/r)^2)$. Expanding both factors,
$$
\frac{\nu_{\rm ob}}{\nu_{\rm em}}
=\left(1-\frac{R_S}{2r}\right)
\left(1-\frac{R_S}{4r}\right)+O\!\left(\frac{R_S^2}{r^2}\right)
=1-\frac{3R_S}{4r}+O\!\left(\frac{R_S^2}{r^2}\right).
$$
Solving for the emitted <frequency> gives
$$
\boxed{\nu_{\rm em}=\nu_{\rm ob}\left[1+\frac{3R_S}{4r}+O\!\left(\frac{R_S^2}{r^2}\right)\right].}
$$
The gravitational part is $R_S/(2r)$ and the transverse kinematic <time dilation> contributes another $R_S/(4r)$; the absence of line-of-sight motion does not remove this second effect.
As an exact check, the local speed on a <timelike circular Schwarzschild geodesic> satisfies $v^2/c^2=\mu/(r-2\mu)$, where $\mu=GM/c^2=R_S/2$. Therefore the <face-on frequency shift of a circular Schwarzschild source> is
$$
\frac{\nu_{\rm ob}}{\nu_{\rm em}}=\sqrt{1-\frac{3\mu}{r}}
=\sqrt{1-\frac{3R_S}{2r}},
$$
whose expansion agrees with the boxed answer. If the gas is supported or accelerated by pressure or magnetic stresses rather than following a <geodesic>, its <orbital speed> is not fixed by the mass and radius alone; the appropriate first-order result is instead $\nu_{\rm em}\approx\nu_{\rm ob}[1+GM/(rc^2)+v^2/(2c^2)]$, with $v$ supplied independently.
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