= Solution
{c}
Approximate Earth by an isolated, nonrotating, spherically symmetric body. By <Birkhoff's theorem>, the exterior <vacuum> <metric tensor> is the <Schwarzschild metric>. With $x^0=ct$ it is
$$
\boxed{ds^2=-f(r)c^2dt^2+\frac{dr^2}{f(r)}+r^2(d\theta^2+\sin^2\theta\,d\phi^2),\qquad
f(r)=1-\frac{2GM}{rc^2}.}
$$
Here $r$ is the <areal radius>, $\theta$ and $\phi$ are spherical angles, $t$ is <Schwarzschild time> normalized to the <proper time> of a static observer at infinity, $M$ is Earth's <mass>, $G$ the <gravitational constant>, and $c$ the <speed of light>. The nonzero diagonal <metric tensor> components are $g_{00}=-f$, $g_{rr}=f^{-1}$, $g_{\theta\theta}=r^2$, $g_{\phi\phi}=r^2\sin^2\theta$.
The apple is a <test particle>; neglect its effect on the <metric tensor>, air resistance, Earth's rotation and nonspherical structure, and external <tidal forces>. Near Earth's surface, $GM/(R_Ec^2)\simeq7\times10^{-10}$, $h/R_E\ll1$, and the falling <speed> remains much smaller than $c$. Thus the <Newtonian limit> of this <metric tensor> is appropriate, while the exact radial <geodesic> is available as a check. A static observer under the tree has <proper time> $T=\sqrt{f(R_E)}\,t$, up to an additive constant. Such an observer is supported by the ground and has nonzero <proper acceleration>; the released apple instead has zero <proper acceleration>.
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