Four. (20 points) It is known that in the Cartesian coordinate system, there is a disk (including the circumference) $G$ with a center at point $A(2008,525)$ and a radius of 1. The graph of the parabola $y^{2}=2 p x$ intersects the disk $G$. Find the range of values for $p$ (accurate to 0.001).
Let the coordinates of point $P(x, y)$ on the boundary of the disk $G$ be
$$
\left\{\begin{array}{l}
x=2008+\cos \theta, \\
y=525+\sin \theta
\end{array} \quad(0 \leqslant \theta<2 \pi) .\right.
$$

Substituting into the parabola equation $y^{2}=2 p x$, we get
$$
(525+\sin \theta)^{2}=2 p(2008+\cos \theta) \text {. }
$$

Thus, $p=\frac{(525+\sin \theta)^{2}}{2(2008+\cos \theta)}(0 \leqslant \theta<2 \pi)$.
Using a graphing calculator, we obtain
$$
p_{\min }=68.368306, p_{\max }=68.895664 \text {. }
$$

To ensure the parabola passes through $G$, we round up $p_{\text {min }}$ and round down $p_{\max }$. Therefore, the range of $p$ is $68.369 \leqslant p \leqslant 68.895$.