L.Dutch calculates the acceleration of gravity just outside a thin hollow polystyrene shell as
$$ g_s = 4\pi GT\rho $$
where $G$ is the universal constant of gravity, $T$ is the thickness of the shell, and $\rho$ is the density of the polystyrene. He observes that this is independent of $R$, the radius of the sphere.
However this is not the entire story, because the shell has to withstand its own weight without crumbling, and this is not just a function of the local acceleration of gravity. The shell is basically a self-supporting $360^\circ$ arch, and in constant gravity larger archs (of the same cross section) can support smaller weights.
The force of gravity per unit area of the shell is
$$ \tfrac 12 g_s T \rho $$
(where the $\tfrac12$ is because the gravitational strength drops to $0$ just inside the hollow shell due to the shell theorem).
If we draw an equator around the shell, the integral of the southbound component of gravity everywhere on the north hemisphere works out nicely as
$$ \tfrac 12 g_s T \rho \cdot \pi R^2 $$
(there are some nasty-looking trigonometric factors as we write down the integral, but they all happen to cancel out before we integrate, if we parameterize the north hemisphere by its perpendicular projection on the equatorial plane. Note that the new factor is just the area of the circle cut out of the equatorial plane by the shell!)
All this force needs to be transmitted between north and south by lateral compression of the shell across the equator. The area of this cross section is
$$ 2\pi R \cdot T $$
so the average lateral pressure must be
$$ \frac{\tfrac12g_s T \rho \pi R^2}{2\pi R T } = \pi GTR\rho^2 $$
Now there is an $R$ factor again, so we cannot make the shell arbitrarily large without it crushing itself from the sides. The exponents here make sense too:
- $\rho^2$ makes sense because gravity increases by the product of two masses, and making each segment of shell twice as massive increases the forces fourfold.
- $R$ makes sense because each meter of equator needs to support the entire wedge of shelf going from itself up to the north pole.
- $T$ makes sense because making the shell twice as thick makes the forces four times as large, but on the other hand also distribute the forces over twice as much cross section.
So according to this calculation we should use the light class of EPS (the compressive strength scales slower than the square of the density). We should also make the shell thin -- at least until it becomes so thin that it starts to buckle.
I think the above calculation actually takes care of making the shell not buckle under its own gravity [at least that's what I thought until I ran the numbers below] -- but if we make it too lightweight, it might start buckling under external forces, such as gusts of stellar wind or the exhaust from the camera crews' rocket engines.
To be on the safe side, let's dimension the shell such that the force of its self-gravity is about 1 mN/m². Then at being hit by the long-distance puff of a solar flare (which can reach ram pressures of several μPa, based on their measured effect on Earth's magnetosphere -- the ordinary solar wind is a few orders of magnitude weaker than that) should still be negligible compared to the self-gravity.
We can then solve
$$ 2\pi GT^2\rho^2 = 1\;\rm mPa$$
for $T$ to find a thickness of about $96\;\rm m$. Let's make it an even 100 meters for back-of-the-envelope purposes.
Now we can solve
$$ \pi GTR\rho^2 = 12\;\rm p{.}s{.}i{.} $$
for $R$, getting about
$$ \large R \approx 15 \times 10^6 \;\rm km $$
which is just about a tenth of an astronomical unit, or 40 times the distance to the moon.
It looks like there is room enough for a cautious prop designer to increase $T$ by a safety factor of 10 and still build a 1:1 stand-in for Jupiter if he wants (assuming he can source enough polystyrene).
