If we assume that the vacuum level inside the evacuated tube, P, is 5 Pa and the air temperature, T, is 20°C (293.15 K), the air density inside the tube can be estimated using the ideal gas law:

$$\rho = \frac{P}{R T} = \frac{5}{287.05 \times 293.15} = 5.94 \times 10^{-5}\ \text{kg/m}^3$$

where R is the specific gas constant for dry air (287.05 J/(kg·K)).

The screw flights rotate with angular velocity ω. Based on a maximum flight-tip speed of 530 m/s and a flight-tip radius of 0.5 m, the angular velocity is:

$$\omega = \frac{v}{r} = \frac{530}{0.5} \approx 1{,}060\ \text{rad/s}$$

Each screw flight can be modeled as a flat radial blade spanning from the screw shaft radius, $r_i = 0.228\ \text{m}$, to the flight-tip radius, $r_o = 0.5\ \text{m}$. The local tangential velocity varies with radius according to $v(r) = \omega r$.

For a differential blade strip of width $dr$, the drag force is:

$$dD = \tfrac{1}{2}\rho C_d (\omega r)^2 b\,dr$$

and the corresponding differential torque is:

$$d\tau = r\,dD = \tfrac{1}{2}\rho C_d b \omega^2 r^3 dr$$

Integrating from $r_i$ to $r_o$ gives the torque for one blade:

$$\tau_\text{blade} = \frac{\rho C_d b \omega^2}{8}\left(r_o^4 - r_i^4\right)$$

The total length of the launcher sections containing spinning screws is 773 km + 75 km = 848 km. There are two counter-rotating screws, and most sections have four flights per screw. Treating each flight as contributing one blade face, the total effective blade span is:

$$b_\text{total} (N) = 2 \times 4 \times 848{,}000\ \text{m}$$

The total aerodynamic torque is therefore:

$$\tau = \frac{\rho C_d \omega^2}{8} \, b_\text{total} \left(r_o^4 - r_i^4\right)$$

and the corresponding power required to overcome aerodynamic drag is:

$$P = \tau \omega = \frac{\rho C_d \omega^3}{8} \, b_\text{total} \left(r_o^4 - r_i^4\right)$$

Substituting numerical values and using a conservative drag coefficient of $C_d = 1.2$ yields a total aerodynamic power of 4.3 gigawatts.

This estimate represents an intentional upper bound based on a still-air assumption. In practice, the rotating screws will entrain and co-rotate the rarefied gas within the evacuated tube, substantially reducing relative velocity and therefore aerodynamic drag. Additionally, at pressures on the order of a few pascals, continuum drag models are expected to overestimate losses as the flow approaches the slip and free-molecular regimes.

This upper-bound estimate could be further refined by performing a 2D cross-section CFD simulation of the evacuated tube containing twin counterrotating screws and the guideway within the evacuated tube. This approach would capture gas entrainment and partial co-rotation within the tube, providing a more accurate estimate of effective drag and power loss.

The specified vacuum level can be lowered to reduce the screws' power losses due to drag. For example, if it were set to 1 Pa, then the power loss upper bound estimate would only be 0.86 GW.