Ramjet thrust

As I fill in NASA's Ice Gun gaps, here is NASA's ramjet equation: thrust is ṁV_x-ṁV₀+(p_x-p₀)A_x.

It turns out that by now I know a lot of this, if A_x be exactly a Dragon's cross-section 4π m². ṁ₀=3040kg/s; ṁ_x=3648kg/s. Initial V₀=1000m/s. I was despairing of finding V_x... but now I find it here as I plug in temperatures 200K and 2000K out: 3160m/s.

The gross thrust minus ram drag make, then, for 11527680 - 3040000 newtons. So F = 8,487,680 + 4πδp but NASA assure me we can normally neglect δp.

Again that's the initial value. One of our checkpoints was V₀=4400 m/s. ṁ₀ is now 13376kg/s, ṁ_x=16051.2. For that V_x=13914m/s. 223336396.8-58854400: F = 164,481,996.8 N. The force, as you see, rises. Moreover on the assumption of pure hydrogen, much of the initial mass is an oxidiser (you DID notice that the mass going out was more than the mass going in...?). So, the mass that needs to be moved is at first decreasing at 608kg/s.

8487 kilonewtons at an acceleration 32 m/s² can move 265 tonnes, to start its 1000 km odyssey. At 96.8 m/s² this falls to 87 tonnes aiming at 100 km. I expect that to lessen G force on the cargo (which is why we are doing all this), NASA'll lighten up on the temperature of the jet. That is - the jet uses less oxidiser as it has less.