OneTreePhil OP t1_iwqri5u wrote on November 17, 2022 at 5:27 PM

Reply to comment by mfb- in What would the pressure be like in a body of water in free fall? by OneTreePhil

How did you get 4mm/s2?

mfb- t1_iwqtws6 wrote on November 17, 2022 at 5:42 PM

I misread the size and used a radius of 14 km, but reducing it to 7 miles only changes the acceleration to 3 mm/s^(2). Calculation.

The mass is 4/3 pi r^3 rho, the acceleration is a = Gm/r^2 = 4/3 pi r rho G. That's not limited to the surface, it applies everywhere inside the sphere, so we get a linear relationship. The pressure gradient is a*rho, integrated over the radius it's 1/2 r a_surface rho = 2/3 pi r^2 rho^2 G = 18 kPa with the fixed radius.