Submitted by OneTreePhil t3_yx9z3q in askscience
mfb- t1_iwqtws6 wrote
Reply to comment by OneTreePhil in What would the pressure be like in a body of water in free fall? by OneTreePhil
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.
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