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A spherical mass m with radius r of a known fluid with density ρ floats in the vacuum. Calculate the pressure P at the distance x from the center due its own gravity.
Also... at the distance x from the center due its own gravity.
We are on different wavelengths, here. I'm picturing a sphere of liquid in space where g = 0 and P = ρgh is zero. Looking for the pressure inside the sphere "due its OWN gravity". Remember it "floats in vacuum" whereas if it were at the surface of the Earth, g would pull it down to smash on the bottom of its container.P=ρgh
I think he means that g is the 'local' value of gravitational acceleration due to the mass below x. The problem though, is that 'g' will not remain constant from x to the top of the surface of the sphere.We are on different wavelengths, here. I'm picturing a sphere of liquid in space where g = 0 and P = ρgh is zero. Looking for the pressure inside the sphere "due its OWN gravity". Remember it "floats in vacuum" whereas if it were at the surface of the Earth, g would pull it down to smash on the bottom of its container.
V'→ spherical volume with radius xThe mass from the radius x:
V'=4πx³/3
m'=4πx³ρ/3
The pressure at the distance x:
P=ρgh
P=ρGm'(r-x)/x²
P=ρ²G4πx(r-x)/3
The 'g' is not constant over the distance (r - x).P → hydrostatic pressure
P=ρgh
P=ρ²G4πx(r-x)/3
Nope. Think of a thin straw sticking up from the bottom to the top. It'll work fine, and you won't have to deal with the surface area at each radial distance.Wouldn't the spherical shell with thickness dR be easier to work with?
A thin "column" would have to be wider at the top than at the bottom, wouldn't it?