At the center there is no net, but inside the sphere the net effect is the same as if all the mass were concentrated at the center. (Suppose that there was a normal net effect everywhere at and just above the surface and zero net effect just below the surface: that would require that the gravitational field not be continuous, which is not the case)
At the center there is no net, but inside the sphere the net effect is the same as if all the mass were concentrated at the center.
Remember that this is a spherical shell, not a sphere. When calculating the force of gravity inside the Earth, for example, you ignore all mass at higher radius that the location you’re interested in, but not the mass at lower radius.
Why does this work? Imagine opposite cones that originate at the center of the spherical shell. The intersection of that cone with the shell will have surface area that increases with r^2, but the inverse square law decreases with r^2, and so the gravitational effect only depends on the angle of the cone, regardless of distance- but the two cones are pointed in opposite directions, and so cancel out. (This is obvious at the center, but works just as well elsewhere inside the spherical shell.)
that would require that the gravitational field not be continuous, which is not the case
For an infinitely thin shell, the drop in gravitation is infinitely steep. But shells with real thickness will have a gradual dropoff that corresponds to that thickness.
It’s not clear to me where that model came from. We shouldn’t expect the density of celestial bodies to be uniform unless they’re made of something incompressible, and it’s important to separate out the net force of gravity and the weight of mass above you. In steady state, all of the mass in the celestial body is being pulled towards the center by gravity, pushed upwards by the mass below it, and has to push upwards on the mass above it (with the net force being 0). I haven’t done the math to see what the pressure function would look like for a gaseous celestial body, and it seems like the full calculation will have lots of complications, but we can note that the mass below you has to push up harder than the mass above you is pushing down, suggesting the pressure is highest at the center.
delta-P is the first derivative of pressure; it would have to be zero at the center for there to be a pressure maximum at zero.
I would expect a gaseous body to have a roughly spherically symmetric mass distribution, which is all we need. Treat it as an infinite number of infinitely thin spheres each of uniform density, and we can do calculus on it.
We can also do this though experiment with a perfect liquid of uniform density; at least it will have a surface that we can stop at. Pressure is still highest at the center and reality is continuous, meaning dP/dR is zero at the center and approaches zero as R approaches zero.
Surface gravity of a sphere of constant density and radius R is proportional to R? Mass is proportional to volume (R^3) and surface gravity is proportional to mass/R^2, or R^3/R^2, or R.
Okay, I’ve got a new respect for the problems involved with using barometric pressure to measure altitude, and the advantages of using barometric pressure directly for navigational purposes at high altitudes.
At the time of writing, I thought I was paraphrasing Newton. Downbranch I realized that the behavior of being attracted to the center of a shell while inside the shell would, in a body composed entirely of fluid, yield a maximum first derivative of pressure at the center, and no maximum of pressure within the body.
At the center there is no net, but inside the sphere the net effect is the same as if all the mass were concentrated at the center. (Suppose that there was a normal net effect everywhere at and just above the surface and zero net effect just below the surface: that would require that the gravitational field not be continuous, which is not the case)
Remember that this is a spherical shell, not a sphere. When calculating the force of gravity inside the Earth, for example, you ignore all mass at higher radius that the location you’re interested in, but not the mass at lower radius.
Why does this work? Imagine opposite cones that originate at the center of the spherical shell. The intersection of that cone with the shell will have surface area that increases with r^2, but the inverse square law decreases with r^2, and so the gravitational effect only depends on the angle of the cone, regardless of distance- but the two cones are pointed in opposite directions, and so cancel out. (This is obvious at the center, but works just as well elsewhere inside the spherical shell.)
For an infinitely thin shell, the drop in gravitation is infinitely steep. But shells with real thickness will have a gradual dropoff that corresponds to that thickness.
… So pressure in a gaseous celestial body doesn’t increase linearly; delta-p falls off to zero in the center?
It’s not clear to me where that model came from. We shouldn’t expect the density of celestial bodies to be uniform unless they’re made of something incompressible, and it’s important to separate out the net force of gravity and the weight of mass above you. In steady state, all of the mass in the celestial body is being pulled towards the center by gravity, pushed upwards by the mass below it, and has to push upwards on the mass above it (with the net force being 0). I haven’t done the math to see what the pressure function would look like for a gaseous celestial body, and it seems like the full calculation will have lots of complications, but we can note that the mass below you has to push up harder than the mass above you is pushing down, suggesting the pressure is highest at the center.
delta-P is the first derivative of pressure; it would have to be zero at the center for there to be a pressure maximum at zero.
I would expect a gaseous body to have a roughly spherically symmetric mass distribution, which is all we need. Treat it as an infinite number of infinitely thin spheres each of uniform density, and we can do calculus on it.
We can also do this though experiment with a perfect liquid of uniform density; at least it will have a surface that we can stop at. Pressure is still highest at the center and reality is continuous, meaning dP/dR is zero at the center and approaches zero as R approaches zero.
Surface gravity of a sphere of constant density and radius R is proportional to R? Mass is proportional to volume (R^3) and surface gravity is proportional to mass/R^2, or R^3/R^2, or R.
Okay, I’ve got a new respect for the problems involved with using barometric pressure to measure altitude, and the advantages of using barometric pressure directly for navigational purposes at high altitudes.
Can you please use questions instead of confident contradiction when you’re up against Newton?
At the time of writing, I thought I was paraphrasing Newton. Downbranch I realized that the behavior of being attracted to the center of a shell while inside the shell would, in a body composed entirely of fluid, yield a maximum first derivative of pressure at the center, and no maximum of pressure within the body.