# Why is pressure in the outermost layer of a star lower than at its center?

I have done the math and I have obtained the hydrostatic pressure in a star is lower at the outermost layer of a star than in its center, where the pressure is actually maximum. Although the equations tell me this is right, I can't provide an intuitive explanation as to why this happens. Considering the only force keeping the star together is gravitational in nature, then shouldn't pressure be null in the center, where there is no mass to exert any force in nearby layers, and maximum in the outermost layer, where there is the most mass enclosed?

• Well, the outer layers still exist and "want to" get into the center of gravity. Thus the pressure. Commented Jul 1 at 21:31
• Commented Jul 2 at 8:20
• I don't know anything about physics, but it seems intuitive to me that the pressure is lower in the outermost layer, because there isn't anything above it to exert downward pressure. Commented Jul 3 at 17:50
• When you go swimming in the ocean, as you dive deeper you can feel the pressure in your ears build up. Commented Jul 4 at 21:25

Considering the only force keeping the star together is gravitational in nature,

Gravity is also what's keeping our oceans stuck to the Earth's surface. Where's the pressure at its highest, near sea level or at the bottom of the Mariana Trench?

You're only thinking of how the matter at the center of the star does not experience gravity in and of itself. This part is correct, but it does not provide a full picture.

What's you're forgetting about is that this central matter has a bunch of other matter stacked on top of it, which does have weight and exerts its weight on that central matter.
Ignore fluid/gas dynamics for what I just said. Even if matter moves around in the star, whatever matter is currently at the center is going to be "carrying" all the matter further away from the center. That "carry weight" is what's causing the increased pressure.

In a way, you're asking why the bottom people in a human pyramid carry more weight than the person at the top of the pyramid. The key difference is between having weight and carrying it. Pressure is derived from the latter, not the former.

There is a lot of mass above the centre, pushing down on it. On the surface there is no mass above. Hence you get high pressure at the centre and zero at the surface.

This is an exercise in hydrostatics. Imagine the sun consisting of a sphere of water. The deeper you go beneath the surface, the more water weight you have bearing down on you from above, and the inability of a liquid like water to support static shear stresses means that that weight gets translated into pressure acting on the surface area of an object immersed in it.

At the surface, the depth equals zero and the hydrostatic pressure is likewise zero.

Perhaps the confusion in the question arises because you are thinking of a rigid spherical shell of constant density, and you know that inside such a shell the gravitational field is zero?

A ball of gas is not rigid (and on planetary scales, neither is a ball of rock, as pointed out by Peter in the comments). Therefore, the inner layers must be at sufficiently high pressure to prevent the outer layers from collapsing in. Your statement that "the only force keeping the star together is gravitational in nature" is a bit misleading, because the structure of the star is determined both by gravitational forces and the forces due to pressure.

• Yes and in that sense the difference between the two problems is interesting. Commented Jul 2 at 14:18
• Interesting idea that the pressure inside a rigid body could indeed be zero because the layers around it act like vaults in architecture, carrying all the load. But note that "rigid" on the planetary scale is not obvious: What's rigid on the human scale, e.g. rock, behaves essentially like a fluid on the planetary scale which is why they take on a spherical shape. Indeed, this is one criterion of a planet. Commented Jul 3 at 12:58
• @Peter-ReinstateMonica Good point, I've edited my answer. Commented Jul 4 at 1:44

I believe you are confusing "force of gravity" with "pressure." The gravitational pull (or local acceleration $$g$$) is indeed zero at the center of the Sun, and a maximum elsewhere in the Sun (wherever $$GM/r$$ is greatest, depending on the radial distrubution of matter in the Sun).

If you had an indestructible suit and traveled to the Sun, you would fall and sink through the plasma before coming to rest in the center, and float with no net gravity. This is the same as what would happen on Earth.

But the pressure $$P$$ at any point is the summation of all the gravity $$g$$ and density $$\rho$$ for all regions above that point. To say it mathematically:

$$P(r) = \int^{R}_r \rho g ~dr$$

where $$R$$ is the (surface) radius of the Sun.

If we set $$r=R$$ (at the surface), then the integral goes from $$\int_R^R$$, or zero distance, and so $$P(R)=0$$, whereas at the center, $$r=0$$ so the integral goes from $$\int_0^R$$, and $$P(0)$$ is at a maximum.

• I don't think you'd fall all the way to the centre because the density is very high there, ~200 g/cm^3. Commented Jul 2 at 4:43
• It depends greatly on the nature of the unphysical suit I've proposed, but yes, fair point. If you could propel yourself to the exact center, I think you would be suspended in unstable equilibrium against buoyancy in all directions, but still measure no local gravity. Commented Jul 2 at 4:51
• Interesting: "g has its maximum at the Sun's surface". That must depend on the density gradient, mustn't it? With a very low-density surface layer and a dense core, gravity may continue to increase below the "surface" (which is also, in the case of the Sun, a questionable term, adding to my argument). Commented Jul 3 at 13:03
• And indeed: The maximum gravity should be far inside the Sun, around 0.16 R, based on this Sun data set. Screw the gas, get me to the metallic core! ;-) Commented Jul 3 at 13:40
• Fair point. I may have overlooked that detail Commented Jul 3 at 14:43

Imagine you are sandwiched between two moon like objects each with mass M. The forces acting on you is $$F_{centre} = \frac{2 \times G\times M\times M}{D^2}$$ where D is the distance from the centre of one moon to the other. The pressure exerted on you would be very large. Your mass is not even in the equation because in this context it is negligible. This is an extremely crude approximation of being at the centre of a larger moon. If on the other hand you were on the surface of the moon with a combined mass of the original two moons, the force acting on you due the surface gravity of the moon is given by $$F_{surface} =\frac{2^{4/3} \times M\times m}{D^2}$$. This force is much less because $$m<.

More intuitively, imagine a hole that is dug to the centre of the Earth. If you were placed at that bottom of that hole and all the removed rubble was replaced on top of you, you would be feeling a lot more pressure than if you were standing on top the replaced rubble. Or on the surface of the Earth, would you be better off standing on top of a one tonne block of concrete or have the block of concrete placed on top of you?