B Gravity at the Core of the Sun: How Strong is it?

[Kingpin1880]

I'm not 100% sure if this question has been asked/answered or not but I was curious about this subject and I fear I'm not knowledgeable enough to figure out this question for myself. I know that the gravity at the surface of the sun is roughly 28g, but as gravity gets more intense the closer to the centre of mass of the celestial body in question; how strong would gravity be closer to / within the core?
I'll be happy with a rough estimate if it's a bigger question than I think it is; I fell out of love with the study of science after high school, but I still think it a fascinating subject.
(here's hoping I didn't sound pretentious... also; not sure what "prefix" to use so sorry if this question's in the wrong place)

 

[BOAS]

I'm not 100% sure if this question has been asked/answered or not but I was curious about this subject and I fear I'm not knowledgeable enough to figure out this question for myself. I know that the gravity at the surface of the sun is roughly 28g, but as gravity gets more intense the closer to the centre of mass of the celestial body in question; how strong would gravity be closer to / within the core?
I'll be happy with a rough estimate if it's a bigger question than I think it is; I fell out of love with the study of science after high school, but I still think it a fascinating subject.
(here's hoping I didn't sound pretentious... also; not sure what "prefix" to use so sorry if this question's in the wrong place)

Edit - This post needs to be prefaced with the fact that it does not take the Sun's varying density into account.

There is a famous result (called the Shell Theorem) which says amongst other things, that only the mass inside a sphere contributes to the force of gravity felt at the surface of that sphere.

So as you move closer to the core of the sun, this imaginary sphere is getting smaller, containing less and less mass and thus the force of gravity is decreasing.

Due to the spherical symmetry of the setup, the attractive forces from outside the sphere, when added up cancel each other exactly.

Another result of the shell theorem is that a spherically symmetric mass behaves as if it were a point mass located at it's center of mass. This is convenient when you are outside the volume of the mass, but is obviously not applicable when inside the object.

https://en.wikipedia.org/wiki/Shell_theorem

 

[DrStupid]

So as you move closer to the core of the sun, this imaginary sphere is getting smaller, containing less and less mass and thus the force of gravity is decreasing.

Did you consider the increased density?

 

[mfb]

Yes - This does not change my reasoning.

It should. The mass decreases, but radius goes down as well.
As an example, within 0.1 of the solar radius, we have 0.077 of the mass (where a constant density would just give 0.001). That leads to a gravitational acceleration of 0.077/0.1² = 7.7 times the surface gravity.
Data source

I attached a graph (x-axis is radius, y-axis is g relative to g at the surface). Over most of the sun's radius, gravitational acceleration is stronger than at the surface.

The same effect, just weaker, is present in Earth as well. As you go down through the outer mantle, gravitational acceleration increases.

 

[BOAS]

It should. The mass decreases, but radius goes down as well.
As an example, within 0.1 of the solar radius, we have 0.077 of the mass (where a constant density would just give 0.001). That leads to a gravitational acceleration of 0.077/0.1² = 7.7 times the surface gravity.
Data source

I attached a graph (x-axis is radius, y-axis is g relative to g at the surface). Over most of the sun's radius, gravitational acceleration is stronger than at the surface.

The same effect, just weaker, is present in Earth as well. As you go down through the outer mantle, gravitational acceleration increases.

Yes, I realized my error. Sorry for deleting that post - I was hoping I was fast enough.