<p>You seem to be confusing the simple mathematical relationship between radius and surface area, and the more complex relationship between mass and size.</p>

<p>If you double the radius of a sphere the surface area quadruples.  This is pure maths, and is not particular to stars. The volume is multiplied by 8.</p>

<p>But in a star, increasing the amount of matter by a factor of 8 will not result in the volume increasing by a factor of 8. The relationship between the mass of a star and it's volume is not simple and depends on factors like age, and the particular combination of elements in the star.</p>

<p>Doubling the radius does quadruple the surface area.</p>

<p>Doubling, or even increasing 8 fold, the mass does not quadruple the surface area.</p>

CC BY-SA 4.0
2019-09-08T20:39:41.643Z
2019-09-09T05:39:02.777Z
2019-09-09T05:39:02.777Z
James K
<p>I am researching about the radius of a star and a its surface area. One question I have is about the effect of changing radii in stars. If for example we have one star with radius <span class="math-container">$r$</span> and another one with radius <span class="math-container">$2r$</span>, we know that the area would mathematically be <span class="math-container">$A$</span> for the first star and <span class="math-container">$4A$</span> for the second star since <span class="math-container">$A=4 \pi r^2$</span>. However, I intuitively think that this wouldn't be the case since if I doubled the radius, the surface area will quadruple. But, that means that there will be more molecules and whatever 'materials' the star is made of, meaning that there would be a greater gravitational force towards the core of the star. Would it thus make sense to say that there would be some sort of 'compression' that occurs which pulls all of this extra mass towards the center (effectively making the star denser), and thus the star with <span class="math-container">$2r$</span> actually has a surface area of <span class="math-container">$3\frac{1}{2}A$</span> rather than <span class="math-container">$4A$</span>? </p>

<p>(So technically I'm asking if <span class="math-container">$A\neq 4\pi r ^2$</span> in case of stars because more mass (due to the extra radius) would mean that the gravitational force towards the core would be greater, making the star denser in the process, but also have less surface area than what doubling the radius would mathematically correspond to.)  </p>

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2019-09-09T07:22:09.313Z
A curious thing about stars is as they get more massive they become less dense.  The internal heat pushes the outer layers further outward and more massive stars generate much more internal heat.   Granted end of life stars don't follow this neat equation, but the densest main sequence stars are the small ones, the smaller red dwarfs or brown dwarfs if you count them as stars.   The higher mass main sequence blue stars have much lower density and considerably lower surface gravity.
userLTK
<star><stellar-structure>
About Radius and Surface Area
Anurag
# About Radius and Surface Area

(Asked by: username_0 on 2019-09-08)

I am researching about the radius of a star and a its surface area. One question I have is about the effect of changing radii in stars. If for example we have one star with radius $r$ and another one with radius $2r$, we know that the area would mathematically be $A$ for the first star and $4A$ for the second star since $A=4 \pi r^2$. However, I intuitively think that this wouldn't be the case since if I doubled the radius, the surface area will quadruple. But, that means that there will be more molecules and whatever 'materials' the star is made of, meaning that there would be a greater gravitational force towards the core of the star. Would it thus make sense to say that there would be some sort of 'compression' that occurs which pulls all of this extra mass towards the center (effectively making the star denser), and thus the star with $2r$ actually has a surface area of $3\frac{1}{2}A$ rather than $4A$?

(So technically I'm asking if $A\neq 4\pi r ^2$ in case of stars because more mass (due to the extra radius) would mean that the gravitational force towards the core would be greater, making the star denser in the process, but also have less surface area than what doubling the radius would mathematically correspond to.)

**Tags:** star, stellar-structure

> **Comments:**
> **username_1**: A curious thing about stars is as they get more massive they become less dense.  The internal heat pushes the outer layers further outward and more massive stars generate much more internal heat.   Granted end of life stars don't follow this neat equation, but the densest main sequence stars are the small ones, the smaller red dwarfs or brown dwarfs if you count them as stars.   The higher mass main sequence blue stars have much lower density and considerably lower surface gravity.  

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## 1 answers

### Answer by username_2 on 2019-09-08. Score: 6 | Accepted answer

You seem to be confusing the simple mathematical relationship between radius and surface area, and the more complex relationship between mass and size.

If you double the radius of a sphere the surface area quadruples. This is pure maths, and is not particular to stars. The volume is multiplied by 8.

But in a star, increasing the amount of matter by a factor of 8 will not result in the volume increasing by a factor of 8. The relationship between the mass of a star and it's volume is not simple and depends on factors like age, and the particular combination of elements in the star.

Doubling the radius does quadruple the surface area.

Doubling, or even increasing 8 fold, the mass does not quadruple the surface area.


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