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nacoconut
2010-Dec-03, 01:33 AM
I am having a tough time with this. I'm not a math person at all. Maybe I am doing it right, but probably not. Any help to guide me in the right direction would be very helpful. Thank you.

I am trying to figure out the density of the sun if its radius expanded to 100 times its current size.

Here's what I have:
D= m/v v=4/3Pir^3 Mass of Sun: 1.99x10^30kg Radius of Sun: 695, 500

V= (4/3Pi)(695,500)^3 = 1.409x10^18

x 100, for the radius increase:
v= 1.41x10^20
D= 1.99x10^30/1.41x10^20 = new density of 1.41x10^10 (g/km?) if per km, I changed it to cm and the number is still higher than the sun's current density....

I'm very confused!

Solfe
2010-Dec-03, 02:26 AM
For the mass, you have kg. For the volume, I believe you have km which will be cubed by the formula. 1 kg is a liter of volume, which is obvious much smaller than a km cubed. You need to have units that are equal.

When I have the feeling a unit is off, I can't help but notice, I am always right no matter how bad my math is. :)

astromark
2010-Dec-03, 02:53 AM
' nacoconut ' The solar mass does not increase if no mass is added. By doubling the size of the sun you are reducing its density...
What you are asking is purely a mathematical formula... Increase a hundred times and it still is the same object. Its just less dense and still weighs the same.

Hornblower
2010-Dec-03, 03:06 AM
If the radius in increased by a factor of 100 while not adding any mass, the average density will decrease to 1/1,000,000 of the present value. That is expected to happen when it is in a late red giant stage some billions of years from now.

Remember, density is inversely proportional to volume, which in turn is proportional to radius cubed.

nacoconut
2010-Dec-03, 03:27 AM
Solfe, I always get mixed up with those... thank you for pointing that out :)

hornblower, Okay, so, if the density of the sun is 1.4grams/cm^3 I should just divide that by 1,000,000 which brings me to the density of 1.4x10^-6, or .0000014 g/cm^3
is this correct? If so, where does that percent decrease come from exactly? Like I said, bad with math, but I would really like to completely understand how to arrive at this conclusion. The number makes sense to me since you brought up the stage that the Sun will be in when it reaches this density. It is more graspable when I have something to imagine, rather than just having numbers to look at.

To all: Thanks for your help!

nacoconut
2010-Dec-03, 03:32 AM
Thanks everyone.

hornblower, given that the current density of the Sun is 1.4g/cm^3, the new density, which is found by dividing by 1,000,000 would be: 1.4x10^-6 or .0000014. Is this correct? If so, where does that percentage difference come from? I would really like to completely understand this, like I said, bad with math! Thanks for mentioning the red giant stage, if the number is correct, I can imagine why this would work as I can visualize an actual concept instead of numbers.

Shaula
2010-Dec-03, 07:09 AM
Density = Mass / Volume
Volume = (4.pi.Radius^3) / 3 or to simplify Volume = k R^3 where k is a constant

So if D1 is the first density and D2 is the second we get D1 = K / R1^3 (where K is a constant equal to 3M/4pi) and D2 = K / R2^3

Now take the ratio so we can write:
D1/D2 = (K . R2^3) / (K . R1^3)
which simplifies to:
D1/D2 = (R2 / R1)^3

This means that we can work out D2 given D1 and the relative ratio of the two radii:
D2 = D1 . (R1 / R2)^3
You said that R2 = 100.R1 so the equation became
D2 = D1 (R1 / (100.R1))^3
D2 = D1 / (100^3)
D2 = D1 / 1000,000

To confirm:
Ms = 2e30kg
Rs = 6.955e8m
so Vs = (4/3) pi Rs^3 = 1.4e27
Making D1 = 2e30/1.4e27 = 1.4e3 kg/m3 = 1.4g/cm3
100Rs = 6.955e10m
making the new Vs = 1.4e33
and the new density
D2 = 2e30/1.4e33 = 1.4e-3 kg/m3 = 1.4e-6g/cm3
which is 1000,000 times lower as required

Basically in words the density is inversely proportional to the cube of the radius so keeping everything else the same and changing the radius makes the density change by a factor of 1 / (proportional change in radius^3)

Does that help a bit? I'm not the best explicatorator!

pzkpfw
2010-Dec-03, 07:31 AM
(

... 1 kg is a liter of volume ...

You mean the old 1 kg = 1 L of water? A kg of Gold wouldn't be a litre of volume...

)

Hornblower
2010-Dec-03, 01:25 PM
Addendum to my previous post: Let's review the actual number crunching you did in the opening post.

I am having a tough time with this. I'm not a math person at all. Maybe I am doing it right, but probably not. Any help to guide me in the right direction would be very helpful. Thank you.

I am trying to figure out the density of the sun if its radius expanded to 100 times its current size.

Here's what I have:
D= m/v v=4/3Pir^3 Mass of Sun: 1.99x10^30kg Radius of Sun: 695, 500

V= (4/3Pi)(695,500)^3 = 1.409x10^18

x 100, for the radius increase:
v= 1.41x10^20D= 1.99x10^30/1.41x10^20 = new density of 1.41x10^10 (g/km?) if per km, I changed it to cm and the number is still higher than the sun's current density....

I'm very confused!

My bold for reference. There is the major error you made when substituting the new radius. You simply increased the expression for the new v by a factor of 100, when it should have been the cube of the new radius, or a factor of a million. That threw your result off by a factor of 10,000. I did not see it the first time I read the post, having lost it in the visual clutter created by the actual numbers of kilograms and kilometers that were flying around.

There is no need in this type of exercise to crunch all of those numbers. If you have a density value that you trust for the actual Sun, just divide it by 1 million to get the density for the expanded star.

Cougar
2010-Dec-03, 01:46 PM
There is no need in this type of exercise to crunch all of those numbers.

You are SO right! The original question was:


I am trying to figure out the density of the sun if its radius expanded to 100 times its current size.

Increase the radius 100 times => the volume increases 1003 times => the density decreases by that same factor.

So whatever the density was, it is now 1003 times less. End of story!

DO NOT give me this CRAP "I'm not a math person..." There is no subject that is any more straightforward! "Math is hard" is an insidious MYTH! REJECT THE MYTH!

Cougar
2010-Dec-03, 01:55 PM
Of course, we're talking average density here. The Sun is more dense toward its center. In other words, the density varies with the radius. And it's non-linear to boot, since there are, like, shells of different elements and pressures. Yes, let's stay with average density!

antoniseb
2010-Dec-03, 03:07 PM
Cougar beat me to this. In the likely case that the Sun expands to 100 times its current radius, it is likely that the core of the Sun will become more dense during that time, as it runs out of easy Hydrogen to fuse, and compresses enough to start turning Helium into Carbon. The core temperature goes way up, and the outer envelope of the Sun becomes more rarified, and will have less than a millionth of the current envelope density.

Shaula
2010-Dec-03, 08:01 PM
There is no need in this type of exercise to crunch all of those numbers. If you have a density value that you trust for the actual Sun, just divide it by 1 million to get the density for the expanded star.
No there isn't but it helps to understand why a factor of a hundred converts into a million if you have seen the full version at least once.

nacoconut
2010-Dec-03, 08:08 PM
Cougar, I agree with what you said about math... to an extent. "I am not a math person" simply means that it takes a bit more for me to grasp mathematical ideas, than some who can instantly understand them. I am more of a visual person, so numbers are a bit confusing to me as I cannot see them in action as you may be able to. As you pointed out, it was incorrect to use so many numbers, I also see that I have made this far more complicated that it needed to be.

Thank you for your input.

AndreasJ
2010-Dec-03, 08:37 PM
No there isn't but it helps to understand why a factor of a hundred converts into a million if you have seen the full version at least once.

I suspect this is one of those individual modes of learning things.

Back in elementary school, I used to be puzzled why some of my classmates did various unnecessary steps in similar situations. But they usually got the correct answers and some of them grew up to be engineers, so they must've been doing something right.