A practical (astronomical) math problem

[tim-from-pa]

This does not require deep math to solve, but an understanding of terms and spatial relationships in order to do so.

From Wikipedia we read about the sun:

The Sun is the star at the center of the Solar System. It is almost perfectly spherical and consists of hot plasma interwoven with magnetic fields.<sup class="reference" id="cite_ref-12">[12]</sup><sup class="reference" id="cite_ref-13">[13]</sup> It has a diameter of about 1,392,684 km,<sup class="reference" id="cite_ref-arxiv1203_4898_5-2">[5]</sup> about 109 times that of Earth, and its mass (about 2×10³⁰ kilograms, 330,000 times that of Earth) accounts for about 99.86% of the total mass of the Solar System.

And from another article about the earth, we read:

Mean density 5.515 g/cm³

Here's the questions:

1) From just the information quoted, the sun's mean density is what percentage of the earth's density? (we realize that density varies greatly depending where in the sun the matter is located) Show your work.

2) If we have a pool large enough, will the sun sink or float? Show your work.

NOTE: Density= mass per unit volume. Mean density in this problem assumes equal distribution (density) of matter. Water is defined as a density of one gram per centimeter cubed, or simply one.

The idea of this exercise is to stir up curiosity as one reads data about something. How does it compare to another object? What is the relationship? Or don't we care?

 

[Knotical]

A qualifying question: How porous is the sun?

 

[Claudya]

1) From just the information quoted, the sun's mean density is what percentage of the earth's density? (we realize that density varies greatly depending where in the sun the matter is located) Show your work.

Diameter of Sun = 190x Diameter of earth.
so volume of Sun = 3591364,002 x Volume of earth
(V = π·d3/6)

Mass of Sun = 330k x Mass of Earth

Density = Mass/ Volume

so Density of Sun = 330,000 x Mass of Earth/3591364 Volume of Earth

= roughly 0.1xDensity of Earth

I'm kinda unhappy with that result. I'll do the math again on paper.

2) If we have a pool large enough, will the sun sink or float? Show your work.

Nope. The roughly 6000 Kelvin of the sun's surface would immediately vaporise the water.

 

[tim-from-pa]

Diameter of Sun = 190x Diameter of earth.
so volume of Sun = 3591364,002 x Volume of earth
(V = π·d3/6)

Mass of Sun = 330k x Mass of Earth

Density = Mass/ Volume

so Density of Sun = 330,000 x Mass of Earth/3591364 Volume of Earth

= roughly 0.1xDensity of Earth

I'm kinda unhappy with that result. I'll do the math again on paper.


Nope. The roughly 6000 Kelvin of the sun's surface would immediately vaporise the water.

I'll let you continue to figure this out, but you definitely are on the right track. One hint. Try to think a little more simplistically regarding the ratio of the volumes. Both earth and sun are spheres, right? ;)

 

[Claudya]

We have:
Diameter Sun = 1.4 * 10^6 km
Mass Sun = 2 * 10^30 kg

Density Earth = 5.515 g/cm³

We need:
- the density of the sun in g/cm³ (same units would be neat if we want to compare)
- before we can calculate the sun's density we need to know the sun's volume first, we only know the diameter (unless we cheat and look up the sun's volume on wikipedia ;))

Volume of the Sun
V = 4/3 * Pi * r³; r = radius = d/2
V Sun = 4/3 * Pi * (1,4 * 10^6 km/2)³
V Sun = 4/3 * Pi * (7*10^5 km)3
V Sun = 143675504024173800 km³ = 1.4 *10^18 km³

Density of the Sun
Density Sun = m Sun/ V Sun

We want that to be in g/cm³, so we gotta transform the unit we have.

m Sun = 2 * 10^30 kg = 2 * 10^33 g. That's the easy one.
V Sun = 1.4 * 10^18 km³ = 1.4 *10^27 m³; because 1 km³ = 1 * 10^9 m³ because we're looking at volume units, not mere distance units.
1.4 *10^27 m³ = 1.4 * 10^33 cm³; because 1 m³ = 1 * 10^6 cm³
So V Sun = 1.4 * 10^33 cm³

Density Sun = 2 * 10^33 g/ 1.4 * 10^33 cm³ = 1.43 g/cm³

Comparing Earth and Sun
Density Earth = 5.515 g/cm³
Density Sun = 1.43 g/cm³

Can you do the percentage stuff yourself? I want to go to bed. -_-

As for the other question I stick with the previously given answer.

 

[Sparkey]

We have:
Diameter Sun = 1.4 * 10^6 km
Mass Sun = 2 * 10^30 kg

Density Earth = 5.515 g/cm³

We need:
- the density of the sun in g/cm³ (same units would be neat if we want to compare)
- before we can calculate the sun's density we need to know the sun's volume first, we only know the diameter (unless we cheat and look up the sun's volume on wikipedia ;))

Volume of the Sun
V = 4/3 * Pi * r³; r = radius = d/2
V Sun = 4/3 * Pi * (1,4 * 10^6 km/2)³
V Sun = 4/3 * Pi * (7*10^5 km)3
V Sun = 143675504024173800 km³ = 1.4 *10^18 km³

Density of the Sun
Density Sun = m Sun/ V Sun

We want that to be in g/cm³, so we gotta transform the unit we have.

m Sun = 2 * 10^30 kg = 2 * 10^33 g. That's the easy one.
V Sun = 1.4 * 10^18 km³ = 1.4 *10^27 m³; because 1 km³ = 1 * 10^9 m³ because we're looking at volume units, not mere distance units.
1.4 *10^27 m³ = 1.4 * 10^33 cm³; because 1 m³ = 1 * 10^6 cm³
So V Sun = 1.4 * 10^33 cm³

Density Sun = 2 * 10^33 g/ 1.4 * 10^33 cm³ = 1.43 g/cm³

Comparing Earth and Sun
Density Earth = 5.515 g/cm³
Density Sun = 1.43 g/cm³

Can you do the percentage stuff yourself? I want to go to bed. -_-

As for the other question I stick with the previously given answer.

You go, Claudia. I'm impressed.

 

[Deborah13]

Me too. Wow girl!

 

[tim-from-pa]

For all the folks reading this here, this is why I love to hear from Claudia. She's very insightful and intelligent. Her answer is absolutely correct and there is no flaw or error in her work. None.

However, that being said, I will show tomorrow a shorter way to figure this out. It comes with experience I suspect, because when I was younger I used to figure out everything in absolute terms ie, calculate the actual values and then compare. I'll show a quicker way how this can be done using relative values (ratios) and all this could have been done using a simpler calculation. In that case, had Claudia done it this way, she would have gotten to bed earlier. Again, this comes with experience playing around with math. And to this very day, there's still stuff I figure the long way when I'm not sure.... again, nothing wrong with that.

Sweet dreams Claudia! :love2

 

[tim-from-pa]

OK, G' mornin' all and I'm back again. I'm going to show a simple (but not implying superior) way to calculate those answers.

We know that the sun and earth are spherical, (or closely so) and that in both cases we use the formula

<style type="text/css">P { margin-bottom: 0.08in; }</style> 4/3 [FONT=Times New Roman, serif]π[/FONT][FONT=Times New Roman, serif] r[/FONT]^{[FONT=Times New Roman, serif]3[/FONT]}

However, let's think this way: Volume sun/Volume earth, i.e. how much greater (ratio) of the earth is the sun's volume? If we divide each formula by itself we see that the

4/3 [FONT=Times New Roman, serif]π[/FONT]

cancels and we are left with the ratio of the radius cubes. Now.... the article states the sun is about 109 times bigger (making the earth a unit radius =1) or 109^3 the volume = 1295029 times that of earth. If the sun was the same density as the earth, then the corresponding mass should likewise be as great, but it is not. The article states it is 330000 times the mass.

So... for problem #1, the fraction of the density would work out to be 330000/109^3
or .2548 a little over 25%.

Now, let's check. (.2548) (5.515)= 1.405g/cm^3, which is the answer to the second question. Claudia's answer is more precise because she is working with exact numerical values whereas I'm using numbers "about" 109 times the size of the sun.The slight difference is from estimation. The sun would sink given water's density is 1 (anything greater sinks) but as Claudia pointed out, not without an immense amount of evaporation and probably big explosions!

When one gets to be a old crony like me, we tend to find shortcuts because we get out of breath even doing long math problems.

But I hope that those here with at least some math background can appreciate both methods and why they both derived the same thing and both ways are perfectly good to use. As a matter of fact, a good mathematician finds alternate ways to work the same problem, thus the answer becomes self-checking when agreement is reached.

Great Job Claudia!