## Sunday, December 4, 2016

### How Much Would The Sun Weigh If Filled With Water?

Recently, I was listening to a radio show called "StarTalk" with the physicist Prof. Neil deGrass Tyson.  The episode was a panel of astrophysicists talking about the study of 'black holes' in our universe -- which was completely fascinating I might add.  During the discussion, there was a large amount of dimensional analysis going on in order to drive home the relevance and complexity associated with studying 'black holes' from Earth.  One statistic popped up out of nowhere:

If the Sun was filled with water, it would weigh nearly as much as it does now!

At the time, I was riding the train back home from vacation and did not have my laptop handy to look up a few values to explore the stated statistic further.  Since then, the statement has been lingering in the back of my head.  Today, I decided to perform a couple of calculations to either verify or debunk that statement.  Below are the result of those calculations.

# How Much Does The Sun Weigh?

In order to tackle such a calculation, a few values need to be known.  First, the weight of the Sun needs to be obtained.  If the weight of the Sun is not known, then a comparison based on an analysis is useless.  Second, the Sun will be approximated as a perfect sphere for calculation purposes.  Third, since the sphere is going to be used as the shape of the Sun, then a formula for the volume needs to be obtained.  Finally, the density of water needs to be known to use as a correlation factor of volume to weight.  Right about now you are probably thinking the following:

How do all of these values and formula come together to confirm the statement above?

I will show you in the paragraphs below.  In order fully grasp the nature of the analysis that is about to unfold, lets cast the Sun into a better perspective.  The weight of the Sun can be obtained from the resourceful "wikipedia" page along with other useful facts such as the description shown below:

The Sun is the star at the center of the Solar System. It is a nearly perfect sphere of hot plasma, with internal convective motion that generates a magnetic field via a dynamo process. It is by far the most important source of energy for life on Earth. Its diameter is about 109 times that of Earth, and its mass is about 330,000 times that of Earth, accounting for about 99.86% of the total mass of the Solar System. About three quarters of the Sun's mass consists of hydrogen (~73%); the rest is mostly helium (~25%), with much smaller quantities of heavier elements, including oxygen, carbon, neon, and iron.

Wow!  The description above sets the tone for the following analysis. First, the numbers involved are going to be VERY large (many zeroes before the decimal place).   The Sun is enormous and dwarfs the Earth easily in its description.  The weight of the Sun is listed on the 'wikipedia' page as the following:

The mass of the Sun above is expressed in 'Scientific Notation' to abbreviate the enormous number.  Typically, scientist use this 'abbreviated notation' to express huge numbers more easily.  At the same time, scientific notation can also be used to express very small numbers (i.e., a billionth of a meter, a nanometer = 1/1,000,000,000 meter).  If the mass of the Sun expressed in scientific notation was written out in long form, the mass would appear as shown below:

Alright.  After viewing the mass of the Sun in long form, the ease of using 'Scientific Notation' is completely understandable.

To start the calculation, the volume of a sphere needs to be known.  Below is the volume of a sphere in equation form:

In order to calculate the weight of the Sun filled with water the following steps need to be taken:

1) Obtain the radius (denoted as 'r') of the Sun

2) Calculate the volume of the Sun

3) Calculate the mass of the Sun from the volume (with the density of water)

4) Compare the calculated mass of the Sun to the stated mass (above) from 'wikipedia'

The steps are quite simple.  Keep in mind though, that in order to compare or calculate values, wthe values need to be in correct 'units' (i.e., 'kilogram,' 'gram,' 'milliLiter,' 'Liter,' or 'cubic meters').  Otherwise, completing calculations and comparing calculated volumes is impossible -- like 'comparing apples with oranges.'

With this in mind, lets start calculating the mass of the Sun filled with water following the steps above.  First, the radius of the Sun needs to be known.  From the 'wikipedia' page, the radius of the Sun is stated to be either 695,700 kilometers or 109 times the radius of the Earth.  Expressed  in long form for the calculation, the radius of the Sun is shown below:

The radius of the Sun can be directly plugged into the equation for the volume of a sphere above to yield the following:

The volume of the Sun is shown above.  In order to calculate the mass from the volume, we need the density of water.  Density is the amount of mass contained in a given volume.  For water, the value of the density is 1.00 gram/milliLiter.  Since there are 1000 milliLiters in a single Liter, then the density for water can be expressed as 1000 gram/Liter.

The expression for density of a given molecular compound can be expressed as the 'mass' per 'volume' as shown below:

The first line shows the equation for the density of a given molecular compound.  In the second line, the equation is re-arranged to yield the 'mass' from the two parameters 'density' and 'volume'.  Plugging in the values from the calculation and reference (density) value, the 'mass' of the Sun can be calculated as shown below:

The calculated mass of the Sun is shown above.  Of course, the value of the mass is enormous as expected -- which is good.  Returning to the statement above -- which motivated the article:

If the Sun was filled with water, it would weigh nearly as much as it does now!

The following question can be asked regarding the calculated mass and the reference mass listed from 'wikipedia':

How do the two masses (theoretical and calculated) compare to each other?

The easiest way to compare two value (in the same units -- i.e. 'kilogram') is to express them as a ratio of each other as shown below:

The result shows that the two values have the same "order of magnitude" -- that is 10 raised to the power of 30.  But the two values are NOT nearly the same -- Why Not?   Read onto find out.

## Plasma Is Denser Than Water

In carrying out the calculation above, there are a number of assumptions listed below:

1) The Sun is shaped as a sphere

2) Density of water at 25 degrees Celsius was used in the calculation

What was not figured in was that the density of an mass can change with temperature.  The Sun has a reaction going on in the core of the sphere.  There are four states of matter: Gas, Liquid, Solid, and Plasma.  In the calculation above, the state of matter used for water was a liquid -- 1.00 gram/milliLiter at 25 degrees Celsius.

According to the 'wikipedia' page for 'plasma,' the density can change significantly with temperature as shown in the image taken below:

What does this mean in the overall calculation?

Why should I (you -- the reader) care?

Understanding the accuracy of an approximation is crucial to the words that are used to describe the comparison of two values.  In the situation of comparing two masses -- the result of the calculation and the theoretical mass from 'wikipedia' -- you can easily see that the two are not comparable.  In order to understand the reason why that is the case or might be -- look toward the density of the molecule in question -- in this case water.

The density inside plasma can vary up to 7 orders of magnitude different from that in room temperature.  Meaning, you can pack more mass into a given space.

### Conclusion...

The above calculation is a great opportunity to illustrate the methodology of carrying out a 'dimensional analysis' problem.  Based on the result of the calculation, the statement above regarding the mass of the Sun filled with water nearly being equal to that of the current weight is not necessarily true.  Although, the two values did have the same 'order of magnitude' -- 10 raised to the power of 30.

Nevertheless, the exercise was fun and displayed the power of 'dimensional analysis.'  Now, you have the ability to carry out the same calculations on your own.  The next time that you are listening to the radio or reading the paper and find an interesting fact, you can verify the two values on your own.

Until next time, Have a great day!

#### 1 comment:

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