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Is the Sun losing mass?

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  Is the Sun losing mass? Simon Roberts <retenshun@gmail.com> - 2018-03-14 00:21 -0700
    Re: Is the Sun losing mass? Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2018-03-14 21:47 +0100
      Re: Is the Sun losing mass? Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2018-03-14 23:50 +0100
    Re: Is the Sun losing mass? Simon Roberts <retenshun@gmail.com> - 2018-03-14 15:51 -0700
      Re: Is the Sun losing mass? Simon Roberts <retenshun@gmail.com> - 2018-03-14 17:15 -0700
        Re: Is the Sun losing mass? Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2018-03-19 03:34 +0100
          Re: Is the Sun losing mass? Volney <volney@volney.invalid> - 2018-03-19 00:04 -0400

#452604 — Is the Sun losing mass?

Is the Sun losing mass? 


I'm not a physicist. 

scenario (correct me if wrong) 

Fusion happens on the Sun. 

Due to this we have a certain intensity (power) of light at some distance from the Sun per unit area here on Earth. 

Considering all this and the equation E = Mc^2 what is the rate of loss of mass of the Sun? 




I did a rough calculation near a year ago. 

I must have made a huge mistake in my calculation since I still see the Sun. 

It's mass hasn't diminished so that it does not exist. 

I don't know. Like I said I'm not a physicist.   

I can balance a check book although I don't. Sometimes I overlook details. 


Verification? (of what, I'm unsure) 


Simon Roberts 

My Dropbox link: 

https://www.dropbox.com/sh/iu33ve2mfli3b9p/AABXNfj1AxTp7FNJJYmJqRAWa?dl=0 

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#452680

Simon Roberts wrote:

> Is the Sun losing mass?

Yes.

> I'm not a physicist.

Me neither.  Just a student for now.
 
> scenario (correct me if wrong)
> 
> Fusion happens on the Sun.

_In_ the Sun; to be precise, only in the *core* of the Sun.
 
> Due to this we have a certain intensity (power) of light at some distance
> from the Sun per unit area here on Earth.

Correct.  And perhaps more importantly, infrared radiation to allow for 
liquid water here, and UV radiation to allow for photosynthesis (if one 
follows the definition that “light” is only [human-]*visible* light).
 
> Considering all this and the equation E = Mc^2 what is the rate of loss of
> mass of the Sun?

Not much.  Google is your friend. [psf 6.1]

BTW, unless one wants to emphasize a larger mass in relation to a smaller 
mass, one uses “m” as the symbol for mass in physics, not “M”.
 
> I did a rough calculation near a year ago.
> 
> I must have made a huge mistake in my calculation since I still see the
> Sun.

Yes.
 
> It's mass hasn't diminished so that it does not exist.

Correct.  And that is not going to happen.  The end of the Sun (as we know 
it) and of all the inner planets, including Earth, will come in a far more 
spectacular way in any case (in about 2 to 5 billion years, short scale).
 
> My Dropbox link:
> 
> https://www.dropbox.com/sh/iu33ve2mfli3b9p/AABXNfj1AxTp7FNJJYmJqRAWa?dl=0

And now I should click through all folders to find your *wrong* calculation?

I really have better things to do.


PointedEars
-- 
Q: Why is electricity so dangerous?  
A: It doesn't conduct itself.

(from: WolframAlpha)

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#452708

Thomas 'PointedEars' Lahn wrote:

> Simon Roberts wrote:
>> Considering all this and the equation E = Mc^2 what is the rate of loss
>> of mass of the Sun?
> 
> Not much.  Google is your friend. [psf 6.1]
> 
> BTW, unless one wants to emphasize a larger mass in relation to a smaller
> mass, one uses “m” as the symbol for mass in physics, not “M”.
>  
>> I did a rough calculation near a year ago.
>> 
>> I must have made a huge mistake in my calculation since I still see the
>> Sun.
> 
> Yes.
>  
>> It's mass hasn't diminished so that it does not exist.
> 
> Correct.  And that is not going to happen.  The end of the Sun (as we know
> it) and of all the inner planets, including Earth, will come in a far more
> spectacular way in any case (in about 2 to 5 billion years, short scale).

FWIW, since I also always tell people that Usenet is not the Web, I thought 
it would be interesting and fun to do the calculation here myself (with a 
little help from Wolfram|Alpha).

The energy released by Sol through nuclear fusion can be roughly obtained 
from her luminosity, which has units of power (watt):

  L☉ ≈ 3.848 × 10²⁶ W

 1 W = 1 J∕s

So the energy released per second is (trivially)

  ∆E ≈ 3.848 × 10²⁶ J.

This corresponds [as a point of note, *exactly* as in Einstein’s 1905 
formulation in “Does the Inertia of a Body Depend upon its Energy-Content?”, 
although he used different symbols, and words [1]] to a loss of mass *per 
second* of

  ∆m = ∆E∕c² ≈ 4.281 × 10⁹ kg,

where c is the speed of light (in vacuum).

This may seem much, but consider that the current mass of Sol is

  M☉ ≈ 1.988435 × 10³⁰ kg.

10⁹ is 21 *magnitudes* less than 10³⁰, or less than *one sextillionth* of 
the current mass (if I know my numerals correctly).

So if Sol would only continue to lose mass through nuclear fusion, and I did 
not miscalculate, it would take

  ∆t = M☉∕((∆m)∕s)
     ≈ 1.988435 × 10³⁰ kg∕(4.281 × 10⁹ kg∕s)
     ≈ 4.667 × 10²⁰ s
     ≈ 1.479 × 10¹³ a

or ca. 14 trillion years (short scale), for her to "evaporate" this way.

However, as astronomer Phil Plait points out [2], Sol also loses mass 
through the solar wind (stream of charged particles) that it emits.  
Ignoring electrons because of their comparably small mass, Plait calculates 
that Sol loses “1.5 million tons of material” (protons) or 1.5 × 10⁹ kg per 
second through that.  Again, that is a lot in human measures – but consider 
what she is that we are talking about.

So it would take Sol "only"

  ∆t ≈ 1.988435 × 10³⁰ kg∕((4.281 + 1.5) × 10⁹ kg∕s)
     ≈ 3.44 × 10²⁰ s
     ≈ 1.09 × 10¹³ a

or ca. 11 trillion years to lose all her mass in both ways.  (No wonder 
she’s still around, yes? ;-))

<http://www.wolframalpha.com/input/?i=luminosity+of+sun>
<http://www.wolframalpha.com/input/?i=3.848%C3%9710%5E26+J+in+kg>
<http://www.wolframalpha.com/input/?i=4.281%C3%9710%5E9+kg+%2B+1.5+*+10%5E6+t>
<http://www.wolframalpha.com/input/?i=mass+of+Sol%2F5.781%C3%9710%5E9+kilograms>
<http://www.wolframalpha.com/input/?i=mass+of+Sol%2F(5.781%C3%9710%5E9+kilograms%2Fs)>

But as I alluded to, and Plait also mentions, she is not going to live a 
thousandth of that time.  Because once all the hydrogen is exhausted in 
about 5 billion (5 × 10⁹) years, helium fusion is going to set in where she 
will grow to be a red giant that extends beyond the orbit of Mars.  Once 
helium fusion is done, there will not be enough mass left to ignite fusion 
again, and the core is going to shrink to leave a white dwarf.

<https://en.wikipedia.org/wiki/Sun#After_core_hydrogen_exhaustion>

Sol’s death (as we know her) might come even earlier as the Andromeda Galaxy 
and the Milky Way are on a collision course and projected to merge in ca. 
_4.5_ billion years (_not_ 2).  Distances between stars are huge, and so it 
is unlikely that two stars would collide, but it remains a possibility.  
Certainly gravitational influence of other astronomical objects such as 
other stars could cause Sol to have a very different fate than a white 
dwarf.

<https://en.wikipedia.org/wiki/Andromeda_Galaxy>

HTH.


PointedEars
___________
[1] 
<http://www.slate.com/blogs/bad_astronomy/2014/07/14/solar_wind_versus_fusion_how_does_the_sun_lose_mass.html>
[2] <http://einsteinpapers.press.princeton.edu/vol2-trans/186>

-- 
Q: What did the nuclear physicist order for lunch?  
A: Fission chips.

(from: WolframAlpha)

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#452709

givens. 

the mass of the Sun as approximately 2E10^30 kg.

the intensity of sunlight near Earth as 13677 watts/m^2

the radius of orbit as aproximately 150E10^9 m.

the equation E = mc^2. (different 'E')

watt == (kg)m^2/sec^3.

result.

neglecting all else sun is losing approximately

1 part in 15E10^12 PER YEAR

1  : 15, 000, 000, 000, 000. per 365 days.

not to shabby.

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#452721

'what me worry?'

(version 2)

givens. 

the mass of the Sun as approximately m_sun = 2E+10^30 kg.

the power of sunlight near Earth as 13677 watts/m^2. <- [oops sloppy typo]

[should be '1367' not '13677' the typo does not affect the result].

the radius of orbit as approximately 150E+10^9 m.

the equation E = mc^2 (different 'E').

W == watt == (kg)m^2 / sec^3 = (joules / sec).

result.

neglecting all else the Sun is losing approximately

1 part in 15E+10^12 PER YEAR of its mass, m_sun.

1  : 15, 000, 000, 000, 000 [in 365 days].

1 : 15E+12 translated to 

a change of ( - 130 E+15 kg / year ) approximately but close in magnitude.

not to shabby.

Simon.

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#453187

Simon Roberts wrote:

> 'what me worry?'

Are you talking to yourself?
 
> (version 2)
> 
> givens.

Your English is bad, too.
 
> the mass of the Sun as approximately m_sun = 2E+10^30 kg.

The correct value is M☉ ≈ 1.988435 × 10³⁰ kg.  You must be more careful with 
your numbers: A difference of 0.02 × 10³⁰ = 2 × 10²⁸ already equals 20 
octillion kilograms here (short scale).

If you do not know the correct value, you can calculate it approximately by 
plugging the data of Terra’s (approximately circular) orbit into

   a_r = g

  v²∕r = G M☉∕r²,

where v is Terra’s average orbital speed (you get that from the average 
orbital radius r and the period).
 
> the power of sunlight near Earth as 13677 watts/m^2. <- [oops sloppy typo]

Power is measured in W, not W∕m².

The value is irrelevant unless you use it properly.
 
> [should be '1367' not '13677' the typo does not affect the result].

Yes, your result is *still* wrong.
 
> the radius of orbit as approximately 150E+10^9 m.

The radius of Terra’s orbit only matters if you actually use it.  So far 
there is no indication that you do.  Your calculation lacks all marks of
the required scientific rigor.  GIGO.

AISB, it is easier and better to use Sol’s luminosity (as *measured* by
e.g. NASA and subsequently *defined* as a unit [L☉] by the IAU).
 
> the equation E = mc^2 (different 'E').
> 
> W == watt == (kg)m^2 / sec^3 = (joules / sec).
> 
> result.

*After* I told you already.
 
> neglecting all else the Sun is losing approximately

Which is another reason why this is all wrong.
 
> 1 part in 15E+10^12 PER YEAR of its mass, m_sun.

Wrong, it is approximately 6.795 × 10⁻¹⁴ M☉/a due to stellar nuclear fusion.

<http://www.wolframalpha.com/input/?i=3.848%C3%9710%5E26+J%2Fc%5E2+*+86400+*+365.25%2F(mass+of+Sol)>
 
But note that this is an *approximation*: As Sol loses mass, she does not 
require as much radiation pressure to sustain herself against her self-
gravity, so the energy output decreases as well.

> 1  : 15, 000, 000, 000, 000 [in 365 days].
> 
> 1 : 15E+12 translated to
> 
> a change of ( - 130 E+15 kg / year ) approximately but close in magnitude.
> 
> not to shabby.

The correct value is 1.351 × 10¹⁷ kg/a.  You are off by ca. 5 quadrillion 
kg/a (short scale; ca. 3.7 %).
 
<http://www.wolframalpha.com/input/?i=3.848%C3%9710%5E26+J%2Fc%5E2+*+86400+*+365.25>

(In scientific notation, the base is always less than 10.)


PointedEars
-- 
A neutron walks into a bar and inquires how much a drink costs.
The bartender replies, "For you? No charge."

(from: WolframAlpha)

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#453193

On 3/18/2018 10:34 PM, Thomas 'PointedEars' Lahn wrote:
> Simon Roberts wrote:

>> the mass of the Sun as approximately m_sun = 2E+10^30 kg.
> 
> The correct value is M☉ ≈ 1.988435 × 10³⁰ kg.  You must be more careful with
> your numbers: A difference of 0.02 × 10³⁰ = 2 × 10²⁸ already equals 20
> octillion kilograms here (short scale).

He did write 'approximately'. <1% qualifies for such a discussion.

> Power is measured in W, not W∕m².

That's the solar constant, which is correctly measured in W∕m².

Volney

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