Did I Overlook Eddington Luminosity in Solar Wind Mass Loss Calculation?

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Hi there

Problem is:

Assuming spherical Symmetry, estimate the rate of mass loss from the sun, if at the Earth, the measured velocity of the solar wind is 400km/s and the proton density of the wind is roughly 7 particles/cm^3. Give your answer in solar Masses per year.

My attempted solution:

as proton density is 7particles per cm^3 this equals 7*10^6 particles per m^3.
as the speed is 400km per sec this is 4*10^5m per sec

multiply these values to get proton flux:
(7*10^6)*(4*10^5)=2.8*10^12protons m^-2 s^-1

find the surface area of the relevant sphere, this will have radius of 1AU as this flux is measured at the Earth:

A=4*3.14*(1.49598*10^11)^2=2.810873*10^23m^2

now we can find the total protons passing through the entire surface and of course this equals total protons leaving the sun:

(2.810873*10^23)*(2.8*10^12)=7.87044*10^35 protons s^-1

convert this to kg per second:

say proton mass= 1.6727*10^-27kg

so (7.87044*10^35)*(1.627*10^-27)=1316409794kg s^-1

convert to kg per year:

31536000 seconds per year so:

1316409794*31536000=4.151429928*10^16kg per year

find what fraction this mass is of the mass of sun:

(4.151429928*10^16)/(2*10^30)=2.0757*10^-14

so that means mass lost per year this way is (2.0757*10^-14)solarmasses per year.

**************************

When I first read this question, I thought it would involve some calculations regarding the Eddington Luminosity as this is what governs how much mass is blown away...however, the final method that I used did not need that, so can someone please confirm that I did not miss anything or over simplify?

Also, the question speaks of protons in the solar wind, is it safe to assume (as I did in the solution) that these are the only particles blown away?

Any feedback appreciated.
 
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I didn't check your calculations in detail, but the methodology looks good. And the answer sounds reasonable.

AStaunton said:
When I first read this question, I thought it would involve some calculations regarding the Eddington Luminosity as this is what governs how much mass is blown away...however, the final method that I used did not need that, so can someone please confirm that I did not miss anything or over simplify?
The eddington luminosity does not govern how much mass is blown away for the sun---because the sun is not 'eddington limited.' The eddington limit established an approximate maximum mass of a star, but that mass is about 2 orders of magnitude larger than our sun.

AStaunton said:
Also, the question speaks of protons in the solar wind, is it safe to assume (as I did in the solution) that these are the only particles blown away?
That is not only a good approximation, but also the standard one. Electrons are blown away at about the same rate, but their mass is negligible (2000x smaller than proton). Additionally, the amount of heavier elements (e.g. He) in the solar wind is small compared to H (protons). I'd assume the detailed solar-wind composition is roughly the same as the sun overall (thus you could look up the specific composition).
 
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