This is problem 1-4 from Principles of Stellar Evolution and Nucleosynthesis by Clayton:

Assuming at the Earth a characteristic velocity of 400km/s and density of 10amu/cm$^{3}$ for the solar wind, calculate the rate of mass loss for the sun.

There weren't any formulas about this in the section so I made a stab at it with dimensional analysis.

$$\frac{dM}{dt} = \frac{\rho V}{\Delta t} = \rho v A$$ $$\frac{dM}{dt} = \left( \frac{10 amu}{cm^{3}} \right) \left( \frac{400km}{s} \right) \left( \frac{4 \pi (6.96e10 cm)^{2}}{1} \right) \left( \frac{10^{5}cm}{km} \right) \left(\frac{10^{-24} g}{1 amu} \right) \left( \frac{M_{\odot}}{2 \times 10^{33} g} \right) \left( \frac{3600s}{hr} \right) \left( \frac{24 hr}{day}\right)\left(\frac{365day}{yr}\right)$$ $$\frac{dM}{dt} = 3.84 \times 10^{-19} M_{\odot} / yr $$

However, the answer given in the book is $0.4 \times 10^{-13} M_{\odot} / yr$. So, I'm off by about five magnitudes. Can anyone point out where I went wrong and/or point me in the correct direction?


$$\left( \frac{4 \pi (6.96e10\,\text{cm})^{2}}{1} \right)$$

This is the primary source of your error. Your value of 6.96×1010 cm is the radius of the Sun. The problem specifically said "Assuming at the Earth ...". You need to calculate the flux through the surface of a sphere whose radius is about one astronomical unit rather than one solar radius. The astronomical unit is 149597870700 meters (exactly), or about 1.5×1013 cm. This error alone makes your value low by a factor of about 50000. The remaining factor of two results mostly from using 10-24 grams per amu.

Dimensional analysis can only take you so far. While your result is dimensionally correct, you didn't think enough about the nature of the problem.


The velocity and the density are at the earth position, so the area term must include the earth-sun distance instead of the solar radius.


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