<p>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.</p>

CC BY-SA 3.0
2016-02-26T06:40:33.813Z
2016-02-26T06:40:33.813Z
xzackli
<blockquote>
  <p>$$\left( \frac{4 \pi (6.96e10\,\text{cm})^{2}}{1} \right)$$</p>
</blockquote>

<p>This is the primary source of your error. Your value of 6.96×10<sup>10</sup> cm is the radius of the Sun. The problem specifically said "<em>Assuming at the Earth ...</em>". 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 1.5×10<sup>13</sup> cm. This error alone makes your value low by a factor of about 50000. The remaining factor of two results mostly from using 10<sup>-24</sup> grams per amu.</p>

<p>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.</p>

CC BY-SA 3.0
2016-02-26T11:25:21.343Z
2016-02-26T11:55:32.867Z
2016-02-26T11:55:32.867Z
David Hammen
<p>This is problem 1-4 from Principles of Stellar Evolution and Nucleosynthesis by Clayton:</p>

<p>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.</p>

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

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

<p>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?</p>

<star><the-sun><solar-wind>
Rate of Mass Loss from the Solar Wind
Ryan Farber
# Rate of Mass Loss from the Solar Wind

(Asked by: username_0 on 2016-02-26)

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?

**Tags:** star, the-sun, solar-wind


---

## 2 answers

### Answer by username_2 on 2016-02-26. Score: 6 | Accepted answer

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

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.


---

### Answer by username_1 on 2016-02-26. Score: 6

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.


---

