Question

The Earth effectively sits in a low-density H II region made up of the ionized solar wind. The wind
has is expanding constantly at about 400 kms−1 (i.e. the density decreases as r−2) and in the region of the Earth’s orbit, Ne = 10 cm−3. Estimate t and Tb at an observing frequency of 100MHz due to free-free absorption from this wind, at large angles from the Sun.

Answer 1

Answer on Question 55058, Physics / Astronomy | Astrophysics

Question:

The Earth effectively sits in a low-density H II region made up of the ionized solar wind. The wind has is expanding constantly at about $400\mathrm{~kms}^{-1}$ (i.e. the density decreases as $\mathbf{r} - 2$ ) and in the region of the Earth's orbit, $\mathrm{Ne} = 10\mathrm{~cm}^{-3}$ . Estimate t and Tb at an observing frequency of $100\mathrm{MHz}$ due to free-free absorption from this wind, at large angles from the Sun.

Answer:

Assume that we are observing in the anti-solar direction. Then, the emission measure (EM) is given by:

$$\left(\frac {E M}{A U c m ^ {- 6}}\right) = \int_ {1} ^ {\infty} \left(\frac {N _ {e}}{c m ^ {- 3}}\right) ^ {2} d \left(\frac {s}{A U}\right),$$

where the distances are measured in astronomical units (1AU = 4.848 × 10-6 pc). Ne is given by:

$$\left(\frac {N _ {e}}{c m ^ {- 3}}\right) = \frac {1 0}{\left(\frac {s}{A U}\right) ^ {2}}$$ $$\left(\frac {E M}{A U c m ^ {- 6}}\right) = \int_ {1} ^ {\infty} \frac {1 0 0}{x ^ {3}} d x = \frac {1 0 0}{3} \Rightarrow E M = 1. 6 1 \times 1 0 ^ {- 4} p c \times c m ^ {- 6}$$

Answer: EM=1.61×10-6 pc×cm-6

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