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IOAA 2019, Теория — Задача 7. Effect of sunspots on solar irradiance

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Theory problem sheet · 20 т.

Условие

Since 1978, the solar constant has been almost continuously measured by detectors on-board artificial satellites. These accurate measurements revealed that there are seasonal, monthly, yearly, and longer timescale variations in the solar constant. While the seasonal variations have their origins in the periodically varying Earth–Sun distance, the decade-long quasi-cyclic variations mainly depend on the activity cycle(s) of the Sun.

a) Calculate the value of the solar constant at the top of the Earth's atmosphere, when the Earth is 1 au from a perfectly quiet Sun, assuming that Sun emits as a perfect black-body. [4 т.]

b) Calculate the solar constant of this perfectly quiet Sun in early January and early July, and find their ratio. [4 т.]

c) Calculate the solar constant again at 1 au in the presence of a near equatorial sunspot with mean temperature of $T_{\mathrm{sp}} = 3300\ \mathrm{K}$ and diameter of sunspot, $D_{\mathrm{sp}} = 90\ 000\ \mathrm{km}$ . Calculate the ratio - blank Sun to Sun with sunspot.

Assume the sunspot is circular and ignore the effects of its spherical projection. Neglect any other activity features. Also assume that the Sun is rotating fast enough, hence solar irradiance is still isotropic. [7 т.]

d) In reality solar irradiance is no longer isotropic. Calculate the ratio of solar irradiance for the cases when the sunspot is not visible from the Earth to the case when it is fully visible. [5 т.]

Решение

a)

The Sun's total power can be calculated from the Stefan-Boltzmann law:

\text{(7.1)}\quad P = A\sigma T^4,\qquad \text{(1 p)}

where $A$ is the emitting surface, $\sigma$ is the Stefan-Boltzmann constant and $T$ is the (effective) temperature.
The total power emitted homogeneously and isotropically by the Sun ( $P_\odot$ ) distributes uniformly on the surface area of a sphere with radius $d_\oplus = 1\ \mathrm{au}$ . Hence:

\text{(7.2)}\quad I_\odot = \frac{P_\odot}{4\pi d_\oplus^2} = \frac{\sigma 4\pi R_\odot^2 T_\odot^4}{4\pi d_\oplus^2} = \sigma\frac{R_\odot^2 T_\odot^4}{d_\oplus^2},\qquad \text{(2 p)}

where $R_\odot$ and $T_\odot$ denote the radius and effective temperature of the Sun, respectively.
With the numerical values taken from the Table of constants:

I_\odot = \boxed{1365.9\ \mathrm{W\ m^{-2}}}\qquad \text{(1 p)}

b)

In the first days of January the Earth is in perihelion, while in the beginning of July it is in aphelion. Therefore, instead of the mean Earth-Sun distance (i.e. the semi-major axis of the Earth's orbit), now one should substitute the perihelion and aphelion distances, respectively. Therefore, one has to write that:

\text{(7.3)}\quad I_{\odot,\mathrm{perihelion}} = \frac{P_\odot}{4\pi\left(d_\oplus(1-e)\right)^2} = \sigma\frac{R_\odot^2 T_\odot^4}{\left(d_\oplus(1-e)\right)^2} = \frac{I_\odot}{(1-e)^2},\qquad \text{(1 p)}

\text{(7.4)}\quad I_{\odot,\mathrm{aphelion}} = \frac{P_\odot}{4\pi\left(d_\oplus(1+e)\right)^2} = \sigma\frac{R_\odot^2 T_\odot^4}{\left(d_\oplus(1+e)\right)^2} = \frac{I_\odot}{(1+e)^2},\qquad \text{(1 p)}

where $e = 0.016\,710\,22$ is the eccentricity of the Earth's orbit (see the Table of constants).
With numerical values:

in early January: $I_{\odot,\mathrm{perihelion}} = \boxed{1412.7\ \mathrm{W\ m^{-2}}}$ (1 p)

in early July: $I_{\odot,\mathrm{aphelion}} = \boxed{1321.4\ \mathrm{W\ m^{-2}}}$ (1 p)

The ratio of the values: $I_{\odot,\mathrm{perihelion}}/I_{\odot,\mathrm{aphelion}} = \boxed{1.0691}$ . The perihelion irradiation is higher by about 7 %. (1 p)

c)

In this case the total power of the Sun becomes

\text{(7.5)}\quad P_\odot' = \sigma\left[T_\odot^4(A_\odot - A_{\mathrm{sp}}) + T_{\mathrm{sp}}^4 A_{\mathrm{sp}}\right],\qquad \text{(2 p)}

where $A_\odot$ stands for the solar surface, while $A_{\mathrm{sp}}$ is the area of the spot (as a circle).
Thus, the solar constant takes the following form:

\text{(7.6)}\quad I_\odot' = \frac{P_\odot'}{4\pi d_\oplus^2} = \frac{\sigma\pi\left[T_\odot^4(4R_\odot^2 - R_{\mathrm{sp}}^2) + T_{\mathrm{sp}}^4 R_{\mathrm{sp}}^2\right]}{4\pi d_\oplus^2} = \frac{\sigma}{4d_\oplus^2}\left[T_\odot^4(4R_\odot^2 - R_{\mathrm{sp}}^2) + T_{\mathrm{sp}}^4 R_{\mathrm{sp}}^2\right] =

= I_\odot\left[1 - \left(\frac{R_{\mathrm{sp}}}{2R_\odot}\right)^2 + \left(\frac{R_{\mathrm{sp}}}{2R_\odot}\right)^2\left(\frac{T_{\mathrm{sp}}}{T_\odot}\right)^4\right] = I_\odot\left[1 - \left(\frac{R_{\mathrm{sp}}}{2R_\odot}\right)^2\left(1 - \left(\frac{T_{\mathrm{sp}}}{T_\odot}\right)^4\right)\right] = \boxed{0.9991 I_\odot}\quad \text{(2 p)}

With numerical value:

I_\odot' = \boxed{1364.6\ \mathrm{W\ m^{-2}}}\qquad \text{(1 p)}

d)

As we have seen, the ratio is

A = \frac{I_\odot'}{I_\odot} = \frac{1364.64\ \mathrm{W\ m^{-2}}}{1365.92\ \mathrm{W\ m^{-2}}} = \boxed{0.9991},

or, converting it into percentage, the solar constant in the case of the spotted Sun will be lower by 0.09 %.
When this sunspot emerges in the surface of the Sun, solar emission no longer will be isotropic, i.e. uniform in every directions. When the sunspot is fully seen from the Earth, the ratio of the irradiation of the Earth in the spotted and unspotted cases will be equal to the ratio of the solar powers in the two cases, i.e.,

\text{(7.7)}\quad P_{\odot,\mathrm{disk}} = \sigma T_\odot^4 A_\odot = \sigma\pi T_\odot^4 R_\odot^2\qquad \text{(1 p)}

\text{(7.8)}\quad P_{\odot,\mathrm{disk}}' = \sigma\left[T_\odot^4(A_\odot - A_{\mathrm{sp}}) + T_{\mathrm{sp}}^4 A_{\mathrm{sp}}\right] = \sigma\pi\left[T_\odot^4(R_\odot^2 - R_{\mathrm{sp}}^2) + T_{\mathrm{sp}}^4 R_{\mathrm{sp}}^2\right]\qquad \text{(1 p)}

and therefore, the ratio of the irradiations (or the effective solar constants) can be written as:

\text{(7.9)}\quad A' = \frac{P_{\odot,\mathrm{disk}}'}{P_{\odot,\mathrm{disk}}} = \frac{T_\odot^4(R_\odot^2 - R_{\mathrm{sp}}^2) + T_{\mathrm{sp}}^4 R_{\mathrm{sp}}^2}{T_\odot^4 R_\odot^2} = 1 - \left(\frac{R_{\mathrm{sp}}}{R_\odot}\right)^2\left[1 - \left(\frac{T_{\mathrm{sp}}}{T_\odot}\right)^4\right]\qquad \text{(2 p)}

With numerical values:

A' = \boxed{0.9963}\qquad \text{(1 p)}

Thus, when the spot directed towards the Earth the total irradiation of the Earth becomes weaker by 0.37 %.

Оригинал в Архива: TH-problem-final.pdf · официални решения: TH-solution-final.pdf · допълнителни данни: TH-anssheet.pdf