202004240521 Homework 1 (Q5)

Estimate the apparent magnitude of Mercury and Venus when the Earth-planet-Sun forms a right angle. (Assuming the planet reflects 50% of solar light.)

Setup.

Recall the apparent magnitude $m_b$ , the Sun’s luminosity being the reference, is given by the formula

$m_b=-0.23+5\log D-2.5\log (L/L_{\odot})$ ,

where $D$ is the distance (in parsec) from the source to the Earth, $L$ the luminosity of the source, and $L_\odot =3.8\times 10^{33}\,\mathrm{erg\, s^{-1}}$ the solar luminosity.

CASE I of Mercury

According to http://coolcosmos.ipac.caltech.edu/ask/25-How-far-is-Mercury-from-Earth-, the average distance between Mercury and Earth is 0.0000024954 pc. From Wikipedia on Mercury (planet), the mean radius of Mercury $R_{\textrm{Mercury}}$ is $2439.7\pm 1\,\mathrm{km}$ .

The luminosity of Mercury $L_{\textrm{Mercury}}$ due to reflection of sunlight, is related to the solar luminosity $L_\odot =3.8\times 10^{33}\,\mathrm{erg\, s^{-1}}$ by

$L_{\textrm{Mercury}}=\displaystyle{\frac{\pi R^2_{\textrm{Mercury}}}{4\pi D^2}}L_\odot$ ,

where the Sun-Earth distance $D$ is one astronomical unit $1\,\mathrm{AU}=1.5\times 10^{13}\,\mathrm{cm}$ .

Thence one computes

$\begin{aligned} L_{\textrm{Mercury}} & = \displaystyle{\frac{\pi (2.4397\times 10^8)^2}{4\pi (1.5\times 10^{13})^2}}\times 3.8\times 10^{33} \\ & = 2.513124127\times 10^{23}\enspace \mathrm{erg\, s^{-1}} \end{aligned}$

Thus, the apparent magnitude $m_{b,\textrm{mercury}}$ of Mercury is

$\begin{aligned} m_{b,\textrm{mercury}} & =-0.23+5\log (0.0000024954)-2.5\log (6.613484544\times 10^{-11})\\ & = -2.795375004 \end{aligned}$

CASE II of Venus

According to http://coolcosmos.ipac.caltech.edu/ask/52-How-far-away-is-Venus-from-Earth-, the average distance between Venus and Earth is 0.0000012963 pc. From Wikipedia on Venus, the mean radius of Venus $R_{\textrm{Venus}}$ is $6051.8\pm 1\,\mathrm{km}$ .

The luminosity of Venus $L_{\textrm{Venus}}$ due to reflection of sunlight, is related to the solar luminosity $L_\odot =3.8\times 10^{33}\,\mathrm{erg\, s^{-1}}$ by

$L_{\textrm{Venus}}=\displaystyle{\frac{\pi R^2_{\textrm{Venus}}}{4\pi D^2}}L_\odot$ ,

where the Sun-Earth distance $D$ is one astronomical unit $1\,\mathrm{AU}=1.5\times 10^{13}\,\mathrm{cm}$ .

Thence one computes

$\begin{aligned} L_{\textrm{Venus}} & = \displaystyle{\frac{\pi (6.0518\times 10^8)^2}{4\pi (1.5\times 10^{13})^2}}\times 3.8\times 10^{33} \\ & = 1.546358626\times 10^{24}\enspace \mathrm{erg\, s^{-1}} \end{aligned}$

Thus, the apparent magnitude $m_{b,\textrm{venus}}$ of Venus is

$\begin{aligned} m_{b,\textrm{venus}} & =-0.23+5\log (0.0000012963)-2.5\log (4.069364804\times 10^{-10})\\ & = -6.190288956 \end{aligned}$

Correction.

Without heed to the 50% reduction of solar light reflection, this problem need be redone.

CASE I of Mercury

CASE II of Venus

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