IOAA 2015, Теория, кратки задачи, английски — Задача 9

1

Solar luminosity is

$L_\odot = 4\pi R_\odot^2\sigma T_\odot^4$

$= 4\times3.14\times(6.96\times10^{10})^2\,(5.67\times10^{-5})(5780)^4$ ergs/sec

$= 3.9\times10^{33}$ ergs/sec

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2

Flux of the Sun intercepted by Venus is

$L_V = (A_{V(projected)}/A_{(sphere,0.72au)})\times L_\odot$

Also,

$L_V = 4\pi R_V^2\sigma T_V^4$

So,

$4\pi R_V^2\sigma T_V^4 = (\pi R_V^2/4\pi d^2)\times L_\odot = (\pi R_V^2/4\pi d^2)\times4\pi R_\odot^2\sigma T_\odot^4$

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3

Therefore,

$T_V^4 = (R_\odot^2/4d^2)\times T_\odot^4 = [(6.957\times10^{5}\ \mathrm{km})^2/4\times(0.72\times149.6\times10^{6})^2]\times(5780)^4$

$= (483998490000/46407499776000000)\times(5780)^4 = (4.84\times10^{11}/4.64\times10^{16})\times(5780)^4$

$= 1.0429316216908190111241385226915\mathrm{e}{-5}\times1114577187760656$

$= 1.04\times10^{-5}\times1.11\times10^{15}$

$= 11624277939.308134327737156481455 = 1.1624\times10^{10}$

$T_V = 328.5$ K

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4

The flux density of Venus at closest approach to Earth at an observing frequency of 5 GHz:

($1'' = 4.84\times10^{-6}$ rad)

$S_{5GHz} = B_{5GHz}\,\Omega = (2kT_b\nu^2/c^2)\,\Omega \qquad (\rightarrow T_b = T_V)$

$= (2\times1.38\times10^{-16}\times328.5\times(5\times10^{9})^2\times\pi\,(66\times4.84\times10^{-6})^2)/(3\times10^{10})^2$

$= 8.07\times10^{-22}$ ergs s$^{-1}$ cm$^{-2}$ Hz$^{-1}$

$= 80.7$ Jy

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