The angle of maximum elongation for Venus is | vedclass

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(A) The angle of elongation $\varepsilon$ is the angle formed at the Earth between the direction of the planet and the direction of the Sun.When the p

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$0.68$

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(A) The angle of elongation $\varepsilon$ is the angle formed at the Earth between the direction of the planet and the direction of the Sun.When the planet appears at its maximum elongation,the line of sight from Earth is tangent to the planet's orbit. Consequently,the angle subtended by the Sun and Earth at the planet is $90^\circ$.From the geometry of the right-angled triangle formed by the Sun $(S)$,the planet $(P)$,and the Earth $(E)$:$\cos \varepsilon = \frac{	ext{Adjacent}}{	ext{Hypotenuse}} = \frac{r_{PS}}{r_{ES}}$Given $\varepsilon = 47^\circ$ and $r_{ES} = 1 \ AU$ (distance from Earth to Sun),$r_{PS} = r_{ES} \cos 47^\circ$$r_{PS} = 1 \ AU 	imes \cos 47^\circ \approx 1 	imes 0.682 = 0.68 \ AU$.Therefore,the distance of Venus from the Sun is approximately $0.68 \ AU$.

The angle of maximum elongation for Venus is $47^\circ$. The distance of Venus from the Sun in $AU$ is .......

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