Brightest elongation of a planet.

The best time to observe a planet depends on a lot of factors. Important local circumstances include a free unclouded sky and absence of light pollution. The darkness of the background sky is important too.

Here we are going to ignore the above factors, but focus on another important factor: the apparent brightness of the planet.

The apparent brightness depends on these factors:

  • its size
  • its albedo (the percentage of light that is reflected from illuminated surface)
  • its distance rr to the sun
  • its distance ss to the earth
  • its phase (the illuminated fraction of the circular disk that we see from Earth)

For inner planets, there is a trade off between the latter two factors, which both depend on the planet's elongation (i.e. the angle planet-Earth-Sun). So we may look for an elongation with optimum brightness.

Consider a planet orbiting the sun in a circular orbit with radius of 0.5AU0.5 AU. The question is:

At what elongation ψψ is the planet at its brightest?

If cosψ=ab+c\cos ψ = \frac{a}{b+\sqrt{c}} where aa, bb and cc are positive integers and cc is square free, submit a+b+ca+b+c as your answer.

Assumptions

  • the planet is a perfect sphere
  • exactly half of its surface is lit by the sun, and we can see exactly half of its surface from Earth
  • its reflective properties are the same all over its surface, and any sunlight falling on its surface is reflected evenly in all directions
  • assume that Earth moves in a circular orbit (with a radius 1 AU) as well

Remark

Because a greater elongation offers the chance of observing against a darker background, the found value may serve as a lower bound. For Venus (at r = 0.72 AU) we find an elongation of 39.6° (at a phase of 26.7% - a crescent), but in practice larger elongations generally offer the best visibility. For Mercury ( at r = 0.38 AU) even more so: the largest elongation is the best.

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