1. What is the angular diameter of the Sun, as seen from Mercury, a perihelion? At aphelion?
Mercury is at 0.31 AU (= 46.5 million km) from the Sun at perihelion, and 0.47 AU (= 70.5 million km) from the Sun at
aphelion.
 1, 400, 000 
  1.73
 46,500, 000 
Perihelion: angular diameter  57.3 
 1, 400, 000 
  1.14
 70,500, 000 
Aphelion: angular diameter  57.3 
2.The Hubble Space Telescope has a resolution of about 0.05”. What is the size of the smallest feature
can distinguish on the surface of the Moon (distance = 380.000 km)? On Mercury, at closest approach to
Earth?
We use the angular diameter formula, with angular diameter in units of seconds:
angular size = 206,265 (actual size / distance)
actual size = distance × angular size / 206,265
 0.05 
 = 0.092 km = 92 m
 206265 
 0.05 
Mercury: actual size  76,500, 000 
= 19 km
 206265 
Moon: actual size  380, 000 
3. Calculate Venus’s angular diameter, as seen by an observer on Earth, when the planet is
a) at its brightest, distance is 0,47 AU,
b) at greatest elongation, distance is 0,72 AU,
c) at the most distant point in its orbit, distance is 1,7 AU.
The formula for angular diameter in seconds of arc is
angular diameter = 206,000 (actual diameter / distance)
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The true diameter of Venus is 12 × 10 km, and the various distances are given in the first two sections of the
chapter.
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a. When Venus is brightest, Distance = 0.47 AU = 7.05 × 10 km;
 12 104 km 
  35.1 arc seconds
7
 7.05 10 km 
angular diameter  206265 
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b. At greatest elongation, Distance = 0.72 AU = 1.08 × 10 km;
 12 104 km 
  22.9 arc seconds
8
 1.08 10 km 
angular diameter  206265 
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c. At superior conjunction, Distance = 1.7 AU = 2.55 × 10 km;
 12 104 km 
  9.7 arc seconds
8
 2.55 10 km 
angular diameter  206265 