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I'm stuck on a problem

  1. Oct 5, 2006 #1
    Prove that Venus’s orbit is .72 times the size of earths orbit. I’m told that Venus’s greatest elongation is 47 degrees.
     
  2. jcsd
  3. Oct 6, 2006 #2
    Planetary data for Venus
    mean distance from Sun 108,200,000 km (0.72 AU)
    eccentricity of orbit 0.007
    inclination of orbit to ecliptic 3.4°
    Venusian year (sidereal period of revolution) 224.7 Earth days
    maximum visual magnitude –4.6
    mean synodic period* 584 Earth days
    mean orbital velocity 35 km/s
    radius (equatorial and polar) 6,051.8 km
    surface area 4.6 × 108 km2
    mass 4.87 × 1024 kg
    mean density 5.25 g/cm3
    mean surface gravity 860 cm/s2
    escape velocity 10.4 km/s
    rotation period (Venusian sidereal day) 243 Earth days (retrograde)
    Venusian mean solar day 116.8 Earth days
    inclination of equator to orbit 177°
    atmospheric composition carbon dioxide, 96%; molecular nitrogen, 3.5%; water, 0.02%; trace quantities of carbon monoxide, molecular oxygen, sulfur dioxide, hydrogen chloride, and other gases
    mean surface temperature 737 K (867 °F, 464 °C)
    surface pressure at mean radius 95 bars
    mean visible cloud temperature about 230 K (–46 °F, –43 °C)
    number of known moons none

    *Time required for the planet to return to the same position in the sky relative to the Sun as seen from Earth.
     
  4. Oct 6, 2006 #3
    Venus is at greatest elongation when the sun - venus - earth system makes a 90 degree angle at venus's position. Knowing the distance from the earth to the sun is 1AU, and using some trig, sin(47)*1AU ~ 0.73AU. The reason that this is .01 off of given, is because that given value is the mean distance, not the maximum distance which is what was calculated using the given information.
     
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