1)Solution:
We have: V2/V1 = (a2/a1) (T1/T2)By convention we assign '1' to the inner planet (Venus) and '2' to the outer (Earth). We have a2 = 1 AU and for Venus (from Kepler's third law):
T1 = (224.69/365.25) yr. = 0.6151 yr.
a1 = {[T1]2}1/3 = [(0.6151)2]1/3
a1 = 0.723 AU
Therefore:
V2/V1 = (0.7234)(1/0.6151)
V2/V1 = 1.175
(b) According to a Table of Orbital Velocities in Astrometric & Geodetic Data:
V(Venus) = 35.02 km/s
V(Earth) = 29.78 km/s
Take the ratio of the velocities:
V(Venus)/V(earth)
= (35.02 km/s)/ (29.78 km/s) = 1.175
So, Venus' orbital velocity is 1.175 times Earth's which conforms to the result of part (a).
So, Venus' orbital velocity is 1.175 times Earth's which conforms to the result of part (a).
2. The component, Vp cos (φ) doesn't contribute to the observed angular velocity of the
planet because the component vector direction (along line P'E') is oblique to the motion vector (itself tangent to the orbit).
3) If the
angular velocity of the planet as observed from Earth is: - (Vp - V)/
PE and parallel to the orbital motion, then it must also be in a direction opposite to the orbital motion, and hence is retrograde at opposition. (Since Vp < V )
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