Respuesta :
Answer:
Explanation:
Saturn orbital period
p(s)= 29.46years
Average distance
a(s) = 9.54AU
Venus orbital period
p(v) = 0.62 years
a(v) ?
Using Kepler's third law
a³ ∝p²
a³ = kp²
a³/p² = k
Where
a is the distance if planet from sun
T is the period of the planet
Then,
a(Venus)³/p(venus)² = a(saturn)³/p(saturn)²
a(v)³/p(v)² = a(s)³/p(s)²
a(v) ³/ 0.62² = 9.54³/29.46²
a(v) ³/ 0.62² = 1.0004
a(v)³ = 1.0004× 0.62²
a(v)³ = 0.3846
a(v) = cube root(0.3846)
a(v) = 0.727 AU
Answer:
15.66 AU
Explanation:
Using Kepler's law which states that T² ∝ R³ where T = period of planet and R = distance of planet from sun.
So T₁²/T₂² = R₁³/R₂³
where T₁ = orbital period of Saturn = 29.46 years
T₂ = orbital period of Venus = 0.62 years
R₁ = average distance of Saturn to the sun = 9.54 AU
R₂ = average distance of Venus to the sun = ?
From the equation above,
R₂³ = (T₂²/T₁²)R₁³
R₂ = [∛(T₂/T₁)²]R₁
R₂ = [∛(0.62/29.46)²]9.54 AU
R₂ = [∛(0.021)²]9.54 AU
R₂ = [∛0.000443]9.54 AU
R₂ = [0.0762]9.54 AU
R₂ = 0.727 AU