Saturn has an orbital period of 29.46 years and it's average distance to the sun is 9.54 AU if Venus has an orbital period of 0.62 years what is its average distance to the sun in AU

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

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

since you're a real machine, your ama is always less than your ima, because you can never totally eliminate friction.

answer: option c.

explanation: