Answer:
Explanation:
We know that,
Neptune is 4.5×10^9 km from the sun
And given that,
Earth is 1.5×10^8km from sun
Then,
Let P be the orbital period and
Let a be the semi-major axis
Using Keplers third law
Then, the relation between the orbital period and the semi major axis is
P² ∝ a³
Then,
P² = ka³
P²/a³ = k
So,
P(earth)²/a(earth)³ = P(neptune)² / a(neptune)³
Period of earth P(earth) =1year
Semi major axis of earth is
a(earth) = 1.5×10^8km
The semi major axis of Neptune is
a (Neptune) = 4.5×10^9km
So,
P(E)²/a(E)³ = P(N)² / a(N)³
1² / (1.5×10^8)³ = P(N)² / (4.5×10^9)³
Cross multiply
P(N)² = (4.5×10^9)³ / (1.5×10^8)³
P(N)² = 27000
P(N) =√27000
P(N) = 164.32years
The period of Neptune is 164.32years
