Calculate the period of revolution of Neptune around the sun, given that diameter of its orbit is 30 times the diameter of earth’s orbit around the sun, both orbits being asssumed to the circular.

Solution: According to Kepler’s law of periods,

\(\left(\frac{T_N}{T_E}\right)^2 = \left(\frac{r_N}{r_E}\right)^3\)

Given: \(\frac{r_N}{r_E} = 30\) and \(T_E = 1 \text{ year}\)

Then,

\(T_N^2 = T_E^2 \left(\frac{r_N}{r_E}\right)^3 = (1)^2 \times (30)^3 = 27000\)

\(T_N = \sqrt{27000} = 164.3 years\)