Question

The mass of the Sun is 2 × 1030 kg, and the distance between Neptune and the Sun is 30 AU. What is the orbital period of Neptune in Earth years?

Answer 1

Orbital Period:

• To answer this problem, we must employ the gravitational notion and equation. We already know that gravitational force provides the required centripetal force for orbital motion. As a result of matching gravitational and centripetal forces, we may get the equation for orbital time period.

Answer and Explanation:

Given:

• Mass of Sun (M) = 1.99 * 10^30 Kg
• Orbital radius of the Neptune
• (R) = 30 A.U. = 30 * (1.496 * 10^11) m
• R = 44.88 * 10^11 m

Now we know that the time period is given by

T^2 = 4π^2R^3/GM

T^2 = 4π^2*(44.88*10^11)^3/6.67*10^-11*(1.99*10^30)

T = 5.19 * 10^9 s

Now we know that

1 year = 3.154 * 10^7 s

Therefore, T = 5.19*10^9 s/ 3.154 * 10^7 (s/year)

T = 164.4 year