<h2>Orbital Period:</h2>
- 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.
<h2>Answer and Explanation:</h2><h3>Given:</h3>
- 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
B.axon .................................................................
