The planet Mercury travels around the Sun with a mean orbital radius of 5.8×1010 m. The mass of the Sun is 1.99×1030 kg. Use Newton’s version . . .

Question : The planet Mercury travels around the Sun with a mean orbital radius of 5.8×10¹⁰ m. The mass of the Sun is 1.99×10³⁰ kg. Use Newton’s version of Kepler’s third law to determine how long it takes Mercury to orbit the Sun. Give your answer in Earth days.

Doubt by Nisha

Solution :

Orbital Radius of the Mercury around the Sun, r = 5.8×10¹⁰ m
Mass of Sum, M = 1.99×10³⁰ kg

Using Kepler's Third Law of Motion
T²=kr³
T²=(4π²/GM)r³
T²=4π²r³/GM
T²= [4(3.14)²(5.8×10¹⁰)³]/[6.67×10^-11×1.99×10³⁰]
T² = [39.4384×(5.8)³×10^30+11-30]/6.67×1.99
T² = [7694.905×10¹¹]/13.2733
T² = 579.728×10¹¹
T = √(57.9728×10¹²)
T = 7.614×10⁶ sec
T= 7.614×10⁶/(60×60×24)
T = 88.125 days