You would use
P^2 = (2π)^2(D^3)/((G)(M))
where P is the orbital period in seconds, D is the distance in meters, M is the mass in kilograms of Jupiter, and G is the 6.67 x 10^ -11 m^3/ kg sec^2
M Jupiter = 1.9 x 10^27 kg, D = 1.88 x10^9 meters
P^2 = 39.5*(6.65 x 10^27)/(( 6.67 x 10^ -11)(1.9 x 10^27)) = 2.07 x 10^12
P = 1,440,000 seconds = 16.66 days
Jupiter is the largest planet in the solar system that have four large natural satellites, Ganymede, Callisto, and Europa.
Given that:
Distance = 1.88 x 10⁹ m
Mass = 1.9 x 10²⁷ kg
Radius = 2.40 x 10⁶ m
Gravitational constant = 6.67 x 10⁻¹¹ m³ kg/ sec²
Now,
From the formula of orbital periods, it can be calculated as:
P² = [tex]\dfrac{(2 \pi^2) (D^3)}{GM}[/tex]
Substituting the values in the above formula:
P² = [tex]\dfrac{39.5 \times 6.65 \times 10 ^{27}}{(6.67 \times 10^{-11}) (1.9 \times 10^{27})}\\\\[/tex]
P² = [tex]2.07 \times 10^{12}[/tex]
P = 14440000 seconds or approximately 16.66 days.
Thus, the Callisto will require 16.66 days to orbit around the planet.
To know more about circular orbital period, refer to the following link:
https://brainly.com/question/22764231
