Two satellites are in circular orbits around Jupiter. One, with orbital radius r, makes one revolution every 18 h. The other satellite has orbital radius 3.9r. How long does the second satellite take to make one revolution around Jupiter?
To determine the time it takes for the second satellite to make one revolution around Jupiter, we can use Kepler's Third Law of Planetary Motion. According to this law, the square of the orbital period (T) of a satellite is proportional to the cube of the semi-major axis of its orbit (a^3).
Let's denote the orbital radius of the first satellite as r1 and its orbital period as T1. Similarly, we denote the orbital radius of the second satellite as r2 and its orbital period as T2.
For the first satellite:
r1 = r (given)
T1 = 18 hours (given)
For the second satellite:
r2 = 3.9r (given)
T2 = ? (what we need to find)
Using Kepler's Third Law, we can set up the following equation:
(T2^2 / T1^2) = (r2^3 / r1^3)
Substituting the given values:
(T2^2 / (18^2)) = ((3.9r)^3 / r^3)
Simplifying the equation:
T2^2 = (3.9^3 * r^3) / (18^2)
Taking the square root of both sides:
T2 = √((3.9^3 * r^3) / (18^2))
Now, we can calculate the value of T2 by substituting the given value of r:
T2 = √((3.9^3 * r^3) / (18^2))
= √((3.9^3 * r^3) / 324)
Using a calculator, we can evaluate this expression to find the value of T2 in hours.