Final answer:
The new time period of a satellite, after tripling the radius of its orbit, can be calculated using Kepler's Third Law. It is found to be approximately 18.97 hours, which is √{27} times the original period of 7 hours.
Step-by-step explanation:
According to Kepler's Third Law, the square of the orbital period (T) of a satellite is directly proportional to the cube of the radius of its orbit (r). Specifically, T² ≈ r³. To find the new period after increasing the radius of orbit to three times its previous value, we can use the ratio of the periods squared being equal to the ratio of the radii cubed, which is:
(T2/T1)² = (r2/r1)³
Given that the initial period (T1) is 7 hours, and the radius of orbit is increased to three times its previous value (r2 = 3 × r1), the equation becomes:
(T2/7 hours)² = (3³)
Solving for T2 gives us a new period that is about √{27} times the original period, which is approximately 18.97 hours, since 3³ is 27 and the square root of 27 is approximately 5.20.
The new time period of the satellite, when the radius of the orbit is increased to three times its previous value, is therefore approximately 18.97 hours.