Two satellites are monitored as they orbit the Earth; satellite X is four times as far from the Earth's center as is satellite Y. From Kepler's third law one may conclude that the period of revolution of X is what factor times that of Y?Select one:a.2.0b.8.0c.22.6d.1/2
Question
Two satellites are monitored as they orbit the Earth; satellite X is four times as far from the Earth's center as is satellite Y. From Kepler's third law one may conclude that the period of revolution of X is what factor times that of Y?Select one:a.2.0b.8.0c.22.6d.1/2
Solution
Kepler's third law states that the square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit. This can be written as T^2 ∝ r^3, where T is the orbital period and r is the distance from the center of the planet.
Given that satellite X is four times as far from the Earth's center as satellite Y, we can write this as r_x = 4r_y.
We want to find the factor by which the period of revolution of X is greater than that of Y, which we can write as T_x/T_y.
Substituting our values into Kepler's third law, we get:
(T_x/T_y)^2 = (r_x/r_y)^3 (T_x/T_y)^2 = (4r_y/r_y)^3 (T_x/T_y)^2 = 4^3 (T_x/T_y)^2 = 64
Taking the square root of both sides to solve for T_x/T_y, we get:
T_x/T_y = √64 = 8.0
So, the period of revolution of satellite X is 8.0 times that of satellite Y. Therefore, the correct answer is b. 8.0.
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