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The distance between the centers of a small moon and planet in our solar system is 2.0x10^8m. if the moons orbital period is 5.0 x 10^4s, what is the planet

User Haruka
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Final answer:

To find the orbital period of the planet, we can use Kepler's third law of planetary motion, which states that the square of the orbital period is proportional to the cube of the average distance from the Sun.

Step-by-step explanation:

To find the orbital period of the planet, we can use Kepler's third law of planetary motion, which states that the square of the orbital period is proportional to the cube of the average distance from the Sun. Since the distance between the centers of the moon and planet is given as 2.0x10^8m, we can treat this as the average distance from the Sun to the planet. Let's call the orbital period of the planet T and the average distance from the Sun to the planet r. We can write the equation as:

T^2 = (r^3) x (T1^2 / r1^3)

where T1 is the orbital period of the moon and r1 is the distance between the centers of the moon and planet. Now we can plug in the values:

T^2 = (2.0x10^8m)^3 x (5.0x10^4s)^2 / (3.84x10^8m)^3

Using these values, we can solve for T to determine the orbital period of the planet.

User Kannan Mohan
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