# Planet X orbits the star Omega with a "year" that is 180 earth days long. Planet Y circles Omega...

## Question:

Planet X orbits the star Omega with a "year" that is 180 earth days long. Planet Y circles Omega at four times the distance of Planet X. How long is a year on Planet Y in earth days?

## Kepler's Third Law

Kepler's laws of planetary motion describes the motion of planets around Sun. The Kepler's third law state that the square of the time period of a planet in its orbit is directly proportional to the cube of the radius of the orbit of that planet.

## Answer and Explanation:

**Given :**

- The time period of the planet X is, {eq}T_X = 180 \ days {/eq}

- Let the radius of orbit of planet X around the star omega is, {eq}r_X = r {/eq}, the radius of orbit of planet Y around the star omega will be, {eq}r_Y = 4 r {/eq}

Let the time period of the planet Y be, {eq}T_Y {/eq}

Applying the Kepler's third law, we get:

{eq}\begin{align*} \dfrac{T_Y^2 }{T_X^2 } &= \dfrac{r_Y^3 }{r_X^3} \\ \dfrac{T_Y^2 }{180^2 } &= \dfrac{(4r)^3 }{r^3} \\ T_Y^2 &= (180)^2 (4)^3 \\ \color{blue}{ T_Y } &= \color{blue}{ \boxed{ 1440 \ days } } \end{align*} {/eq}

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