# An asteroid revolves around the Sun with a mean orbital radius twice that of Earth's. Predict the...

## Question:

An asteroid revolves around the Sun with a mean orbital radius twice that of Earth's. Predict the period of the asteroid in Earth years. ( {eq}6.38 \ast 10^6 {/eq} radius)

## Kepler's Laws of Planetary Motion:

The motion of the planets around the sun is described by three laws known as Kepler's Laws.

The first law states that all the planets move in elliptical orbits around the sun. The sun is at one of the focuses of the ellipse.

The second law states that the line joining a planet to the sun sweeps out equal area in equal interval of time.

The third law states how the period of revolution of a planet depends on the average distance of the planet from the sun.

We are given:

• The orbital radius of the asteroid is twice the orbital radius of the earth, {eq}r_a=2r_e {/eq}, where {eq}r_e {/eq} is the radius of the earth.

According to Kepler's third law, the square of the orbital period of a planet is proportional to the cube of its orbital radius.

{eq}T^2\propto r^3 {/eq}

For the given asteroid, let the time period be {eq}T_a {/eq}, then we have:

{eq}\begin{align*} \dfrac{T_a^2}{T_e^2}&=\dfrac{r_a^3}{r_e^3}\\ \Rightarrow T_a^2&=T_e^2\times \dfrac{\left ( 2r_a^3 \right )^3}{r_a^3}\\ \Rightarrow T_a^2&=8T_e^2\\ \Rightarrow T_a&=2.8\times T_e\\ &=2.8\times 1\;\rm yr\\ &=\boxed{2.8\;\rm yr} \end{align*} {/eq} 