The period of the earth around the sun is 1 year and its distance is 150 million km from the sun....

Question:

The period of the earth around the sun is 1 year and its distance is 150 million km from the sun. An asteroid in a circular orbit around the sun is at a distance 475 million km from the sun. What is the period of the asteroid s orbit? Answer in units of years

Period of the motion:

The square of the period of the motion of the planet is directly proportional to the cube of the radius of the motion as per Kepler's third law of planetary motion.

Given data:

• Time period of motion around sun {eq}\rm (T_{1}) = 1 \ year {/eq}
• Radius of the earth {eq}\rm (R_{1}) = 150 \ million-km {/eq}
• Radius of the asteroid {eq}\rm (R_{2}) = 475 \ million -km {/eq}

Now, the time period of the asteroid would be

{eq}\begin{align} \rm \dfrac{T_{2}^{2}}{T_{1}^{2}} &= \rm \dfrac{R_{2}^{3}}{R_{1}^{3}} \\[0.3 cm] \rm \dfrac{T_{2}^{2}}{1^{2}} &= \dfrac{475^{2}}{150^{2}} \\[0.3 cm] \rm T_{2} &= 5.64 \ years \\[0.3 cm] \end{align} {/eq}