Suppose there existed a planet that went around the sun twice as fast as the earth. What would be its orbital size as compared to that of the earth ?

Explanation

We are given: Period of revolutiion of earth `=T_(E )=1` year
So period of revolution of the planet, `T_(P)= (T_(E ))/(2)= (1)/(2)` year, Orbital size of earth `r_(E )= 1A`. U., We are to calculate. Orbital size of the planet `r_(p)`
Using Kepler.s third law
`r_(p)= r_(E ) ((T_(p))/(T_(E )))^(2//3) =1 ((1//2)/(1))^(2//3)= ((1)/(2))^(2//3)= 0.63A.U`