#### Question

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?

#### Solution 1

Lesser by a factor of 0.63

Time taken by the Earth to complete one revolution around the Sun,

*T*_{e} = 1 year

Orbital radius of the Earth in its orbit, *R*_{e }= 1 AU

Time taken by the planet to complete one revolution around the Sun, `T_P = 1/2T_e = 1/2` year

Orbital radius of the planet = *R*_{p}

From Kepler’s third law of planetary motion, we can write:

`(R_p/R_e)^3 = (T_p/T_e)^2`

`(R_p/R_e) = (T_p/T_e)^(2/3)`

`=((1/2)/1)^(2/3) = (0.5)^(2/3) = 0.63`

Hence, the orbital radius of the planet will be 0.63 times smaller than that of the Earth.

#### Solution 2

Here, `T_e = 1 "year"`, `T_p = T_e/2 = 1/2` year

`r_c = 1 A.U.`

Using Kepler's third law,we have `r_p = r_c (T_p/T_e)^(2/3)`

`=>r_p = 1("1/2"/1)^(2/3)` = 0.63 AU