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?

Now planet revolves in orbit around sun due to attractive gravitational force between planet and sun, from Kepler’s third law of planetary motion we know that square of time period one complete rotation of planet T around the sun is proportional to the cube of mean distance (average distance) between the planet and sun R

T²∝ R³

Le the Time taken by the earth for one complete revolution be T_e

T_e = 1 Year

Let mean distance of the earth from the sun or orbital radius be R_e

Since the planet is moving twice as fast as earth the time taken by the planet to complete one complete revolution is half of that taken by earth

Let the Time period of revolution of the planet be T_p, then we have

T_P = 1/2T_e = 1/2 year

Let, the orbital radius of this planet be R_P

Now, according to the Kepler’s third law of planetary motion, we have

And

Using both equations we get the relation

Or we can say

Simplifying we get the relation for the orbital radius of the planet as

Since planet took half as much time as earth so we have

T_p/T_e = 1/2

i.e. the radius of the planet is

Therefore, the radius of the orbit of this planet is 0.63 times the radius of the orbit of the Earth.