**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:**

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, $R_{e}=1 \mathrm{AU}$

Time taken by the planet to complete one revolution around the Sun, $T_{p}=\frac{1}{2} T_{e}=\frac{1}{2}$ year

Orbital radius of the planet $=R_{p}$

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

$\left(\frac{R_{P}}{R_{e}}\right)^{3}=\left(\frac{T_{P}}{T_{e}}\right)^{2}$

$\frac{R_{P}}{R_{e}}=\left(\frac{T_{P}}{T_{e}}\right)^{\frac{2}{3}}$

$=\left(\frac{\frac{1}{2}}{1}\right)^{\frac{2}{3}}=(0.5)^{\frac{2}{3}}=0.63$

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