Question

The time period of revolution of planet $X$ around the sun is $8$ times that of $Y$. The distance of $X$ from the sun is how many times greater than that of the sun ?

• A. $ 5\frac{1}{2}$ times
• B. 4 times
• C. 3 times
• D. None of these

Answer 1

Answer: (B)

Kepler's third law provides the relation between time period of revolution with distance $r$.
The square of the period of revolution of any planet around the sun is directly proportional to the cube of the semimajor axis of its elliptical orbit (Kepler's third law)
$T^{2} \propto a^{3}$
Given, $T_{x}=8 T_{y}$
$\therefore \frac{T_{x}^{2}}{T_{y}^{2}} =\frac{R_{x}^{3}}{R_{y}^{3}} $
$\Rightarrow \frac{\left(8 T_{y}\right)^{2}}{T_{y}^{2}} =\frac{R_{x}^{3}}{R_{y}^{3}} $
$\Rightarrow \frac{64}{1} =\frac{R_{x}^{3}}{R_{y}^{3}} $
$\Rightarrow R_{x} =4 R_{y}$
Therefore, distance of planet $x$ is four times that of planet $y$.
Note : Larger the distance of a planet from the sun, larger will be its period of revolution around the sun.