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, $T_x=8T_y$
$\therefore \frac{T_x^2}{T_y^2}=\frac{R_x^3}{R_y^3}$
$\Rightarrow \frac{{\left(8T_y\right)}^2}{T_y^2}=\frac{R_x^3}{R_y^3}$
$\Rightarrow \frac{64}{1}=\frac{R_x^3}{R_y^3}$
$\Rightarrow R_x=4R_y$
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.