The period of revolution of planet $A$ around the sun is 8 times that of $B$. The distance of $A$ from sun…
The period of revolution of planet $A$ around the sun is 8 times that of $B$. The distance of $A$ from sun is how many times greater than that of $B$ from the sun?
2
3
4
5
Solution
If $T_A$ and $T_B$ are the time periods of planet $A$ and $B$ around the sun respectively, then
$
T_A=8 T_B
$
According to Kepler's law,
$
\begin{array}{rlrl}
& & T^2 & \propto R^3 \\
\Rightarrow & \left(\frac{T_A}{T_B}\right)^2 & =\left(\frac{R_A}{R_B}\right)^3 \\
\Rightarrow & & \frac{R_A}{R_B} & =\left(\frac{T_A}{T_B}\right)^{2 / 3}=\left(\frac{8 T_B}{T_B}\right)^{2 / 3} \\
& & =(8)^{2 / 3}=\left(2^3\right)^{2 / 3}=2^2 \\
\Rightarrow & & \frac{R_A}{R_B} & =4 \Rightarrow R_A=4 R_B
\end{array}
$