Comparing Orbital Distances Using Time Periods – Rankers Physics

Keplers Law: Practice Problem & Solution

The period of revolution of planet A around the sun is $8$ times that of B. The distance of A from the sun is how many times greater than that of B from the sun? (1997)
$4$
$5$
$2$
$3$

Short Solution (Best for NEET):

To solve this problem, we apply the core principles of Keplers Law. Understanding the underlying formula is key to arriving at the correct answer below:

According to Kepler's third law, $T^2 \propto r^3$. Given $T_A = 8T_B$, so $\left(\frac{r_A}{r_B}\right)^3 = \left(\frac{T_A}{T_B}\right)^2 = (8)^2 = 64$. Taking the cube root yields $r_A = 4r_B$.

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