Comparing Orbital Distances Using Time Periods

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)

$4$

$5$

$2$

$3$

Solution:

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 $r_A = 4r_B$.

Comparing Orbital Distances Using Time Periods

Solution:

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