Question

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?

• A. $4$
• B. $5$
• C. $2$
• D. $3$

Answer 1

Answer: (A)

Given,

$T_A=8T_B$

According to Kepler's third law,

$(\frac{T_A}{T_B}{)}^2=(\frac{R_A}{R_B}{)}^3$

$\frac{R_A}{R_B}=(\frac{T_A}{T_B}{)}^{\frac{2}{3}}$

$\frac{R_A}{R_B}=(\frac{8T_B}{T_B}{)}^{\frac{2}{3}}$

$\frac{R_A}{R_B}=2^{3\times \frac{2}{3}}=2^2=4$

$R_A=4R_B$

So the correct answer is $4$.