Question

If the distance of the earth from Sun is $1.5\times 10^{6}km$. Then the distance of an imaginary planet from Sun, if its period of revolution is $2.83$ years is :

• A. $6\times 10^{6}km$
• B. $3\times 10^{6}km$
• C. $6\times 10^{7}km$
• D. $3\times 10^{7}km$

Answer 1

Answer: (B)

$T^2\propto R^3\Rightarrow {\left(\frac{T_1}{T_2}\right)}^2={\left(\frac{R_1}{R_2}\right)}^3$
$\Rightarrow {\left(\frac{1}{2.83}\right)}^2={\left(\frac{1.5\times 10^6}{R_2}\right)}^3$
$\Rightarrow R_2={\left[(2.83{)}^2\times {\left(1.5\times 10^6\right)}^3\right]}^{1/3}$
$=8^{1/3}\times 1.5\times 10^6=3\times 10^{6}km$