Gravitation Question 3
Question 3 - 24 January - Shift 2
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:
(1) $6 \times 10^{7} km$
(2) $6 \times 10^{6} km$
(3) $3 \times 10^{6} km$
(4) $3 \times 10^{7} km$
Show Answer
Answer: (3)
Solution:
Formula: Law of periods
$T^{2} \propto R^{3} \Rightarrow(\frac{T_1}{T_2})^{2}=(\frac{R_1}{R_2})^{3}$
$\Rightarrow(\frac{1}{2.83})^{2}=(\frac{1.5 \times 10^{6}}{R_2})^{3}$
$\Rightarrow R_2=[(2.83)^{2} \times(1.5 \times 10^{6})^{3}]^{1 / 3}$
$=8^{1 / 3} \times 1.5 \times 10^{6}=3 \times 10^{6} km$
