Question

time period of a planet whose semi major axis is 3.5 × 10⁹ m is found to be 100 years. find the time period of a planet whose semi major of axis is twice of it.

Answer 1

According to the third Kepler's law,

$\displaystyle \frac{a_1^3}{T_1^2} = \frac{a_2^3}{T_2^2} \;\; \Rightarrow \; \Big(\frac{T_2}{T_1}\Big)^2 = \Big(\frac{a_2}{a_1}\Big)^3$

where T is a period and a is semi-major axis.

It is given that $a_2=2a_1$, so

$\displaystyle \Big(\frac{T_2}{T_1} \Big)^2 = \Big(\frac{2a_1}{a_1} \Big)^3 = 2^3 = 8$

$\displaystyle T_2^2 = 8 T_1^2$

$\displaystyle T_2 = \sqrt{8} \,T_1 = 2.828 \cdot 100 = 282.8 \; \textrm{years}$

Answer: 282.8 years