The period of a planet around the sun is 8 times that of earth. The ratio of radius of planet's orbit to the…
The period of a planet around the sun is 8 times that of earth. The ratio of radius of planet's orbit to the radius of the earth's orbit is .
4
8
16
64
Solution
From Kepler's third law, $T^2 \propto r^3$ or $\frac{T^2}{r^3}=$ Constant Using ratio of time periods of the two planets, $\frac{T_c^2}{T_p^2}=\frac{r_{\mathrm{p}}{ }^3}{T_{\mathrm{p}}{ }^3}$ Given: $T_p=8 T_e$
$\begin{aligned}
& \frac{T_e^2}{\left(8 T_e\right)^2}=\frac{r_t^3}{r_p^3} \Rightarrow \frac{1}{64}=\frac{r_e^3}{r_p^3} \\
\therefore \quad & \frac{r_e}{r_p}=\left(\sqrt[3]{\frac{1}{64}}\right)=\frac{1}{4} \quad \text { NOW, } \mathrm{rp} / \mathrm{re}=4
\end{aligned}$