MHT CET · Physics · Gravitation
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 .
- A 4
- B 8
- C 16
- D 64
Answer & Solution
Correct Answer
(A) 4
Step-by-step Solution
Detailed explanation
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: \(T_p=8 T_e\)
\(\begin{aligned}
& \frac{T_e^2}{\left(8 T_e\right)^2}=\frac{r_t^3}{r_p^3} \Rightarrow \frac{1}{64}=\frac{r_e^3}{r_p^3} \\
\therefore \quad & \frac{r_e}{r_p}=\left(\sqrt[3]{\frac{1}{64}}\right)=\frac{1}{4} \quad \text { NOW, } \mathrm{rp} / \mathrm{re}=4
\end{aligned}\)
\(\begin{aligned}
& \frac{T_e^2}{\left(8 T_e\right)^2}=\frac{r_t^3}{r_p^3} \Rightarrow \frac{1}{64}=\frac{r_e^3}{r_p^3} \\
\therefore \quad & \frac{r_e}{r_p}=\left(\sqrt[3]{\frac{1}{64}}\right)=\frac{1}{4} \quad \text { NOW, } \mathrm{rp} / \mathrm{re}=4
\end{aligned}\)
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