MHT CET · Physics · Gravitation
A satellite \(S_1\) of mass \(m\) is moving in an orbit of radius \(r\). Another satellite \(S_2\) of mass \(2 \mathrm{~m}\) is moving in an orbit of radius \(2 r\). The ratio of time period of satellite \(S_2\) to that of \(S_1\) is
- A 2:1
- B 1:8
- C 1:4
- D \(2 \sqrt{2}: 1\)
Answer & Solution
Correct Answer
(D) \(2 \sqrt{2}: 1\)
Step-by-step Solution
Detailed explanation
According to Kepler's Law of periods:
\(T^2 \propto R^3\)
Therefore,
\(\frac{T_2}{T_1}=\left(\frac{2 r}{r}\right)^{\frac{3}{2}}=\frac{2 \sqrt{2}}{1}\)
\(T^2 \propto R^3\)
Therefore,
\(\frac{T_2}{T_1}=\left(\frac{2 r}{r}\right)^{\frac{3}{2}}=\frac{2 \sqrt{2}}{1}\)
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