JEE Mains · Physics · STD 11 - 7. gravitation
The time period of a satellite of earth is \(24\) hours. If the separation between the earth and the satellite is decreased to one fourth of the previous value, then its new time period will become \(.......\,hours\)
- A \(4\)
- B \(6\)
- C \(12\)
- D \(3\)
Answer & Solution
Correct Answer
(D) \(3\)
Step-by-step Solution
Detailed explanation
\(T ^2 \propto R ^3\) \(\frac{ T _1^2}{ T _2^2}=\frac{ R _1^3}{ R _2^3} \Rightarrow\left(\frac{ T _1}{ T _2}\right)^2=\left(\frac{ R }{\frac{ R }{4}}\right)^3\) \(\frac{ T _1^2}{ T _2^2}=64\) \(T _2^2=\frac{ T _1^2}{64}\) \(T _2=\frac{24}{8}=3\)
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