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JEE Mains · Physics · STD 11 - Gravitation
  • JEE Mains 2023
  • Easy
  • Numerical

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 \(.......\,hours\)

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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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