Q162 / 225 · JEE Main 2003 · Physics
The time period of a satellite of earth is 5 hours. If the separation between the earth and the satellite is increased to 4 times the previous value, the new time period will become
- A10 hours
- B80 hours
- Correct answerC40 hours
- D20 hours
Solution
$\mathrm{T}^2 \propto \mathrm{R}^3 \Rightarrow\left(\frac{\mathrm{T}_1}{\mathrm{~T}_2}\right)^2=\left(\frac{\mathrm{R}_1}{\mathrm{R}_2}\right)^3$
$\Rightarrow\left(\frac{\mathrm{T}_1}{\mathrm{~T}_2}\right)=\left(\frac{\mathrm{R}_1}{\mathrm{R}_2}\right)^{3 / 2}=\left(\frac{1}{4}\right)^{3 / 2} \quad \Rightarrow \frac{\mathrm{T}_2}{\mathrm{~T}_1}=(4)^{3 / 2}=8$
$\Rightarrow \mathrm{T}_2=8 \times \mathrm{T}_1=8 \times 5=40=40$ hours
Hence, the correct answer is option: 'C'.