The time period of a satellite of earth is 5 hours. If the separation between the earth and the satellite is…

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
  1. 10 hours
  2. 80 hours
  3. 40 hours
  4. 20 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

Asked in: JEE Main 2003

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