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The time period a satellite of earth is 5 hours. If the separation between the earth and the satellite in increased to four times the previous value, the new time period of the satellite will be
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40 hours
$\begin{aligned} & \mathrm{T}^2 \alpha \mathrm{r}^3 \\ & \therefore \frac{\mathrm{T}_2^2}{\mathrm{~T}_1^2}=\frac{\mathrm{r}_2^3}{\mathrm{r}_1^3}=(4)^3=64 \\ & \therefore \frac{\mathrm{T}_2}{\mathrm{~T}_1}=8 \text { or } \mathrm{T}_2=8 \mathrm{~T}_1=8 \times 5=40 \text { hours. }\end{aligned}$
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