Physics - Gravitation Question with Solution | TestHub

Using Kepler's III^rdlaw

$T^2\propto r^3$

$\frac{{T^{\prime }}^2}{T^2}=\frac{{\left(r^{\prime }\right)}^3}{r^3}$

$\Rightarrow \frac{T^{\mathrm{\text{'}2}}}{T^2}=\frac{{\left(4R\right)}^3}{{\left(R\right)}^3}$

$T^{\mathrm{\text{'}2}}=64T^2$

$\Rightarrow T^{\mathrm{\text{'}2}}=8T\Rightarrow T\mathrm{\text{'}}=8\times 5=40$$\mathrm{h}$