Answer:
According to Kepler's law of periods, \[{{\left( \frac{{{T}_{2}}}{{{T}_{1}}} \right)}^{2}}={{\left( \frac{{{R}_{2}}}{{{R}_{1}}} \right)}^{3}}\] or \[{{\left( \frac{{{T}_{2}}}{5} \right)}^{2}}={{\left( \frac{4{{R}_{1}}}{{{R}_{1}}} \right)}^{3}}\] or \[{{T}_{2}}=\sqrt{64\times 25}=\mathbf{40h}.\]
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