Answer:
According to Kepler?s third law, \[\frac{T_{1}^{2}}{T_{2}^{2}}=\frac{r_{1}^{3}}{r_{2}^{3}}\] or \[{{r}_{2}}={{r}_{1}}{{\left( \frac{{{T}_{2}}}{{{T}_{1}}} \right)}^{2/3}}\] Here\[{{r}_{1}}=3.8\times 105\,km;\,\,{{T}_{1}}=28\,\,days;\,\,{{T}_{2}}=1\,\,day.\] \[\therefore {{r}_{2}}=3.8\times {{10}^{5}}\times {{\left( \frac{1}{28} \right)}^{2/3}}=4.1\times {{10}^{4}}km\]
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