Physics - Gravitation Question with Solution | TestHub

According to Kepler's third law, the square of the time period of a satellite is directly proportional to the cube of its orbital radius.

$T^2\propto r^3$

${\left(\frac{T_2}{T_1}\right)}^2={\left(\frac{r_2}{r_1}\right)}^3$

Given $T_1=24$ hours and $r_2=\frac{r_1}{4}$.

${\left(\frac{T_2}{24}\right)}^2={\left(\frac{1}{4}\right)}^3$

${\left(\frac{T_2}{24}\right)}^2=\frac{1}{64}$

$\frac{T_2}{24}=\frac{1}{8}$

$T_2=3$ hours.

Answer: 3 hours