QJEEM23Pha777: JEE Main 2023 Physics - Class 11 JEEM23Pha777 Solution: The time period of a satellite of earth is 24 hours. If the…

$\because {T^2} \propto {R^3}$

$\therefore$ ${{T_1^2} \over {T_2^2}} = {{R_1^3} \over {R_2^3}}$

${{{{24}^2}} \over {T_2^2}} = {{R_1^3} \over {{{\left( {{{{R_1}} \over 4}} \right)}^3}}}$

${{{{24}^2}} \over {T_2^2}} = {4^3}$

${T_2} = {{24} \over {{2^3}}} = 3$ hours