Question Number 226601 by mr W last updated on 07/Dec/25
Answered by mr W last updated on 07/Dec/25
$${mass}\:{of}\:{particle}={m} \\ $$$${total}\:{mass}\:{of}\:{earth}={M}_{{E}} \\ $$$${radius}\:{of}\:{earth}={R} \\ $$$${F}=\frac{{GmM}}{{x}^{\mathrm{2}} }=\frac{{GmM}_{{E}} }{{x}^{\mathrm{2}} }\left(\frac{{x}}{{R}}\right)^{\mathrm{3}} =\frac{{GmM}_{{E}} {x}}{{R}^{\mathrm{3}} } \\ $$$${ma}=−\frac{{GmM}_{{E}} {x}}{{R}^{\mathrm{3}} } \\ $$$${a}=−\frac{{GM}_{{E}} {x}}{{R}^{\mathrm{3}} }=−{kx} \\ $$$$\frac{{d}^{\mathrm{2}} {x}}{{dt}^{\mathrm{2}} }+{kx}=\mathrm{0} \\ $$$$\omega=\sqrt{{k}} \\ $$$${T}=\frac{\mathrm{2}\pi}{\omega}=\mathrm{2}\pi\sqrt{\frac{{R}^{\mathrm{3}} }{{GM}_{{E}} }}=\mathrm{2}\pi\sqrt{\frac{{R}}{{g}}} \\ $$$${R}=\mathrm{6371}×\mathrm{10}^{\mathrm{3}} \:{m} \\ $$$${M}_{{E}} =\mathrm{5}.\mathrm{9742}×\mathrm{10}^{\mathrm{24}} \:{kg} \\ $$$${G}=\mathrm{6}.\mathrm{6743}×\mathrm{10}^{−\mathrm{11}} \:\frac{{m}^{\mathrm{3}} }{{kg}\:{s}^{\mathrm{2}} } \\ $$$$\Rightarrow{T}\approx\mathrm{5060}\:{sec}\:\approx\:\mathrm{84}.\mathrm{3}\:{minutes} \\ $$$$ \\ $$$${average}\:{speed}\:=\frac{\mathrm{4}{R}}{{T}}=\frac{\mathrm{2}}{\pi}\sqrt{{gR}}\:\approx\mathrm{6}.\mathrm{1}\:{km}/{s} \\ $$$${max}.\:{speed}\:={R}\omega=\sqrt{{gR}}\:\approx\mathrm{7}.\mathrm{9}\:{km}/{s} \\ $$