Physics - Gravitation Question with Solution | TestHub

A geostationary satellite has an orbital period equal to Earth's rotation period, $T=24\text{ hours}=24\times 3600\text{ s}=86400\text{ s}$. The orbital radius $r$ is the sum of Earth's radius and the height of the satellite: $r=R_E+h=6400\text{ km}+36000\text{ km}=42400\text{ km}=42.4\times 10^6\text{ m}$. The orbital speed $v$ is given by $v=\frac{2\pi r}{T}$. Substituting the values, $v=\frac{2\times 3.14159\times 42.4\times 10^6\text{ m}}{86400\text{ s}}\approx 3082\text{ m/s}$. Converting to km/s, $v\approx 3.082\text{ km/s}$, which is approximately $3.1\text{ km/s}$.