Radius of a geostationary orbit

Problem

Calculate the orbital radius of a geostationary satellite above the Earth ($T = 86\,164$ s, $M_T = 5{,}97\cdot 10^{24}$ kg) using Kepler’s third law.

The ISS’s low orbit and the much wider geostationary orbit (GEO).

Solution

Equating the gravitational force to the centripetal force of circular motion, with $v = 2\pi r/T$, gives Kepler’s third law: $T^2 = \frac{4\pi^2}{G M_T}\, r^3$ From which the orbital radius is isolated: $r = \sqrt[3]{\frac{G M_T\, T^2}{4\pi^2}}$ Calculating the factors: $G M_T = 6{,}67\cdot 10^{-11}\cdot 5{,}97\cdot 10^{24} = 3{,}98\cdot 10^{14}$ m$^3$/s$^2$; $T^2 = (86\,164)^2 = 7{,}42\cdot 10^9$ s$^2$. So: $r = \sqrt[3]{\frac{3{,}98\cdot 10^{14}\cdot 7{,}42\cdot 10^9}{4\pi^2}} = \sqrt[3]{\frac{2{,}96\cdot 10^{24}}{39{,}48}} = \sqrt[3]{7{,}49\cdot 10^{22}}$ $r \approx 4{,}22\cdot 10^7~\text{m}$ Subtracting the Earth’s radius $R_T \approx 6{,}37\cdot 10^6$ m, the altitude above the surface is: $h = r - R_T \approx 4{,}22\cdot 10^7 - 0{,}64\cdot 10^7 \approx 3{,}58\cdot 10^7~\text{m} \approx 35\,800~\text{km}$ $\ev{r \approx 4{,}22\cdot 10^7~\text{m} \ (\approx 35\,800~\text{km altitude})}$

Links

Topics: Gravitation Concepts: Kepler’s laws · Orbits Skills: Symbolic setup Objects: Satellite