Estimating the radius of a geostationary orbit

Problem

Estimate the radius of a geostationary orbit, i.e. the orbit in which a satellite stays above the same point on Earth’s equator. (Use Kepler’s third law and the Moon as a reference, or the period formula directly.) Given: $M_T = 5{,}97\cdot 10^{24}\ \text{kg}$, $R_T = 6{,}37\cdot 10^6\ \text{m}$.

Solution

Geostationary condition. The satellite must have a period equal to Earth’s rotation day: $T = 24\ \text{h} = 24\cdot 3600\ \text{s} = 8{,}64\cdot 10^4\ \text{s}$

Set-up with Kepler’s third law. From gravity acting as the centripetal force: $T = 2\pi\sqrt{\frac{r^3}{GM_T}} \quad\Longrightarrow\quad r = \left(\frac{GM_T\,T^2}{4\pi^2}\right)^{1/3}$

Numerical substitution. With $GM_T = 6{,}67\cdot 10^{-11}\cdot 5{,}97\cdot 10^{24} \approx 3{,}98\cdot 10^{14}\ \text{m}^3/\text{s}^2$: $r = \left(\frac{3{,}98\cdot 10^{14}\cdot (8{,}64\cdot 10^4)^2}{4\pi^2}\right)^{1/3} = \left(\frac{3{,}98\cdot 10^{14}\cdot 7{,}46\cdot 10^{9}}{39{,}5}\right)^{1/3} = \left(7{,}5\cdot 10^{22}\right)^{1/3}$

$r \approx 4{,}2\cdot 10^7\ \text{m} \approx 42\,000\ \text{km}$

Altitude above the surface. Subtracting Earth’s radius: $h = r - R_T \approx 42\,000 - 6\,400 \approx 36\,000\ \text{km}$

$\ev{h \sim 36\,000\ \text{km}}$

This is the altitude of telecommunications and weather satellites, which appear fixed in the sky because they rotate in sync with the Earth.

Links

Topics: Gravitation Concepts: Kepler’s laws · Uniform circular motion · Orbits Skills: Fermi estimates · Symbolic set-up Objects: Satellite