Period of a satellite at 400 km

Problem

A satellite orbits in a circular orbit at $400\ \text{km}$ above the Earth’s surface. Calculate its orbital period ($M_T = 5{,}97\cdot 10^{24}\ \text{kg}$, $R_T = 6{,}37\cdot 10^6\ \text{m}$, $G = 6{,}67\cdot 10^{-11}\ \text{N}\,\text{m}^2/\text{kg}^2$).

Solution

Orbital radius. The altitude must be measured from Earth’s centre: $r = R_T + h = 6{,}37\cdot 10^6 + 0{,}40\cdot 10^6 = 6{,}77\cdot 10^6\ \text{m}$

Set-up. Gravity provides the centripetal force; Kepler’s third law follows: $\frac{GM_T m}{r^2} = m\,\frac{4\pi^2 r}{T^2} \quad\Longrightarrow\quad T = 2\pi\sqrt{\frac{r^3}{GM_T}}$

Numerical substitution. With $GM_T = 6{,}67\cdot 10^{-11}\cdot 5{,}97\cdot 10^{24} \approx 3{,}98\cdot 10^{14}\ \text{m}^3/\text{s}^2$ and $r^3 = (6{,}77\cdot 10^6)^3 \approx 3{,}10\cdot 10^{20}\ \text{m}^3$: $T = 2\pi\sqrt{\frac{3{,}10\cdot 10^{20}}{3{,}98\cdot 10^{14}}} = 2\pi\sqrt{7{,}79\cdot 10^{5}} \approx 2\pi\cdot 883\ \text{s}$

$\ev{T \approx 5{,}55\cdot 10^3\ \text{s} \approx 92\ \text{min}}$

A low orbit lasts about an hour and a half: this is the order of magnitude for satellites in low Earth orbit (LEO), including the International Space Station.

Links

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