Physics · Ch 6 — Gravitation

Satellites, Orbital Speed and Time Period

6.4.1

Satellites, Orbital Speed and Time Period

A satellite stays in orbit because Earth's own gravity supplies exactly the centripetal force needed to keep it curving around the planet rather than flying off in a straight line.

Orbital speed. For a satellite of mass $M_s$ moving in a circular orbit of radius $(R_e+h)$ , equating the required centripetal force to the gravitational force gives

\frac{M_sv^2}{R_e+h}=\frac{GM_sM_e}{(R_e+h)^2} \;\Rightarrow\; v=\sqrt{\frac{GM_e}{R_e+h}}. \qquad (6.57,\,6.58)

As the orbital height $h$ increases, the required orbital speed $v$ decreases -- satellites farther out move more slowly.

Time period. Since speed is distance travelled divided by time, and the satellite covers the full circumference $2\pi(R_e+h)$ in one period $T$ ,

v=\frac{2\pi(R_e+h)}{T} \;\Rightarrow\; T=2\pi\sqrt{\frac{(R_e+h)^3}{GM_e}}. \qquad (6.59,\,6.60)

Squaring both sides shows $T^2\propto(R_e+h)^3$ -- exactly Kepler's third law, now applied to artificial satellites instead of planets. For a satellite orbiting very close to the surface ( $h\ll R_e$ ), this reduces to

T=2\pi\sqrt{\frac{R_e}{g}} \qquad (6.62)

which, using $R_e\approx6.4\times10^6\ \text{m}$ and $g=9.8\ \text{m s}^{-2}$ , gives a period of about 85 minutes -- close to the actual period of satellites in low Earth orbit. …

Figure 6.20A satellite in circular orbit at height h

What this figure shows. The Earth is drawn with a satellite orbiting in a circle at height h above the surface, so its orbital radius measured from Earth's centre is R_e + h; this figure is the geometric basis for deriving both the satellite's orbital speed v = sqrt(GM_e/(R_e+h)) and its time period, since the circumference 2pi(R_e+h) divided by v giv …

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