A geostationary satellite moves around the Earth in a…

A geostationary satellite moves around the Earth in a circular orbit. (a) What is the radius of its orbit? (b) What are its velocity and acceleration?

Step 1: Idea

\"Geostationary\" means that the satellite always stays above the same point on the Earth. For that it must (1) orbit in the equatorial plane, (2) move in the direction of the Earth's rotation, and (3) have exactly the Earth's rotation period. Condition (3) fixes the radius through Kepler's third law.

Step 2: (a) Radius

The Earth's true rotation period (relative to the stars) is the sidereal day, $T = 86\,164\ \mathrm{s}$. Gravity provides the centripetal force: $\frac{4\pi^2}{T^2}r = \frac{\gamma M}{r^2}$.

\[ r = \sqrt[3]{\frac{\gamma MT^2}{4\pi^2}}. \]

\[ r = \sqrt[3]{\frac{6.67 \cdot 10^{-11} \cdot 5.97 \cdot 10^{24} \cdot 86\,164^2}{4\pi^2}}, \]

\[ r = \sqrt[3]{7.49 \cdot 10^{22}}. \]

\[r \approx 4.22 \cdot 10^7\ \mathrm{m} \approx 6.6\,R_Z\]

The height above the surface is $42\,200 - 6370 \approx 35\,800\ \mathrm{km}$.

Step 3: (b) Velocity and acceleration

The speed follows from the circumference and the period; the acceleration is centripetal and equals the Earth's field strength at that distance: $v = \frac{2\pi r}{T} = \frac{2\pi \cdot 4.22 \cdot 10^7}{86\,164}$.

\[ a = \frac{v^2}{r} = \frac{\gamma M}{r^2}. \]

\[\begin{gathered} v \approx 3.07\ \mathrm{km/s} \\ a \approx 0.224\ \mathrm{m/s^2} \end{gathered}\]

The satellite moves much more slowly than a low orbiter (7.9 km/s), and the Earth's gravity there is about 44 times weaker than at the surface.

Check

$ g(R_Z/r)^2 $ $= 9.81/6.62^2 $ $= 0.224\ \mathrm{m/s^2}\ \checkmark $

With $T = 86\,400\ \mathrm{s}$ instead of the sidereal day: $4.23 \cdot 10^7\ \mathrm{m}$, 3.08 km/s, a difference only in the third digit $\checkmark$