Determine the period of an artificial Earth satellite moving…

Determine the period of an artificial Earth satellite moving at a height $h = nR_Z$ ($n = 1, 2, 3, \ldots$), where $R_Z$ is the Earth's radius.

Step 1: Idea

We use the period formula from Problem 19 and express the orbit radius in Earth radii. It is convenient to replace $\gamma M$ by $gR_Z^2$, so that only the well-known values $g$ and $R_Z$ appear.

Step 2: General formula

The height is measured from the surface, so the orbit radius is $r = R_Z + nR_Z = (1 + n)R_Z$. Then

\[ T = 2\pi\sqrt{\frac{r^3}{\gamma M}} = 2\pi\sqrt{\frac{(1 + n)^3R_Z^3}{gR_Z^2}}, \]

\[ T = 2\pi\sqrt{\frac{R_Z}{g}}\,(1 + n)^{3/2}. \]

The first factor is the period of an orbit at the surface (Problem 19): \[\begin{gathered} T_n = T_0\,(1 + n)^{3/2} \\ T_0 = 2\pi\sqrt{\frac{R_Z}{g}} \approx 84.4\ \mathrm{min} \end{gathered}\]

Step 3: Values

For $n = 1, 2, 3$ the factors are $2^{3/2} = 2.83$, $3^{3/2} = 5.20$ and $4^{3/2} = 8$: $T_1 = 84.4 \cdot 2.83\ \mathrm{min}$.

\[ T_2 = 84.4 \cdot 5.20\ \mathrm{min}, \]

\[ T_3 = 84.4 \cdot 8\ \mathrm{min}. \]

\[\begin{gathered} T_1 \approx 3.98\ \mathrm{h} \\ T_2 \approx 7.31\ \mathrm{h} \\ T_3 \approx 11.25\ \mathrm{h} \end{gathered}\]

Doubling the orbit radius ($n = 1$) makes the period $2\sqrt2 \approx 2.8$ times longer: the path is twice as long and the satellite is $\sqrt2$ times slower.

Check

Geostationary orbit ($T = 23.93\ \mathrm{h}$): $ (1 + n)^{3/2} = 17.0 $, so $1 + n = 6.6$ (Problem 24) $\checkmark$