The Earth's mean density is rho = 5500 kg/m^3 and its radius…

The Earth's mean density is $\rho = 5500\ \mathrm{kg/m^3}$ and its radius $R = 6.37 \cdot 10^6\ \mathrm{m}$. At what height does an artificial satellite orbit if its period is $T = 90\ \mathrm{min}$ (circular orbit)? Also show that $\left[\sqrt[3]{\gamma\rho T^2/3\pi}\right] = 1$.

Step 1: Idea

From the period we find the orbit radius $r$ (Kepler's third law); the height is then $H = r - R$. The Earth's mass is expressed through its density and radius.

Step 2: Orbit radius

Gravity provides the centripetal force: $m\frac{4\pi^2}{T^2}r = \gamma\frac{Mm}{r^2}$.

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

Insert $M = \tfrac43\pi R^3\rho$: $r^3 = \frac{\gamma T^2}{4\pi^2}\cdot\tfrac43\pi R^3\rho = \frac{\gamma\rho T^2}{3\pi}R^3$.

\[ r = R\sqrt[3]{\frac{\gamma\rho T^2}{3\pi}}. \]

Step 3: Numbers

With $T = 90\ \mathrm{min} = 5400\ \mathrm{s}$: $\frac{\gamma\rho T^2}{3\pi} = \frac{6.67 \cdot 10^{-11} \cdot 5500 \cdot 5400^2}{3\pi} = 1.135$.

\[ \sqrt[3]{1.135} = 1.0431. \]

\[ r = 1.0431\,R = 6.645 \cdot 10^6\ \mathrm{m}, \]

\[ H = r - R = 0.0431\,R. \]

\[H \approx 275\ \mathrm{km}\]

The orbit is only about 4% of the Earth's radius above the surface, just above the densest part of the atmosphere. (The same height would result for any planet with the same density: the size of the planet cancels.)

Step 4: The bracket relation

Square brackets denote the unit of a quantity. The relation says that $\gamma\rho T^2/3\pi$ is a pure number with no unit, which is necessary because it multiplies $R$ to give a length. Checking the units:

\[ \left[\gamma\rho T^2\right] = \frac{\mathrm{N\,m^2}}{\mathrm{kg^2}} \cdot \frac{\mathrm{kg}}{\mathrm{m^3}} \cdot \mathrm{s^2} = \frac{\mathrm{kg\,m\,s^{-2}} \cdot \mathrm{s^2}}{\mathrm{kg\,m}} = 1. \]

(The value itself is 1.043, not 1; the relation is about units only.)

Check

Back-substitution: $ 2\pi\sqrt{r^3/\gamma M} $ with $M = 5.95 \cdot 10^{24}\ \mathrm{kg}$, $r = 6.645 \cdot 10^6\ \mathrm{m}$: $T = 5400\ \mathrm{s}\ \checkmark$