Question

A satellite is in a circular orbit around Earth with period $T=90$ minutes. Take Earth's radius $R_E=6370$ km, Earth's mass $M_E=5.98\times10^{24}$ kg, and $G=6.67\times10^{-11}\ \text{N\,m}^2\text{/kg}^2$. Find the altitude above mean sea level (km).

Hint:

For circular orbits, use $T\propto r^{3/2}$. After finding $r$, subtract Earth's radius to get altitude.

Answer 1

Step 1: Orbital radius from period.
For a circular orbit \[ T=2\pi\sqrt{\frac{r^3}{\mu}}, \mu=GM_E. \] Hence \[ r=\Big(\mu\,(T/2\pi)^2\Big)^{1/3}. \]

Step 2: Substitute.
\[ \mu=6.67\times10^{-11}\times 5.98\times10^{24}=3.986\times10^{14}\ \text{m}^3\!/\text{s}^2, \] \[ T=90\times60=5400\ \text{s}, r=\Big(3.986\times10^{14}\,(5400/2\pi)^2\Big)^{1/3}=6.654\times10^6\ \text{m}. \]

Step 3: Altitude.
\[ h=r-R_E=6.654\times10^6-6.370\times10^6=2.840\times10^5\ \text{m} \Rightarrow \boxed{h\approx 284\ \text{km}}. \]