Satellites: Orbits and Energy — Question 7

Question 7

Matching a rotation period is necessary but not sufficient for a stationary sky position

A spherical planet has $\mu=3.986\times10^{14}\,\mathrm{m^3/s^2}$, radius $R=6.378\times10^6\,\mathrm m$, and rotation period $T=86164\,\mathrm s$. Ignore perturbations.

1. Find the circular orbit radius and altitude for this period.

2. State the additional geometric and directional requirements for a satellite to remain above one fixed surface longitude and latitude.

Question 7 - Solution

Matching a rotation period is necessary but not sufficient for a stationary sky position

(a) Derivation and calculation

Match circular orbital angular speed to planetary rotation: $$\frac{\mu}{r^2}=\left(\frac{2\pi}{T}\right)^2r, \qquad r=\left(\frac{\mu T^2}{4\pi^2}\right)^{1/3}.$$ $$\boxed{r=\left[\frac{(3.986\times10^{14})(86164)^2}{4\pi^2}\right]^{1/3} \simeq4.2164\times10^7\ \mathrm m}.$$ $$\boxed{h=r-R\simeq3.5786\times10^7\ \mathrm m}.$$

(b) Result and interpretation

A fixed surface-relative position requires $$e=0,\quad i=0,\quad \Omega_{\mathrm{orbit}}=\Omega_{\mathrm{planet}}>0.$$ The orbit is circular, equatorial, and prograde. Then $$\frac{d}{dt}(\phi_{\mathrm{satellite}}-\phi_{\mathrm{surface}})=0,\qquad \lambda_{\mathrm{satellite}}=0.$$ Inclination produces latitude motion; eccentricity produces nonuniform angular motion; retrograde rotation cannot keep the longitude fixed.

Physical check

The calculated radius is measured from the center, while altitude subtracts the planetary radius.