Irodov 1.224: When the Satellite Outpaces the Ground

Problem Irodov 1.224

A satellite moves in a circular equatorial orbit of radius $R = 2.00\times10^4$ km, traveling from west to east, the same sense as the Earth’s own rotation. An observer fixed at a point on the equator sees the satellite pass overhead once every $\tau = 11.6$ hours. Given the value of the gravitational constant and the Earth’s own period of axial rotation, find the mass of the Earth.

The satellite's circular equatorial orbit of radius R about the Earth, moving west to east with speed v. — The satellite’s circular equatorial orbit of radius $R$ about the Earth, moving west to east with speed $\vect v$ .

Given

$R = 2.00\times10^4$ km
$\tau = 11.6$ h
$\gamma$ known
$T_E$ known

Find

Find the mass

M

of the Earth

Solution

Step 1The orbital period the satellite would have on its own

Newton’s second law for the satellite, with gravity supplying the centripetal force at radius $R$ , reads

\frac{\gamma M m}{R^2} = m\omega^2 R, \qquad \omega = \frac{2\pi}{T},

where $m$ is the satellite’s mass and $T$ is its own orbital period. Canceling $m$ and substituting $\omega = 2\pi/T$ ,

\begin{align*}\frac{\gamma M}{R^2} &= \left(\frac{2\pi}{T}\right)^2 R \\ \gamma M &= \frac{4\pi^2 R^3}{T^2}\end{align*}

so that the Earth’s mass follows from the orbit’s radius and period alone.

M = \frac{4\pi^2 R^3}{\gamma T^2}

(1)

$T$ is not given directly; it must be recovered from the sighting interval $\tau$ , which is the next step.

Step 2Relating

T

to the sighting interval

The satellite moves west to east, the same sense as the Earth’s own rotation, so a ground observer sees it drift slowly ahead of the planet beneath it. Let $T_E$ be the Earth’s own period of axial rotation. The satellite’s angular speed is $2\pi/T$ and the ground point’s is $2\pi/T_E$ , both in the same sense, so their difference is the rate at which the satellite gains on the ground point:

\omega_{\text{rel}} = \frac{2\pi}{T} - \frac{2\pi}{T_E}.

The satellite laps the ground point, and so reappears overhead, once every time it gains a full turn on it, so $\omega_{\text{rel}} = 2\pi/\tau$ :

\frac{1}{\tau} = \frac{1}{T} - \frac{1}{T_E}.

Solving for $T$ ,

\begin{align*} \frac{1}{T} &= \frac{1}{\tau} + \frac{1}{T_E}, \\ T &= \frac{\tau T_E}{T_E + \tau} \tag{2}\end{align*}

Step 3Combining the two relations

Substituting the orbital period (2) into (1) replaces the unknown $T$ by the two given times,

M = \frac{4\pi^2 R^3}{\gamma}\left(\frac{T_E+\tau}{\tau T_E}\right)^2 ,

and pulling $1/T_E$ out of the bracket separates the Kepler factor from the correction the sighting interval brings,

\begin{align*}\frac{T_E+\tau}{\tau T_E} &= \frac{1}{T_E}\left(1+\frac{T_E}{\tau}\right) \\ M &= \frac{4\pi^2 R^3}{\gamma T_E^2}\left(1+\frac{T_E}{\tau}\right)^2 .\end{align*}

Here $T_E$ is the sidereal rotation period, approximated as $24$ h, and the other three quantities convert to consistent SI units as

T_E = 24\text{ h} = 8.64\times10^4\text{ s}, \qquad R = 2.00\times10^4\text{ km} = 2.00\times10^7\text{ m},

\tau = 11.6\text{ h} = 4.176\times10^4\text{ s}, \qquad \gamma = 6.67\times10^{-11}\ \text{N}\,\text{m}^2\,\text{kg}^{-2}.

Numerically,

1 + \frac{T_E}{\tau} = 1 + \frac{8.64\times10^4}{4.176\times10^4} = 3.069, \qquad \left(1+\frac{T_E}{\tau}\right)^2 = 9.42,

\frac{4\pi^2 R^3}{\gamma T_E^2} = \frac{4\pi^2(2.00\times10^7)^3}{(6.67\times10^{-11})(8.64\times10^4)^2} = 6.34\times10^{23}\ \text{kg}.

Multiplying the two factors together,

M = (6.34\times10^{23}\ \text{kg})(9.42) = 5.97\times10^{24}\ \text{kg}.

M = \dfrac{4\pi^2 R^3}{\gamma T_E^2}\left(1+\dfrac{T_E}{\tau}\right)^2 = 6.0\times10^{24}\ \text{kg}

(3)

Check

As $\tau \to \infty$ the factor $(1+T_E/\tau)^2 \to 1$ and $M \to 4\pi^2R^3/(\gamma T_E^2)$ , exactly Kepler’s third law for $T=T_E$ : the satellite’s own period equals the Earth’s, it is geostationary, and it never again passes overhead, consistent with $\tau\to\infty$ . And $T = \tau T_E/(T_E+\tau) \approx 2.82\times10^4$ s $\approx 7.8$ h, less than $T_E$ as the sign in Step 2 assumed.

Remark

The relation $1/\tau = 1/T - 1/T_E$ is the same synodic-period construction that gives the length of the lunar month or the interval between planetary oppositions: the observed repeat interval is set by the difference of two angular rates, never by either alone. A west to east (prograde) satellite below geostationary radius, where $T<T_E$ as here, is sighted less often than its own orbital period ( $\tau>T$ here, $11.6$ h against $7.8$ h) because the Earth’s rotation chases it in the same sense; above geostationary radius that same prograde satellite would instead be sighted more often than $T$ , since $T>T_E$ there flips the sign of $1/T-1/T_E$ . A retrograde satellite at the same radius would be sighted more often than $T$ , since the two rotations would then add instead of subtract.