Irodov 1.225: Retrograde Orbit Against a Spinning Earth

Problem Irodov 1.225

A satellite moves in a circular orbit of radius $R = 1.00\times10^4$ km in the Earth’s equatorial plane, traveling from east to west, opposite to the Earth’s own sense of rotation. Find the satellite’s velocity and acceleration as measured in a reference frame that is fixed to, and rotates with, the Earth.

The satellite m rides a ring of radius R (dimensioned from the Earth's center out to the ring's near-empty limb), moving against the Earth's own eastward spin Ω; its velocity v, tangent to the orbit, carries it west. — The satellite $m$ rides a ring of radius $R$ (dimensioned from the Earth’s center out to the ring’s near-empty limb), moving against the Earth’s own eastward spin $\vect\Omega$ ; its velocity $\vect v$ , tangent to the orbit, carries it west.

Given

$R = 1.00\times10^4$ km, circular equatorial orbit, east to west

Find

the speed

v'

and the acceleration

w'

of the satellite in the frame fixed to the Earth

Solution

Step 1Orbital speed and the Earth’s own rotation rate

Gravity supplies the centripetal force for the circular orbit, where $M$ is the Earth’s mass; the Earth itself rotates uniformly about its own axis with period $T$ .

\frac{\gamma M m}{R^2} = \frac{m v^2}{R} \;\Longrightarrow\; v = \sqrt{\frac{\gamma M}{R}}, \qquad \Omega = \frac{2\pi}{T}. \tag{1}

Step 2Speeds add for a retrograde orbit

The Earth-fixed frame rotates rigidly at $\Omega$ , so a point of that frame at the satellite’s radius $R$ moves eastward at $\Omega R$ ; the satellite’s own velocity $v$ is westward, the opposite sense. Its speed relative to the rotating frame is therefore the sum of the two speeds, not their difference.

v' = v + \Omega R = \sqrt{\frac{\gamma M}{R}} + \frac{2\pi R}{T}. \tag{2}

Step 3The relative acceleration is an ordinary centripetal one

Both the satellite’s own angular rate $\omega=v/R$ and the frame’s rate $\Omega$ are constant, so the satellite’s angular rate relative to the frame, $\omega+\Omega$ , is constant too: as seen from Earth-fixed axes the satellite traces a circle of the same fixed radius $R$ at that constant rate, which is genuine uniform circular motion in the rotating frame.

w' = \frac{v'^2}{R}

(3)

Step 4Substitute and evaluate

Substituting (2) into (3) and factoring out $v^2/R$ puts the result in terms of the same quantities.

\begin{align*}w' &= \frac{(v+\Omega R)^2}{R} \\ &= \frac{v^2}{R}\left(1+\frac{\Omega R}{v}\right)^2\end{align*}

Each of the two factors is now written in the given quantities, using $v^2 = \gamma M/R$ from (1),

\begin{align*}\frac{v^2}{R} &= \frac{\gamma M}{R^2} \\ \frac{\Omega R}{v} &= \frac{2\pi R}{T}\sqrt{\frac{R}{\gamma M}}\end{align*}

so that the acceleration reads

w' = \frac{\gamma M}{R^2}\left[1+\frac{2\pi R}{T}\sqrt{\frac{R}{\gamma M}}\right]^2.

The numbers entering, with the mean solar day taken for the Earth’s rotation period, are

T = 8.64\times10^4\ \mathrm{s}, \qquad \gamma M = 3.98\times10^{14}\ \mathrm{m^3/s^2}, \qquad R = 1.00\times10^7\ \mathrm{m},

from which the two speeds are

\begin{align*}v &= \sqrt{\frac{\gamma M}{R}} = \sqrt{\frac{3.98\times10^{14}}{1.00\times10^7}}\ \mathrm{m/s} = 6.31\times10^3\ \mathrm{m/s} \\ \Omega R &= \frac{2\pi R}{T} = \frac{2\pi(1.00\times10^7)}{8.64\times10^4}\ \mathrm{m/s} = 0.727\times10^3\ \mathrm{m/s}\end{align*}

and adding them, then squaring and dividing by $R$ , gives the two answers.

\begin{align*}v' &= v + \Omega R = (6.31+0.727)\times10^3\ \mathrm{m/s} = 7.04\times10^3\ \mathrm{m/s} \\ w' &= \frac{v'^2}{R} = \frac{(7.04\times10^3\ \mathrm{m/s})^2}{1.00\times10^7\ \mathrm{m}} \approx 4.9\ \mathrm{m/s^2}\end{align*}

v' = \sqrt{\gamma M/R} + 2\pi R/T = 7.0\ \mathrm{km/s}

(4)

w' = \frac{\gamma M}{R^2}\left[1+\frac{2\pi R}{T}\sqrt{\frac{R}{\gamma M}}\right]^2 = 4.9\ \mathrm{m/s^2}

(5)

Check

As $\Omega\to0$ (a non-rotating Earth) (2) gives $v'\to v$ and (3) gives $w'\to v^2/R=\gamma M/R^2$ , the ordinary inertial-frame orbital speed and centripetal acceleration, as it must once the frame stops rotating. Numerically $\gamma M/R^2=3.98\ \mathrm{m/s^2}$ and $w'=4.9\ \mathrm{m/s^2}$ is indeed larger, consistent with the frame’s own rotation adding to, not subtracting from, the satellite’s apparent speed and curvature.

Remark

A satellite orbiting the other way, west to east with the Earth’s own spin, would instead have $v'=|v-\Omega R|$ , and $v'=0$ exactly when $v=\Omega R$ : a satellite fixed in the Earth’s sky. Solving $\sqrt{\gamma M/R}=\Omega R$ for that radius gives $R=(\gamma M/\Omega^2)^{1/3}\approx4.22\times10^4$ km, the geostationary radius, more than four times this problem’s $R$ . That $R$ here is so much smaller than the geostationary radius is exactly why $v\gg\Omega R$ and the retrograde correction is a modest 12 percent on top of the orbital speed rather than a comparable term.