Irodov 1.218: The Synodic Rhythm of Close Orbits – Amit & Deepali Sinha Foundation

Problem Irodov 1.218

Two satellites orbit the Earth on circular paths that lie in the same plane. One orbits at radius $r = 7000\text{ km}$ ; the other orbits at the smaller radius $r-\Delta r$ , with $\Delta r = 70\text{ km}$ . Find the time interval between successive instants at which the two satellites come closest to each other.

The two circular orbits, radii r and r-Δ r and not to scale, at an instant when the satellites are radially aligned: the closest-approach configuration when both circle the same way. The inner satellite is the faster, v_1<v_2. — The two circular orbits, radii $r$ and $r-\Delta r$ and not to scale, at an instant when the satellites are radially aligned: the closest-approach configuration when both circle the same way. The inner satellite is the faster, $v_1<v_2$ .

Given

$r = 7000\text{ km}$
$\Delta r = 70\text{ km}$

Find

the time interval between closest approaches

Solution

Step 1Angular velocity of a circular satellite orbit

For a satellite of any mass on a circular orbit of radius $\rho$ , Newton’s second law in the radial direction reads

\frac{\gamma M m}{\rho^2} = m\,\omega^2(\rho)\,\rho

\omega(\rho) = \sqrt{\frac{\gamma M}{\rho^3}}

(1)

where $M$ is the Earth’s mass.

Step 2The two satellites drift apart at a rate fixed by

\Delta r

The two orbits differ only slightly, so the difference in angular velocity is well approximated by a derivative. Differentiating (1) and evaluating the change over $\Delta r$ ,

\begin{align*}\frac{d\omega}{d\rho} &= -\frac{3}{2}\sqrt{\gamma M}\,\rho^{-5/2} = -\frac{3}{2}\frac{\omega(\rho)}{\rho} \\ \Delta\omega &= \omega(r-\Delta r)-\omega(r) \approx -\frac{d\omega}{d\rho}\bigg|_r\Delta r\end{align*}

\Delta\omega = \frac{3}{2}\frac{\omega(r)}{r}\,\Delta r

(2)

Step 3The satellites realign periodically

Two satellites at a constant relative angular rate return to the same relative angular position, and so to the same minimum-distance configuration, once every synodic period $\Delta t = 2\pi/(\text{relative rate})$ . The stem does not fix whether the two satellites circle the Earth the same way or in opposite senses, and the relative rate differs between the two cases. For the same sense the satellites separate at the rate (2); for opposite senses they close the gap between them at the sum of the two rates, which (2) also rewrites in terms of $\omega(r)$ alone,

\omega(r-\Delta r) + \omega(r) = \bigl(\omega(r)+\Delta\omega\bigr) + \omega(r) = 2\omega(r) + \Delta\omega .

Dividing $2\pi$ by each of the two relative rates gives the two recurrence times.

\begin{align*}\Delta t_{\text{same}} &= \frac{2\pi}{\Delta\omega} = \frac{2\pi}{\dfrac{3}{2}\dfrac{\omega(r)}{r}\,\Delta r} = \frac{2\pi}{\omega(r)\left(\dfrac{3\Delta r}{2r}\right)} \\ \Delta t_{\text{opp}} &= \frac{2\pi}{2\omega(r)+\Delta\omega} = \frac{2\pi}{\omega(r)\left(2+\dfrac{3\Delta r}{2r}\right)}\end{align*}

The two differ only by the extra $2$ in the denominator, so a single relative-rate parameter $\delta$ carries both branches at once:

\delta = \begin{cases} 0, & \text{same sense} \\ 2, & \text{opposite senses} \end{cases}

(3)

With that parameter the two lines collapse to one,

\Delta t = \frac{2\pi}{\omega(r)\left(3\Delta r/(2r)+\delta\right)} .

Finally (1) supplies the prefactor,

\frac{2\pi}{\omega(r)} = 2\pi\sqrt{\frac{r^3}{\gamma M}} = \frac{2\pi}{\sqrt{\gamma M}}\,r^{3/2},

and substituting it leaves the recurrence time in the given quantities alone.

\Delta t \approx \frac{2\pi}{\sqrt{\gamma M}}\cdot\frac{r^{3/2}}{3\Delta r/(2r)+\delta}

(4)

Step 4Evaluate numerically

The recurrence time (4) is evaluated with the Earth’s mass $M_\oplus$ and the given $r=7.0\times10^6\ \text{m}$ , $\Delta r = 7.0\times10^4\ \text{m}$ , building up the shared factor first.

\begin{align*}\gamma M_\oplus &= (6.67\times10^{-11})(5.96\times10^{24}) = 3.98\times10^{14}\ \text{m}^3/\text{s}^2 \\ \frac{2\pi}{\sqrt{\gamma M_\oplus}} &= 3.15\times10^{-7}\ \text{s}\cdot\text{m}^{-3/2} \\ r^{3/2} &= 1.85\times10^{10}\ \text{m}^{3/2} \\ \frac{2\pi r^{3/2}}{\sqrt{\gamma M_\oplus}} &= (3.15\times10^{-7})(1.85\times10^{10}) = 5.83\times10^{3}\ \text{s} \\ \frac{3\Delta r}{2r} &= \frac{3(7.0\times10^4)}{2(7.0\times10^6)} = 1.5\times10^{-2}\end{align*}

Dividing this shared factor by $3\Delta r/(2r)+\delta$ evaluates both cases from Step 3.

\begin{align*}\Delta t_{\text{same}} &= \frac{5.83\times10^{3}\ \text{s}}{1.5\times10^{-2}} \approx 3.9\times10^5\ \text{s} \approx 4.5\ \text{days} \\ \Delta t_{\text{opp}} &= \frac{5.83\times10^{3}\ \text{s}}{1.5\times10^{-2}+2} \approx 2.9\times10^3\ \text{s} \approx 0.80\ \text{hour}\end{align*}

The book prints $0.84$ hour for the opposite-sense case; the gap is a slip in the key’s own arithmetic, not in this derivation (recorded in the answer record’s note).

\begin{gathered}\Delta t \approx 3.9\times10^5\ \text{s} \approx 4.5\ \text{days} \quad (\delta=0)\\ \Delta t \approx 2.9\times10^3\ \text{s} \approx 0.80\ \text{hour} \quad (\delta=2)\end{gathered}

(5)

Here $\delta=0$ is the same-sense case and $\delta=2$ the opposite-senses one.

Check

Dimensionally, $[\sqrt{\gamma M}] = \text{m}^{3/2}/\text{s}$ , so $r^{3/2}/\sqrt{\gamma M}$ is a time and the dimensionless denominator leaves $\Delta t$ a time, as required. In the same-sense case ( $\delta=0$ ), $\Delta r\to0$ merges the two orbits and $\Delta t\to\infty$ : satellites on identical orbits never separate in phase, which is exactly what an infinite recurrence time means. In the opposite-sense case ( $\delta=2$ ), by contrast, $\Delta r\to0$ leaves $\Delta t\to2\pi/(2\omega(r))=T/2$ finite: two satellites on the same orbit but circling oppositely still meet twice per orbit, at opposite ends of a diameter, half the orbital period apart.

Remark

$\Delta t=2\pi/\Delta\omega$ is the same construction as the synodic period between two planets on slightly different orbits (Earth and Mars, say): whenever two bodies circle a common center at nearly equal but distinct angular rates, the time between repeated alignments is set by the difference of the rates, not by either rate alone. In the co-rotating branch ( $\delta=0$ ) that difference is tiny here, $\Delta r/r$ being only about one percent, so the recurrence time runs far longer than either orbital period: $\Delta t_{\text{same}}/T=\omega(r)/\Delta\omega=2r/(3\Delta r)\approx67$ .