IPhO 1992 · Theory problem 1

A Rotating Satellite

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The figure shows a satellite which is circling the Earth in an approximately circular orbit in the Earth’s equatorial plane. The satellite consists of a massless central body P and four small peripheral bodies B. The four bodies B each have mass \(m\): they are fastened to P by means of long thin wires of length \(r\) that do not stretch. All these five bodies, P and the four bodies B, are coplanar with the equatorial plane, and they can rotate within this plane. The four radial wires are linked to each other by further thin wires which keep the angles between the radial wires constant at \(90^\circ\).

Figure 1

The link wires are included in the system in order to prevent oscillatory movement of the individual bodies B which would otherwise make the analysis of the movements extremely complicated. All the bodies B rotate around P at the same angular velocity, which is \(\omega\) with respect to the fixed stars. Thus, the satellite behaves as a rigid body.

Analyze the following questions for the general case, considering all possible situations, including both senses of rotation of the bodies B. Also obtain numerical values for certain of the quantities found in questions (1) and (2)—the quantities required and the necessary numerical data are listed at the end of the problem.

1) The drawing shows the satellite in the position where for the various wires, \(\mathbf r\) is parallel, anti-parallel or perpendicular to \(\mathbf R\). (The vector \(\mathbf r\) runs from body P to a body B and has length \(r\); the vector \(\mathbf R\) runs from the centre of mass of the Earth to the body P.)

Determine the force exerted by a radial wire on one of the bodies B in each of these four positions. These positions correspond approximately to the maximum and minimum forces.

2) Inside the four bodies B there are four identical machines, powered by solar energy, connected to the radial wires. Each machine pulls the wire in, towards B, for a short time whenever there is near maximum force in the wire (as indicated in the previous question), and lets the same length of wire out again when the tension is at a minimum. The length of wire that is pulled in and let out is \(1\%\) of the mean length of the radial wire. The mean length does not change with time.

What is the net power converted by one machine, averaged over one rotation of the satellite?

The net power is defined as \(\dfrac{W_1-W_2}{T}\), where \(W_1\) is the work that the machine performs on the wire when pulling it in, \(W_2\) is the work that the wire performs on the machine when it is reeled out and \(T\) is the period of rotation.

3) Discuss the changes in the motion of the satellite that are caused by the action of the machines. In particular, analyze any changes that may occur in each of the situations listed in the table overleaf.

Fill in the table with your results and comments, and don’t forget to hand it in.

Data:

Numerical answers are required in the following situation:

The radius of the orbit of the central body is given by \(R=R_E+500\ \mathrm{km}\).

The mean length of the radial wires is \(r=100\ \mathrm{km}\).
Thus the diameter of the satellite system is \(200\ \mathrm{km}\).

The bodies B have masses \(m=1000\ \mathrm{kg}\).

Initially the four bodies B rotate, as referred to the stars, around the central body P at \(10\) revolutions/hour.

The masses of the wires are negligible and the central body P is massless.

Advice:

Consider both senses of rotation for \(\omega\).
Exact solutions are not expected. Results with \(5\%\) accuracy are fully acceptable.
Ignore the gravitational effect of the moon and the sun.

Useful data:

Quantity Value
Mass of Earth \(M_E=5.97\times10^{24}\ \mathrm{kg}\)
Gravitational constant \(G=6.673\times10^{-11}\ \mathrm{m^3\,kg^{-1}\,s^{-2}}\)
Radius of Earth at the equator \(R_E=6378\ \mathrm{km}\)
Denote the product \(M_EG\) by \(K\). \(K=3.983\times10^{14}\ \mathrm{m^3\,s^{-2}}\)

Country code:

ANSWER TABLE

Fill in this table as part of your answer. Write down equalities or inequalities and/or short explanations where necessary.

The quantity indicated below … increases if … decreases if … stays unchanged if … stays unchanged in all situations.
orbital velocity of the satellite Yes ☐ No ☐
radius \(R\) of the orbit of the satellite Yes ☐ No ☐
angular velocity \(\omega\) of the satellite Yes ☐ No ☐
gravitational potential energy of the satellite Yes ☐ No ☐

Could the satellite reach a higher orbit as a result of the work done by the machines?

Yes ☐
No ☐

Could the satellite reach an arbitrarily high orbit, practically leaving the gravitational influence of Earth? Why?

Answer: ……………………………………………………………………

Official English text of the problem.

Answers

Take the orbital direction as positive, and define \[ \Omega=\sqrt{\frac{K}{R^3}}>0,\qquad \epsilon=\frac{\Delta r}{r}=0.01. \] The spin \(\omega\) is signed: positive for prograde rotation and negative for retrograde rotation.

  1. To leading order in \(r/R\), the wire forces, directed towards P, have magnitudes \[ \boxed{F_{\parallel}=F_{\mathrm{anti}}\simeq mr(\omega^2+2\Omega^2)} \] and \[ \boxed{F_{\perp,1}=F_{\perp,2}\simeq mr(\omega^2-\Omega^2)}. \] For either specified spin direction, \[ \boxed{F_{\max}\simeq30.7\ \mathrm{kN},\qquad F_{\min}\simeq30.3\ \mathrm{kN}}, \qquad F_{\max}-F_{\min}\simeq367\ \mathrm N. \] A complete rotation with taut flexible wires requires \(|\omega|\geq\Omega\), to this accuracy.

  2. The mean net power supplied by one machine is \[ \boxed{\overline P_1= \frac{3\epsilon mr^2\Omega^2}{\pi}\,|\omega-\Omega|}. \] Numerically, \[ \boxed{\overline P_1=1.91\ \mathrm{kW}\quad\text{(prograde)}}, \qquad \boxed{\overline P_1=2.17\ \mathrm{kW}\quad\text{(retrograde)}}. \] At synchronous rotation, \(\omega=\Omega\), there is no repeated pumping cycle and the mean power is zero.

  3. The completed table for the secular rigid-body model is:

Quantity Increases if Decreases if Stays unchanged if Unchanged in all situations?
Orbital speed \(v\) \(\omega>\Omega\) \(\omega<\Omega\) \(\omega=\Omega\) No
Orbital radius \(R\) \(\omega<\Omega\) \(\omega>\Omega\) \(\omega=\Omega\) No
Signed spin angular velocity \(\omega\) \(\omega>\Omega\) \(\omega<\Omega\) \(\omega=\Omega\) No
Gravitational potential energy \(U\) \(\omega<\Omega\) \(\omega>\Omega\) \(\omega=\Omega\) No

Thus retrograde rotation becomes faster in magnitude. Formally, a prograde spin with \(0<\omega<\Omega\) slows and subsequently reverses; flexible-wire feasibility in this interval is qualified below.

Higher orbit: Yes, on the ascending branch, in particular with sufficiently rapid retrograde rotation.

Arbitrarily high orbit: Yes within the ideal Newtonian model. Increasingly negative spin angular momentum compensates the increasing orbital angular momentum. There is no finite upper radius, although reaching infinity would require infinite time and unbounded spin energy.

Solution

We use the small parameters \[ \frac rR=\frac{100}{6878}\simeq0.0145,\qquad \epsilon=0.01. \] We retain the leading gravitational-gradient forces and the leading work in the small wire stroke. The orbit is treated as circular on each pumping cycle, with its radius and the mean spin changing slowly. The mean moment of inertia is \[ I\simeq4mr^2. \]

1) Forces in the four radial wires

Let \(\mathbf e_R\) point from Earth towards P, and let \(\boldsymbol\rho=r\mathbf e\) locate a peripheral body relative to P. Expanding Earth’s gravitational acceleration about P gives \[ \mathbf g(\mathbf R+\boldsymbol\rho)-\mathbf g(\mathbf R) \simeq \frac{K}{R^3} \left[3(\boldsymbol\rho\cdot\mathbf e_R)\mathbf e_R-\boldsymbol\rho\right]. \] This is the tidal acceleration.

At the four specified positions, this acceleration is radial relative to P, to the retained order. If \(\theta\) is the angle between \(\mathbf e\) and \(\mathbf e_R\), its radial component is \[ \delta g_r=\Omega^2r(3\cos^2\theta-1). \] The peripheral body needs radial acceleration \(-\omega^2r\) relative to P. Taking outward from P as positive, Newton’s second law gives \[ -m\omega^2r=m\delta g_r-F, \] where \(F\) is the inward wire force. Hence \[ F=mr\left[\omega^2+\Omega^2(3\cos^2\theta-1)\right]. \]

For the outward and inward bodies, \(\cos^2\theta=1\): \[ \boxed{F_{\parallel}=F_{\mathrm{anti}}\simeq mr(\omega^2+2\Omega^2)}. \] Earth’s differential attraction tends to separate these bodies from P, increasing the required tension.

For the two transverse bodies, \(\cos^2\theta=0\): \[ \boxed{F_{\perp,1}=F_{\perp,2}\simeq mr(\omega^2-\Omega^2)}. \] Here the tidal force points towards P and supplies some of the centripetal force.

Consequently, \[ \boxed{\Delta F=F_{\max}-F_{\min}=3mr\Omega^2 =\frac{3Kmr}{R^3}}. \] The small difference between the nearer and farther bodies is beyond this first-order approximation.

Using the supplied data, \[ R=6.878\times10^6\ \mathrm m,\qquad \Omega=\sqrt{\frac{3.983\times10^{14}}{(6.878\times10^6)^3}} =1.1064\times10^{-3}\ \mathrm{rad\,s^{-1}}, \] and \[ |\omega|=\frac{10(2\pi)}{3600} =1.74533\times10^{-2}\ \mathrm{rad\,s^{-1}}. \] The two force scales are \[ mr\omega^2=3.04617\times10^4\ \mathrm N,\qquad mr\Omega^2=122.412\ \mathrm N. \] Therefore, \[ F_{\max}\simeq3.07066\times10^4\ \mathrm N,\qquad F_{\min}\simeq3.03393\times10^4\ \mathrm N, \] giving the rounded answers \(30.7\ \mathrm{kN}\) and \(30.3\ \mathrm{kN}\).

The sense of rotation does not affect these forces, since they depend on \(\omega^2\).

A physical restriction follows from \(F_{\min}\geq0\): \[ |\omega|\geq\Omega. \] If \(|\omega|<\Omega\), the transverse position would require a compressive radial force. A flexible wire cannot provide it, so the assumed rigid rotation cannot persist through a full cycle.

2) Net power of one machine

The stroke is \[ \Delta r=\epsilon r=1000\ \mathrm m. \] To first order in this stroke, the work supplied during one inward stroke and recovered during the following outward stroke are \[ W_{\mathrm{in}}\simeq F_{\max}\Delta r,\qquad W_{\mathrm{out}}\simeq F_{\min}\Delta r. \] Thus one in–out cycle supplies \[ \Delta W=(F_{\max}-F_{\min})\Delta r =3\epsilon mr^2\Omega^2. \] The centrifugal contributions cancel to this order. Numerically, \[ \Delta W\simeq3.67236\times10^5\ \mathrm J. \]

The Earth–satellite line rotates at \(\Omega\) relative to the stars. Therefore the wire’s orientation relative to that line changes at \[ \dot\theta=\omega-\Omega. \] Each wire encounters two maxima and two minima per relative revolution. The relative-revolution period is \[ T_{\mathrm{rel}}=\frac{2\pi}{|\omega-\Omega|}, \] so the net work of one machine during that period is \(2\Delta W\). Hence \[ \boxed{ \overline P_1=\frac{2\Delta W}{T_{\mathrm{rel}}} =\frac{3\epsilon mr^2\Omega^2}{\pi}|\omega-\Omega| }. \]

For prograde rotation, \(\omega=|\omega|\): \[ T_{\mathrm{rel}}=384.37\ \mathrm s, \qquad \overline P_1= \frac{2(3.67236\times10^5)}{384.37} \simeq\boxed{1.91\ \mathrm{kW}}. \]

For retrograde rotation, \(\omega=-|\omega|\): \[ T_{\mathrm{rel}}=\frac{2\pi}{|\omega|+\Omega}=338.54\ \mathrm s, \] \[ \overline P_1\simeq\boxed{2.17\ \mathrm{kW}}. \]

At \(\omega=\Omega\), the orientation relative to Earth is fixed. There are no repeated passages through tension maxima and minima, so the sustained mean power is zero.

3) Changes in the satellite’s motion

Write the total satellite mass as \(M=4m\). To the retained accuracy, its orbital speed, angular momentum, mechanical energy and gravitational potential energy are \[ v=\sqrt{\frac KR}, \] \[ L=M\sqrt{KR}+I\omega, \] \[ E=-\frac{KM}{2R}+\frac12I\omega^2, \qquad U=-\frac{KM}{R}. \]

Each gravitational force points towards Earth’s centre and therefore exerts no torque about that centre. The machines exert internal forces. Thus, neglecting radiation pressure, \[ \boxed{\dot L=0}. \] Differentiating, \[ \frac{M}{2}\sqrt{\frac KR}\,\dot R+I\dot\omega=0, \] or \[ \boxed{\dot R=-\frac{2I}{MR\Omega}\dot\omega =-\frac{2r^2}{R\Omega}\dot\omega}. \] An increase in signed spin angular momentum therefore reduces the orbital radius.

The four machines supply total mechanical power \(P_{\mathrm{tot}}=4\overline P_1\). Differentiating the energy and substituting angular-momentum conservation gives \[ P_{\mathrm{tot}} =\frac{KM}{2R^2}\dot R+I\omega\dot\omega =I(\omega-\Omega)\dot\omega. \] Thus \[ \boxed{P_{\mathrm{tot}}=I(\omega-\Omega)\dot\omega}. \] Since the operating machines supply positive net power, \[ \operatorname{sgn}(\dot\omega)=\operatorname{sgn}(\omega-\Omega). \]

More explicitly, the cycle-averaged rates are \[ \boxed{ \dot\omega=\frac{3\epsilon\Omega^2}{\pi} \operatorname{sgn}(\omega-\Omega) } \] and \[ \boxed{ \dot R=-\frac{6\epsilon r^2\Omega}{\pi R} \operatorname{sgn}(\omega-\Omega) }. \]

These relations establish every entry in the answer table:

  • Faster prograde rotation, \(\omega>\Omega\): the spin increases, the orbit contracts, the orbital speed increases, and the gravitational potential energy decreases.
  • Retrograde rotation, \(\omega<0\): the signed spin decreases, so its retrograde magnitude increases. The orbit expands, the orbital speed decreases, and the gravitational potential energy increases.
  • Formally, \(0<\omega<\Omega\): the prograde spin slows while the orbit expands. Continued operation would reverse the spin. This interval is subject to the flexible-wire limitation found in part (1).
  • Synchronous rotation, \(\omega=\Omega\): there is no repeated actuation, hence no secular change caused by the machines.

In particular, the orbital speed and potential-energy signs follow directly from \[ \dot v=-\frac{v}{2R}\dot R,\qquad \dot U=\frac{KM}{R^2}\dot R. \]

To interpret the table’s angular-velocity row as an unsigned rotation rate: \(|\omega|\) increases for both faster prograde rotation and retrograde rotation; it decreases for the formal slower-prograde case until the spin reverses.

Possibility of reaching higher and arbitrarily high orbits

A sufficiently rapidly counter-rotating satellite has \(\omega<-\Omega\), so its wires remain in tension and \(\dot R>0\). It can therefore reach a higher orbit.

Angular-momentum conservation does not impose a finite upper radius. Between the initial state and a later circular orbit, \[ M\sqrt{KR}+I\omega =M\sqrt{KR_0}+I\omega_0. \] Since \(I/M=r^2\), \[ \boxed{ \omega(R)=\omega_0-\frac{\sqrt K}{r^2} \left(\sqrt R-\sqrt{R_0}\right) }. \] As \(R\) grows, the increasingly positive orbital angular momentum is compensated by increasingly negative spin angular momentum. The solar-powered machines supply the required energy.

The rate of ascent becomes progressively smaller. On the ascending branch, \[ \dot R=\frac{6\epsilon r^2\sqrt K}{\pi}R^{-5/2}, \] which integrates to \[ \boxed{ R(t)^{7/2}=R_0^{7/2} +\frac{21\epsilon r^2\sqrt K}{\pi}\,t }. \] Every finite target radius is reached after a finite time in this ideal model, but \(R\to\infty\) requires \(t\to\infty\). Moreover, the required spin energy grows without bound: \[ \frac12I\omega^2\sim\frac{2mK}{r^2}R \qquad(R\to\infty). \]

Thus arbitrarily high finite orbits are possible under the stated idealizations, while infinite-distance escape is only an asymptotic limit of this circular-orbit pumping process.

The solution was checked against the official solution.