IPhO 1990 · Theory problem 2

Electric experiments in the magnetosphere of the earth

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In May 1991 the spaceship Atlantis will be placed in orbit around the earth. We shall assume that this orbit will be circular and that it lies in the earth's equatorial plane.

At some predetermined moment the spaceship will release a satellite S, which is attached to a conducting rod of length \(L\). We suppose that the rod is rigid, has negligible mass, and is covered by an electrical insulator. We also neglect all friction. Let \(\alpha\) be the angle that the rod makes to the line between the Atlantis and the centre of the earth. (see Fig. 1).

S also lies in the equatorial plane.

Assume that the mass of the satellite is much smaller than that of the Atlantis, and that \(L\) is much smaller than the radius of the orbit.

Figure 1

Figure 1 The spaceship Atlantis (A) with a satellite (S) in an orbit around the earth. The orbit lies in the earth’s equatorial plane. The magnetic field (B) is perpendicular to the diagram and is directed towards the reader.

\(\mathrm{a}_1\) - Deduce for which value(s) of \(\alpha\) the configuration of the spaceship and satellite remain unchanged (with respect to the earth)? In other words, for which value(s) of \(\alpha\) is \(\alpha\) constant?

\(\mathrm{a}_2\) - Discuss the stability of the equilibrium for each case.

Suppose that, at a given moment, the rod deviates from the stable configuration by a small angle. The system will begin to swing like a pendulum.

b - Express the period of the swinging in terms of the period of revolution of the system around the earth.

In Fig. 1 the magnetic field of the earth is perpendicular to the diagram and is directed towards the reader. Due to the orbital velocity of the rod, a potential difference arises between its ends. The environment (the magnetosphere) is a rarefied, ionised gas with a very good electrical conductivity. Contact with the ionised gas is made by means of electrodes in A (the Atlantis) and S (the satellite). As a consequence of the motion, a current, \(I\), flows through the rod.

\(\mathrm{c}_1\) - In which direction does the current flow through the rod? (Take \(\alpha=0\))

Data:

- the period of the orbit \(T=5{,}4\cdot10^3\ \mathrm{s}\)
- the length of the rod \(L=2{,}0\cdot10^4\ \mathrm{m}\)
- magnetic field strength of the eart at the height of the satellite \(B=5{,}0\cdot10^{-5}\ \mathrm{Wb.m^{-2}}\)
- the mass of the shuttle Atlantis \(m=1{,}0\cdot10^5\ \mathrm{kg}\)

Next, a current source inside the shuttle is included in the circuit, which maintains a net direct current of 0.1 A in the opposite direction.

\(\mathrm{c}_2\) - How long must this current be maintained to change the altitude of the orbit by 10 m.

Assume that \(\alpha\) remains zero. Ignore all contributions from currents in the magnetosphere.

  • Does the altitude decrease or increase?

Official English text of the problem.

Answers

  • a₁: The rod must be radial or tangential. To leading order in \(L/R\), the directed angles are \[ \boxed{\alpha=0,\ \frac{\pi}{2},\ \pi,\ \frac{3\pi}{2}\pmod{2\pi}}. \]
  • a₂: The radial equilibria, \(\alpha=0,\pi\), are stable. The tangential equilibria, \(\alpha=\pi/2,3\pi/2\), are unstable.
  • b: The small-oscillation period is \[ \boxed{T_{\mathrm{swing}}=\frac{T}{\sqrt3}\simeq3.1\times10^3\ \mathrm{s}}. \]
  • c₁: Conventional current flows from the satellite S to Atlantis A.
  • c₂: The altitude increases, and the required time, in the circular-orbit approximation, is \[ \boxed{t=\frac{\pi m\,\Delta h}{IBLT} \simeq5.8\times10^3\ \mathrm{s}\simeq1.6\ \mathrm{h}}. \]

Solution

a₁ — Equilibrium orientations

Let \(R\) be the orbital radius of Atlantis, \(\mu=GM_{\mathrm E}\), and \[ \Omega=\frac{2\pi}{T}. \] Since Atlantis follows a circular orbit, \[ \Omega^2=\frac{\mu}{R^3}. \] The satellite’s much smaller mass allows us to neglect its effect on Atlantis’s motion.

Use a frame centred on Atlantis and rotating with the orbit. Let \(\mathbf e_x\) point towards the earth’s centre, and let \(\mathbf e_y\) point to the right in Fig. 1. The satellite’s displacement from Atlantis is \[ \boldsymbol{\rho} =L\cos\alpha\,\mathbf e_x+L\sin\alpha\,\mathbf e_y. \]

For a general small displacement \(x\mathbf e_x+y\mathbf e_y\), expansion of the earth’s gravitational field to first order in \(L/R\) gives \[ \mathbf g_S-\mathbf g_A =\frac{\mu}{R^3} \left(2x\,\mathbf e_x-y\,\mathbf e_y\right). \] The additional centrifugal acceleration is \[ \Omega^2\left(x\,\mathbf e_x+y\,\mathbf e_y\right). \] Their sum is therefore \[ \mathbf a_{\mathrm{eff}}=3\Omega^2x\,\mathbf e_x. \]

The unit vector perpendicular to the rod, in the direction of increasing \(\alpha\), is \[ \mathbf e_\alpha=-\sin\alpha\,\mathbf e_x+\cos\alpha\,\mathbf e_y. \] The rod’s force acts along the rod. The Coriolis force also acts along the rod, because the relative velocity \(L\dot\alpha\,\mathbf e_\alpha\) is perpendicular to it. Neither contributes to the tangential equation.

Consequently, \[ L\ddot\alpha =\mathbf a_{\mathrm{eff}}\cdot\mathbf e_\alpha =-3\Omega^2L\sin\alpha\cos\alpha, \] or \[ \boxed{\ddot\alpha+3\Omega^2\sin\alpha\cos\alpha=0}. \]

A constant angle requires \[ \sin\alpha\cos\alpha=0, \] so \[ \boxed{\alpha=\frac{n\pi}{2},\qquad n\in\mathbb Z}. \] Thus the rod is either radial or tangential. These angles use the stated approximation \(L\ll R\).

a₂ — Stability

Write \[ \alpha=\alpha_0+\delta,\qquad |\delta|\ll1, \] where \(\alpha_0\) is an equilibrium angle. Linearizing the equation of motion gives \[ \ddot\delta+3\Omega^2\cos(2\alpha_0)\,\delta=0. \]

For the radial configurations, \(\alpha_0=0,\pi\), \[ \ddot\delta+3\Omega^2\delta=0. \] The acceleration restores the rod towards equilibrium, so both radial configurations are stable.

For the tangential configurations, \(\alpha_0=\pi/2,3\pi/2\), \[ \ddot\delta-3\Omega^2\delta=0. \] This equation admits an exponentially growing displacement, \[ \delta\propto e^{\sqrt3\,\Omega t}, \] so both tangential configurations are unstable.

b — Period of small oscillations

Near either stable equilibrium, the equation found above is that of simple harmonic motion: \[ \ddot\delta+\omega_{\mathrm{swing}}^2\delta=0, \qquad \omega_{\mathrm{swing}}=\sqrt3\,\Omega. \] Hence \[ \boxed{T_{\mathrm{swing}} =\frac{2\pi}{\sqrt3\,\Omega} =\frac{T}{\sqrt3}}. \] Using the given orbital period, \[ T_{\mathrm{swing}} =\frac{5.4\times10^3}{\sqrt3}\ \mathrm{s} \simeq\boxed{3.1\times10^3\ \mathrm{s}}. \]

c₁ — Direction of the induced current

The magnetic force on a positive charge moving with the rod is \[ \mathbf F_B=q\,\mathbf v\times\mathbf B. \] In Fig. 1, \(\mathbf v\) points left and \(\mathbf B\) points out of the page. Therefore \(\mathbf v\times\mathbf B\) points upwards.

For \(\alpha=0\), S lies below A. The motional emf therefore drives positive charges along the rod from S towards A: \[ \boxed{\text{Conventional current flows from S to A.}} \]

c₂ — Time required and direction of altitude change

The current source reverses the current, so it flows from A to S, downwards in the figure. The magnetic force on the rod is \[ \mathbf F=I\,\boldsymbol{\ell}\times\mathbf B, \] where \(\boldsymbol{\ell}\) points along the current. Downwards crossed with out of the page points left, along the orbital velocity.

Since the rod is perpendicular to the field, \[ F=IBL =(0.1)(5.0\times10^{-5})(2.0\times10^4) =\boxed{0.10\ \mathrm N}. \] This forward force supplies energy to the orbit.

Treat the weak thrust as producing a gradual change of a nearly circular orbit. The total orbiting mass is approximately the given shuttle mass \(m\). For a circular orbit, \[ v^2=\frac{\mu}{R}, \] and the total mechanical energy is \[ E=\frac12mv^2-\frac{\mu m}{R} =-\frac{\mu m}{2R}. \] The magnetic force supplies power \(Fv\), so \[ Fv=\frac{dE}{dt} =\frac{\mu m}{2R^2}\frac{dR}{dt}. \] Substituting \(v=\Omega R\) and \(\mu=\Omega^2R^3\), \[ F\Omega R =\frac{m\Omega^2R}{2}\frac{dR}{dt}, \] which gives \[ \boxed{\frac{dR}{dt}=\frac{2F}{m\Omega} =\frac{FT}{\pi m}}. \]

The change of altitude equals the change of orbital radius. Over a rise of only \(10\ \mathrm m\), we may treat \(B\), \(F\), and \(T\) as constant. Thus \[ t=\frac{\pi m\,\Delta h}{FT} =\frac{\pi m\,\Delta h}{IBLT}. \] Numerically, \[ t= \frac{\pi(1.0\times10^5)(10)} {(0.10)(5.4\times10^3)} \ \mathrm s =5.82\times10^3\ \mathrm s. \] Therefore, \[ \boxed{t\simeq5.8\times10^3\ \mathrm s \simeq97\ \mathrm{min}\simeq1.6\ \mathrm h}. \]

Because the force adds orbital energy, \(E\) becomes less negative and \(R\) increases: \[ \boxed{\text{The altitude increases by }10\ \mathrm m.} \]

The solution was checked against the official solution.