Съдържание

IPhO 1990, Теория и експеримент — Задача 2. Electric experiments in the magnetosphere of the earth.

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Всички материали от IPhO 1990

THE 21st INTERNATIONAL PHYSICS OLYMPIAD - 1990 · 10 т.

Внимание: Бележка към темата

Complete combined archive booklet, attributed to Hans Jordens, contains three theory questions and two experiments with keys. Printed errors and gaps are preserved: the neutron-star key changes the flattening denominator and uses a square-law radius inference; the LED key references missing graphs, has printed inconsistent table values and one unassigned final value; magnet-speed tables print m/s but their values match cm/s. Native source values and all grading remain unchanged.

Условие

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.

Atlantis A with orbital velocity V, satellite S, rod L, angle alpha, Earth surface and outward magnetic field B. — 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.

a₁- 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? [1 т.]

a₂- Discuss the stability of the equilibrium for each case. [1 т.]

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. [5 т.]

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.

c₁- In which direction does the current flow through the rod? (Take $\alpha=0$ ) [1 т.]

Data:

Quantity	Value
the period of the orbit	$T=5{,}4\cdot10^3\,\mathrm s$
length of the rod	$L=2{,}0\cdot10^4\,\mathrm m$
magnetic field strength of the earth 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 $0.1\,\mathrm A$ in the opposite direction.

c₂- How long must this current be maintained to change the altitude of the orbit by $10\,\mathrm m$ .
Assume that $\alpha$ remains zero. Ignore all contributions from currents in the magnetosphere.

Does the altitude decrease or increase? [2 т.]

Решение

a₁-

Since $m_2\ll m_1$ , the Atlantis travels around the earth with a constant speed. The motion of the satellite is composed of the circular motion of the Atlantis about the earth and (possibly) a circular motion of the satellite about the Atlantis.
For $m_1$ we have:

m_1\Omega^2R=\frac{Gm_1m_a}{R^2}\quad\rightarrow\quad\Omega^2=\frac{Gm_a}{R^3}

Atlantis and satellite force geometry; m1 m2 R L cos alpha alpha Fc Fg.

For $m_2$ we have:

m_2L\ddot\alpha=-(F_g-F_c)\sin(\alpha)=-\left(\frac{Gm_2m_a}{(R-L\cos(\alpha))^2}-m_2\Omega^2(R-L\cos(\alpha))\right)\sin(\alpha)

Using the approximation:

\frac1{(R-L\cos(\alpha))^2}\approx\frac1{R^2}+\frac{2L\cos(\alpha)}{R^3}

and equation (1), one finds:

L\ddot\alpha=-\left(\frac{Gm_a}{R^2}+\frac{2Gm_a}{R^3}L\cos(\alpha)-\frac{Gm_a}{R^3}R+\frac{Gm_a}{R^3}L\cos(\alpha)\right)\sin(\alpha)

so:

\ddot\alpha+3\Omega^2\sin(\alpha)\cos(\alpha)=0\qquad (2)

If $\alpha$ is constant: $\ddot\alpha=0$

\begin{aligned}\sin(\alpha)=0&\quad\rightarrow\quad\alpha=0;\quad\alpha=\pi\\\cos(\alpha)=0&\quad\rightarrow\quad\alpha=\pi/2;\quad\alpha=3\pi/2\end{aligned}

a₂-

The situation is stable if the moment $M=m_2L\ddot\alpha L=m_2L^2\ddot\alpha$ changes sign in a manner opposed to that in which the sign of $\alpha-\alpha_0$ changes:

$\alpha$	sign( $\alpha-\alpha_0$ ), left	sign( $\alpha-\alpha_0$ ), right	sign( $M$ ), left	sign( $M$ ), right
$0$	−	+	+	−
$\pi/2$	−	+	−	+
$\pi$	−	+	+	−
$3\pi/2$	−	+	−	+
$2\pi$	−	+	+	−

The equilibrium about the angles $0$ en $\pi$ is thus stable, whereas that around $\pi/2$ and $3\pi/2$ is unstable.

b-

For small values of $\alpha$ equation (2) becomes:

\ddot\alpha+3\Omega^2\alpha=0

This is the equation of a simple harmonic motion.
The square of the angular frequency is:

\omega^2=3\Omega^2

so:

\omega=\Omega\sqrt3\quad\rightarrow\quad T_1=\frac{2\pi}{\omega}=\frac13\sqrt3\left(\frac{2\pi}{\Omega}\right)\approx0{,}58T_0

c₁-

According to Lenz's law, there will be a current from the satellite (S) towards the shuttle (A).

c₂-

For the total energy of the system we have:

U=U_{kin}+U_{pot}=\frac12m\Omega^2R^2-\frac{Gmm_a}{R}=-\frac12\frac{Gmm_a}{R}

A small change in the radius of the orbit corresponds to a change in the energy of:

\Delta U=\frac12\frac{Gmm_a}{R^2}\Delta R=\frac12m\Omega^2R\Delta R

In the situation under $c_1$ energy is absorbed from the system as a consequence of which the radius of the orbit will decrease.
Is a current source inside the shuttle included in the circuit, which maintains a net current in the opposite direction, energy is absorbed by the system as a consequence of which the radius of the orbit will increase.
From the assumptions in $c_2$ we have:

\Delta U=F_lv t=BIL\Omega R t=\frac12m\Omega^2R\Delta R\quad\rightarrow\quad t=\frac12\frac{m\Omega\Delta R}{BIL}

Numerical application gives for the time: $t\approx5{,}8\cdot10^3\,\mathrm s$ ; which is about the period of the system.

Marking breakdown:

Part	Criterion	Marks
$a_1$		1
$a_2$		1
b	Atlantis in uniform circular motion	0,5
b	calculation of the period $\Omega$	0,5
b	equation of motion of the satellite	2,5
b	equation of motion for small angles	0,5
b	period of oscillations	1
$c_1$		1
$c_2$	calculation of the time the current has to be maintained	1,5
$c_2$	increase or decrease of the radius of the orbit	0,5

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