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APhO 2000, Теория — Задача 1. Eclipses of the Jupiter’s Satellite

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Всички материали от APhO 2000

1st ASIAN PHYSICS OLYMPIAD — THEORETICAL COMPETITION · 25 април 2000 г.

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Условие

A long time ago before scientists could measure the speed of light accurately, O Römer, a Danish astronomer studied the time eclipses of the Jupiter’s satellite. He was able to determine the speed of light from observed periods of a satellite around the planet Jupiter. Figure 1 shows the orbit of the earth E around the sun S and one of the satellites M around the planet Jupiter. (He observed the time spent between two successive emergences of the satellite M from behind Jupiter).

A long series of observations of the eclipses permitted an accurate evaluation of the period of M. The observed period T depends on the relative position of the earth with respect to the frame of reference SJ as one of the coordinate axes. The average time of revolution is $T_0$ = 42h 28 m 16s and maximum observed period is ( $T_0$ + 15)s.

Circular orbit of the earth E around the sun S at the centre, with vertical and horizontal diameters drawn; the radius SE makes angle θ with the horizontal axis. A line from E runs to Jupiter J far to the right on the horizontal axis, making a small angle φ with it; a small circle around J shows the orbit of the satellite M. — Figure 1 : The orbits of the earth E around the sun and a satellite M around Jupiter J.

The average distance of the earth E to the Sun is $R_\mathrm{E}$ = 149.6 x $10^{6}$ . The maximum distance is $R_\mathrm{E,max}$ = 1.015 $R_\mathrm{E}$ . The period of revolution of the earth is 365 days and of Jupiter is 11.9 years. The distance of the satellite M to the planet Jupiter $R_\mathrm{M}$ = 422 x $10^{3}$ km.

a. Use Newton’s law of gravitation to estimate the distance of Jupiter to the Sun. Determine the relative angular velocity $\omega$ of the earth with respect to the frame of reference Sun-Jupiter (SJ). Calculate the speed of the earth with respect to SJ. [1 т.]

b. Take a new frame which Jupiter is at rest with respect to the Sun. Determine the relative angular velocity $\omega$ of the earth with respect to the frame of reference Sun-Jupiter ( SJ). Calculate the speed of the earth with respect to SJ. [1 т.]

c. Suppose an observed saw M begin to emerge from the shadow when his position was at $\theta_k$ and the next emergence when he was at $\theta_{k+1}$ , k = 1,2,3,… From these observations he got the apparent periods of revolution T ( $t_k$ ) as a function of time $t_k$ from Figure 1 and then use an approximate expression to explain how the distance influences the observed periods of revolution of M. Estimate the relative error of your approximate distance. [3 т.]

d. Derive the relation between d ( $t_k$ ) and T ( $t_k$ ). Plot period T ( $t_k$ ) as a function of time of observation $t_k$ . Find the positions of the earth when he observed maximum period, minimum period and true period of the satellite M. [2 т.]

e. Estimate the speed of light from the above result. Point out sources of errors of your estimation and calculate the order of magnitude of the error. [2 т.]

f. We know that the mass of the earth = 5.98 x $10^{24}$ kg and 1 month = 27d 7h 3m. Find the mass of the planet Jupiter.

Отговори

Решение

Внимание: Непълно решение

The official solutions contain no solution for part f (mass of Jupiter).

a. ( Total Point : 1 ) Assume the orbits of the earth and Jupiter are circles, we can write the centripetal force = equal gravitational attraction of the Sun.

G\frac{M_E M_s}{R_E^2} = \frac{M_E V_E^2}{R_E}

G\frac{M_J M_S}{R_J^2} = \frac{M_J V_J^2}{R_J}

(0.5 point)

where

G = universal gravitational constant

$M_\mathrm{S}$ = mass of the Sun

$M_\mathrm{E}$ = mass of the Earth

$M_\mathrm{J}$ = mass of the Jupiter

$R_\mathrm{E}$ = radius of the orbit of the Earth

$V_\mathrm{E}$ = velocity of the Earth

$V_\mathrm{J}$ = velocity of Jupiter

Hence

\frac{R_J}{R_E} = \left(\frac{v_E}{v_J}\right)^2

We know

T_E = \frac{2\pi}{\omega_E} = \frac{2\pi R_E}{v_E},\ \mathit{and}

T_J = \frac{2\pi}{\omega_J} = \frac{2\pi R_J}{v_J}

we get

\frac{T_E}{T_J} = \frac{\frac{R_E}{v_E}}{\frac{R_J}{v_J}} = \left(\frac{R_E}{R_J}\right)^{3/2}

R_J = 779.8\times 10^6\ km

( 0.5 point )

b. ( Total Point: 1 ) The relative angular velocity is

\begin{aligned} \omega = \omega_E - \omega_J &= 2\pi\left(\frac{1}{365} - \frac{1}{11.9\times 365}\right) \\ &= 0.0157\ rad/day \end{aligned}

( 0.5 point )

and the relative velocity is

\begin{aligned} v &= \omega R_E = 2.36\times 10^6 km/day \\ &= 27.3\times 10^3\ km \end{aligned}

( 0.5 point )

c. ( Total Point: 3 ) The distance of Jupiter to the Earth can be written as follows

\begin{aligned} \mathrm{d}(t) &= \mathrm{R}_J - \mathrm{R}_E \\ \mathrm{d}(t).\mathrm{d}(t) &= (\mathrm{R}_J - \mathrm{R}_E).(\mathrm{R}_J - \mathrm{R}_E) \end{aligned}

(1.0 point)

\begin{aligned} \mathrm{d}(t) &= \left(\mathrm{R}_\mathrm{J}^2 + \mathrm{R}_\mathrm{E}^2 - 2R_E R_J\cos\omega t\right)^{\frac{1}{2}} \\ &\approx R_J\left(1 - 2\left(\frac{R_E}{R_J}\right)\cos\omega t + ...\right)^{\frac{1}{2}} \\ &\approx R_J\left(1 - \frac{R_E}{R_J}\cos\omega t + ...\right) \end{aligned}

Dotted circular orbit of the earth around the sun S with two nearby positions of the earth E joined to S; the small angle between them is labelled ωT₀ and the angle of SE from the horizontal is θ. Lines from both earth positions run to Jupiter J (with satellite M on a small circle) on the horizontal axis; the angle φ at the axis, an auxiliary dotted line making angle β with the axis, and a small angle α at E are marked. — Figure 1: Geometrical relationship to get Δd(t)

The relative error of the above expression is the order of

\left(\frac{R_E}{R_J}\right)^2 \approx 4\%

The observer saw M begin to emerge from the shadow when his position was at d(t) and he saw the next emergence when his position was at d(t +T0)/ Light need time to travel the distance $\Delta d = d(t+T_0) - d(t)$ so the observer will get apparent period T instead of the true period $T_o$ .

\begin{aligned} \Delta d &= R_E(\cos\omega t - \cos\omega(t+T_0)) \\ &\approx R_E\omega T_0\sin\omega t \end{aligned}

(1.0 point)

because $\omega T_0 \approx 0.03, \sin\omega t + ..., \cos\omega T_0 \approx 1 - ...$

We can also get this approximation directly from the geometrical relationship from Figure 1.

(1.0 point)

or we can use another method.

From the figure above we get

\begin{aligned} \beta &= (\phi + \alpha) \\ \frac{\omega T_0}{2} + \beta + \theta &= \frac{\pi}{2} \end{aligned}

(1.0 point)

\begin{aligned} \Delta d &\approx \omega T_0 R_E\cos\alpha \\ &\approx \omega T_0 R_E\sin\left(\omega t + \frac{\omega T_0}{2} + \phi\right) \end{aligned}

\omega T_0 \approx 0.03\ \mathrm{and}\ \phi \approx 0.19

(1.0 point)

d. ( Total Point: 2)

\begin{aligned} T - T_0 &\approx \frac{\Delta d(t)}{c};\ c = \mathrm{velocity\ of\ light} \\ T &\approx T_0 + \frac{\Delta d(t)}{c} = T_0 + \frac{R_E\omega T_0\sin\omega t}{c} \end{aligned}

(1.0 point)

e. Total Point : 2 from

T_{\max} = T_0 + \frac{R_E\omega T_0}{c}

we get

\frac{R_E\omega T_0}{c} = 15

Hence

C=2.78 x $10^{5}$ km/s

(1.0 point)

Оригинал в Архива: th_problems.pdf · официални решения: th_solutions.pdf