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Условие Отговори Решение

USAPhO 1997, Четвъртфинал — Задача 4

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Всички материали от USAPhO 1997

United States Physics Team — 1997 Creative Response Portion of Exam 1 · 20 т.

Указания

4 Questions, 60 Minutes

Show all work, as partial credit may be earned.

Communicate! The grader will not attempt to read your mind.

A hand-held calculator may be used. Its memory must be cleared of data and programs. Calculators may not be shared.

Possibly useful approximations:

(1 + x)^n \approx 1 + nx \quad \text{for } |x| \ll 1

\cos\theta \approx 1 - \theta^2/2! \quad \text{for } \theta \ll 1

\sin\theta \approx \theta \quad \text{for } \theta \ll 1

Условие

Consider the gravitational force $F$ on a planet due to the Sun when that force includes a small perturbation $\Gamma$ giving a departure from Newton's law of gravitation,

F = (1 + \Gamma)\,(GMm/R^2)

where $M$ is the Sun's mass, $m$ is the planet's mass, $G$ is Newton's constant, $R$ is the distance between the center of the Sun and the center of the planet, and we take $\Gamma$ to be a constant $\ll 1$ . (For example, general relativity provides such a perturbation.) Approximate the orbit as circular.

(a) Show that the planet's period is $T \approx T_o(1 - \tfrac{1}{2}\Gamma)$ , where $T_o$ would be the planet's period if there were no perturbation. [7 т.]

(b) In one revolution, a planet moving under the purely Newtonian force would travel through the angle $2\pi$ . But with the perturbation, in the same amount of time the planet travels through an additional angle $\delta$ . Calculate $\delta$ in terms of $\Gamma$ . [7 т.]

(c) The general theory of relativity contributes the perturbation $\Gamma = 6v^2/c^2$ to Newton's law of gravitation, where $v$ is the speed of the planet relative to the Sun, and $c$ is the speed of light. The precession angle $\delta$ of part (b) may be approximated as $\delta \approx (6\pi/c^2)(GM/R)$ . Demonstrate this claim that $\delta \approx (6\pi/c^2)(GM/R)$ , by using your result of part (b) and the given perturbation. [2 т.]

(d) Using the result $\delta \approx (6\pi/c^2)(GM/R)$ , and using the data below, calculate the numerical value of $\delta$ for the planet Mercury. Express your answer in seconds of arc per century. This is our simple model's prediction of general relativity's contribution to the precession of Mercury's orbit. [4 т.]

Data:

Period of Mercury's orbit = 88 days

Radius of Mercury's orbit = $5.8\times10^{10}$ m

$G = 6.7\times10^{-11}\ \mathrm{Nm^2/kg^2}$

Mass of Sun = $2.0\times10^{30}$ kg

$c = 3.0\times10^{8}$ m/s

3600" = 1 degree.

[This problem adapted from A.P. French, Newtonian Mechanics, Norton & Co., NY, 1971.]

Отговори

Решение

(a, 7) From $F = ma$ , since the orbit is circular so that the acceleration is $v^2/R$ ,
$(1 + \Gamma)\,(GMm/R^2) = mv^2/R$ (2)

Using $v = 2\pi R/T$ where $T$ is the period, (2)

$F = ma$ becomes (after some re-arrangement)
$T^2 = (4\pi^2/GM)R^3\,(1 + \Gamma)^{-1}$
Notice that $T_o^2 = (4\pi^2/GM)R^3$ , where $T_o$ is the period of the planet's orbit with purely the Newtonian force. Hence, $T = T_o(1 + \Gamma)^{-1/2}$ (2)

Since $\Gamma \ll 1$ , by the binomial expansion
$T \approx T_o(1 - \tfrac{1}{2}\Gamma)$ (1)

(b, 7)
In elapsed time
$\Delta t = T_o$ , the planet turns through the angle $2\pi + \delta$ (2)

This angle can also be equated to $\omega T_o$ where $\omega$ is the planet's orbital angular velocity, viz.,
$\omega = 2\pi/T$
$\approx (2\pi/T_o)(1 - \tfrac{1}{2}\Gamma)^{-1}$ (2)

by virtue of the binomial expansion again,
$\omega \approx (2\pi/T_o)(1 + \tfrac{1}{2}\Gamma)$ (1)

Therefore, we complete the calculation:
$2\pi + \delta = \omega T_o$
$= (2\pi)(1 + \tfrac{1}{2}\Gamma)$
cancelling the $2\pi$ ,
$\delta = \pi\Gamma$ . (2)

(c, 2)
$\delta = \pi\Gamma$
$= 6\pi v^2/c^2$
From $F = ma$ above, we may approximate $v^2$ with $GM/R$ , (keeping the $\Gamma$ from $F = ma$ would give $\delta$ here to second order in $\Gamma$ ), so that
$\delta \approx 6\pi GM/Rc^2$ (2)

(d, 4)
For one lap around the Sun, from the given numerical data,
$\delta_1 = 6\pi GM/Rc^2$
$= 6\pi\,(6.7\times10^{-11}\ \mathrm{Nm^2/kg^2})(2.0\times10^{30}\ \mathrm{kg})(3.0\times10^{8}\ \mathrm{m/s})^{-2}(5.8\times10^{10}\ \mathrm{m})^{-1}$
$= 4.8\times10^{-7}$ radians (2)

In 100 years, Mercury makes
$(365/88)\times100 \approx 415$ laps around the Sun. (1)

For 415 laps,
$\delta_{415} = 415\,(4.8\times10^{-7}$ radians $)$
$\approx 2\times10^{-4}$ radians
$= (2\times10^{-4})(180/\pi)$ degrees
$= (2\times10^{-4})(180/\pi)\,(3600'')$
$= 41''$ . (1)

Оригинал в Архива: Quarterfinal 1997.pdf · официални решения: Quarterfinal 1997 Solutions.pdf