Съдържание

Условие Отговори Решение

USAPhO 2003, Полуфинал — Задача 1. A1

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Всички материали от USAPhO 2003

AAPT / AIP United States Physics Team 2003 — 2003 Semi-Final Exam

Указания

DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN

Work Part A first. You have 90 minutes to complete all four problems.
After you have completed Part A, you may take a break.
Then work Part B. You have 90 minutes to complete both problems.
Show all your work. Partial credit will be given.
Start each question on a new sheet of paper. Be sure to put your name in the upper right-hand corner of each page, along with the question number and the page number/total pages for this problem. For example,

Doe, Jamie
A1 – 1/3
A hand-held calculator may be used. Its memory must be cleared of data and programs. It may not be used in graphing mode. Calculators may not be shared. You may not use any tables, books, or collections of formulas. You may use a ruler or straight edge.
Questions with the same point value are not necessarily of the same difficulty.
Good luck!

Possibly Useful Information

Gravitational field at the Earth’s surface	$g = 9.8\ \mathrm{N/kg}$
Newton’s gravitational constant	$G = 6.67 \times 10^{-11}\ \mathrm{N{\cdot}m^2/kg^2}$
Coulomb’s constant	$k = 1/4\pi\varepsilon_0 = 9.0 \times 10^{9}\ \mathrm{N{\cdot}m^2/C^2}$
Biot-Savart constant	$k_\mathrm{m} = \mu_0/4\pi = 10^{-7}\ \mathrm{T{\cdot}m/A}$
Speed of light in a vacuum	$c = 3.0 \times 10^{8}\ \mathrm{m/s}$
Boltzmann’s constant	$k_\mathrm{B} = 1.38 \times 10^{-23}\ \mathrm{J/K}$
Avogadro’s number	$N_\mathrm{A} = 6.02 \times 10^{23}\ \mathrm{(mol)^{-1}}$
Ideal gas constant	$R = N_\mathrm{A}k_\mathrm{B} = 8.31\ \mathrm{J/(mol{\cdot}K)}$
Stefan-Boltzmann constant	$\sigma = 5.67 \times 10^{-8}\ \mathrm{J/(s{\cdot}m^2{\cdot}K^4)}$
Elementary charge	$e = 1.6 \times 10^{-19}\ \mathrm{C}$
1 electron volt	$1\ \mathrm{eV} = 1.6 \times 10^{-19}\ \mathrm{J}$
Planck’s constant	$h = 6.63 \times 10^{-34}\ \mathrm{J{\cdot}s} = 4.14 \times 10^{-15}\ \mathrm{eV{\cdot}s}$
Electron mass	$m = 9.1 \times 10^{-31}\ \mathrm{kg} = 0.51\mathrm{MeV/c^2}$
Binomial expansion	$(1 + x)^n \approx 1 + nx$ for $\lvert x\rvert \ll 1$
Small angle approximations	$\sin\theta \approx \theta$ ; $\cos\theta \approx 1 - \tfrac{1}{2}\theta^2$

Условие

A satellite of mass $m$ is in a circular orbit of radius $r$ around a star of mass $M$ .

a. What is the magnitude of the satellite’s momentum $p$ ? Express your answer in terms of given quantities and constants. [5 т.]

In Parts b, c, and d you should express the impulse vector $\vec{I}$ in terms of the satellite’s original momentum vector $\vec{p}$ .

b. We wish to give the satellite a single impulse of minimum possible magnitude, causing it to escape from the star. What impulse $\vec{I}_e$ is necessary? [5 т.]

c. We wish to give the satellite a single impulse of minimum possible magnitude, causing it to crash into the star. What impulse $\vec{I}_{C0}$ is necessary if you neglect the radius of the star? [5 т.]

d. Suppose that the star has a radius $R$ . Now what impulse $\vec{I}_{CR}$ is necessary to crash the satellite into the star? Your answer should also involve the ratio $r/R$ . [10 т.]

Отговори

Решение

a. Setting the mass times centripetal acceleration equal to the gravitational force

m\frac{v^2}{r} = G\frac{Mm}{r^2}.

Solving for speed

v = \sqrt{\frac{GM}{r}},

and multiplying by mass to get the momentum

p = mv = m\sqrt{\frac{GM}{r}}

b. To escape from the star the total energy must be at least zero. Writing the escape speed $v_e$ ,

\tfrac{1}{2}mv_e^2 - G\frac{Mm}{r} = 0.

Solving for $v_e$ , and multiplying by mass to get the needed momentum,

p_e = mv_e = m\sqrt{\frac{2GM}{r}} = \sqrt{2}p.

This is greater than p above, so the impulse must be in the direction of $\vec{p}$ .

\vec{I}_e = \vec{p}_e - \vec{p} = \left(\sqrt{2} - 1\right)\vec{p}

c. Since angular momentum $L = mvr$ must be conserved, the only way to crash into a star with $R = 0$ is with $L = 0$ . The satellite’s momentum $mv_{C0}$ must be slowed to zero.

\vec{I}_{C0} = m\vec{v}_{C0} - \vec{p} = 0 - \vec{p} = -\vec{p}.

d. Letting $v_r$ be the new speed at orbital radius $r$ and $v_R$ be the speed at stellar radius $R$ , angular momentum conservation yields

mv_r r = mv_R R.

Solving for $v_R$

v_R = v_r\frac{r}{R}. \qquad \mathrm{(A1\text{-}1)}

Mechanical energy is also conserved.

\tfrac{1}{2}mv_r^2 - G\frac{Mm}{r} = \tfrac{1}{2}mv_R^2 - G\frac{Mm}{R}

Regrouping terms and canceling $m$ ,

G\frac{M}{R} - G\frac{M}{r} = \tfrac{1}{2}v_R^2 - \tfrac{1}{2}v_r^2.

Substituting (A1-1)

2GM\left(\frac{1}{R} - \frac{1}{r}\right) = v_r^2\left(\frac{r}{R}\right)^2 - v_r^2

\frac{2GM}{rR}(r - R) = \frac{v_r^2}{R^2}\left(r^2 - R^2\right)

Multiplying by $R$ and dividing by $(r - R)$

\frac{2GM}{r} = \frac{v_r^2}{R}(r + R)

Solving for $v_r$ and multiplying by mass to get $p_r$ ,

p_r = mv_r = m\sqrt{\frac{2GMR}{r(r + R)}} = m\sqrt{\frac{GM}{r}}\sqrt{\frac{2}{1 + \frac{r}{R}}} = p\sqrt{\frac{2}{1 + \frac{r}{R}}}

This is less than the original orbital momentum. Finding the impulse,

\vec{I}_{CR} = \vec{p}_r - \vec{p} = \vec{p}\sqrt{\frac{2}{1 + r/R}} - \vec{p} = -\vec{p}\left(1 - \sqrt{\frac{2}{1 + r/R}}\right)

Оригинал в Архива: Semifinal 2003.pdf · официални решения: Semifinal 2003 Solutions.pdf