Съдържание

Условие Отговори Решение

USAPhO 1997, Полуфинал — Задача 6. B2

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Всички материали от USAPhO 1997

United States Physics Team — 1997 Exam 2

Указания

Entia non multiplicanda sunt praeter necessitatem

DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN

Please show all work, and do not hesitate to use explanatory words in addition to equations. Partial credit will be awarded.
Begin each problem on a fresh sheet of paper. In the upper right-hand corner of each page, write your name, problem number, and the page number/total number of pages for this problem (e.g., "A2, 3/4" for "Problem A2, page 3 of four").
You may use a non-programmable calculator, but may not use any tables, books, notes, or collections of formulas.
Work Part A first. You have ninety 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.

Possibly useful information:

Gravitational field on Earth’s surface: $g = 9.8\ \mathrm{N/kg}$

Gravitational constant: $G = 6.67\times10^{-11}\ \mathrm{N\text{-}m^2 / kg^2}$

Coulomb constant: $k = 1 / 4\pi\epsilon_o = 8.99\times10^{9}\ \mathrm{N\text{-}m^2 / C^2}$

Speed of light in vacuum: $c = 3.00\times10^{8}\ \mathrm{m/s}$

Permeability of empty space: $\mu_o$ , where $\mu_o\epsilon_o = 1/c^2$

Ideal gas constant $R = 8.31\ \mathrm{J / mol\text{-}K}$

Boltzmann’s constant: $k_B = 1.38\times10^{-23}\ \mathrm{J/K}$

Avagadro’s number: $N_A = 6.02\times10^{23}$

1 electron volt = $1\ \mathrm{eV} = 1.60\times10^{-19}\ \mathrm{J}$

Planck’s constant: $h = 6.63\times10^{-34}\ \mathrm{J\text{-}s} = 4.14\times10^{-15}\ \mathrm{eV\text{-}s}$

$(1 - x)^{-1} = 1 + x + x^2 + x^3 + \ldots$ for $|x| < 1$ $\qquad$ $\cos\theta \approx 1 - \theta^2/2$ for $\theta \ll 1$

$(1 + x)^n = 1 + nx + (1/2!)n(n-1)x^2 + \ldots$ for $|x| < 1$ $\qquad$ $\sin\theta \approx \theta$ for $\theta \ll 1$

Условие

An unmanned space probe approaches the surface of a planet whose atmosphere is pure CO $_2$ (molecular weight $M = 44$ gm/mol). Upon entering atmosphere, the probe descends straight down at a constant speed $v_o$ , recording information about the atmospheric pressure, which is shown on the attached graph. Unfortunately, a technician forgot to calibrate the pressure-measuring instrument, so that the $P$ -axis on the pressure vs. time graph has no units! (The technician has been reprimanded, but fortunately, as leader of the data analysis team, you have figured out how to use the data anyway.) Upon reaching the planet’s surface, the probe reports the surface temperature to be 400 K, and the gravitational field there to be 9.9 N/kg. The radius of the planet is $5.0\times10^{6}$ m. Model the atmosphere locally as an ideal gas.

Graph titled Pressure vsTime Data From Probe: vertical axis P(arbitrary units) from 0 to 60 with gridlines every 10, horizontal axis t(sec) with gridlines every 500 s, labelled 500, 1500, 2500, 3500; the curve rises from 0 at t = 0, slowly at first, passes about 10 near 1550 s, 30 near 3000 s, reaches 60 at about 4000 s and stays flat at 60 afterwards.

(a) Apply Newton’s Second Law to a small slab of the atmosphere of thickness $\Delta y$ that is in static equilibrium, to show that the change in pressure $\Delta P$ between the top and bottom sides of the slab is given by $\Delta P = \pm\rho g\Delta y$ , where $\rho$ the atmosphere’s density and $g$ is the local gravitational field (the plus or minus sign is determined by the choice for the positive direction of the vertical $y$ -axis; work out the sign in terms of your choice). [5 т.]

(b) Estimate the velocity $v_o$ with which the probe descends to the surface. [15 т.]

(c) Estimate the temperature of the atmosphere 15 km above the surface. Neglect the variation of the planet’s gravitational field between the surface and this height. [15 т.]

(d) Justify the neglect of the variation of the planet’s gravitational field between the surface and 15 km above the surface. Suggestion: Calculate the fractional difference between the gravitational field’s magnitude on the surface and its value 15 km above the surface, and assess whether that difference is significant compared to the accuracy of the estimate made in part (c). [5 т.]

Отговори

Решение

(a) Apply $F = ma$ to a slab of atmosphere of thickness $\Delta y$ , cross-sectional area $A$ , density $\rho$ . Take $y$ positive upwards.

Forces: $P(y+\Delta y)A$ (down), $P(y)A$ (up), $mg$ (down).

F = ma \quad (\text{which} = 0)

P(y)A - P(y+\Delta y)A - (\rho A\Delta y)g = 0

\underbrace{P(y+\Delta y) - P(y)}_{\Delta P} = -\rho g\Delta y

\Delta P = -\rho g\Delta y

[note $\rho = \rho(y)$ , $g = g(y)$ ]

(b) Since $v_o = \dfrac{-\Delta y}{\Delta t}$ (probe going down, $y$ points up)

$\Delta P = -\rho g\Delta y$ gives

\frac{\Delta P}{\Delta t} = \rho g v_o \qquad \ldots(1)

Need $\rho$ : From ideal gas eq. of state $PV = nRT$ ,

P = \frac{n}{V}RT \qquad \text{where } \frac{n}{V} = \frac{\#\,\text{moles}}{\text{vol.}} = \frac{\#\,\text{moles}}{\text{mass}}\,\frac{\text{mass}}{\text{vol.}}

\frac{n}{V} = \frac{1}{M}\rho

\Rightarrow\ P = \frac{\rho}{M}RT

\rho = \frac{PM}{RT} \qquad \ldots(2)

Using $\rho$ from (2) in (1),

\frac{\Delta P}{\Delta t} = \frac{PM}{RT}\,g v_o

v_o = \left(\frac{\Delta P}{\Delta t}\right)\frac{RT}{PMg} \qquad \ldots(3)

From graph, probe lands on surface at $t = 4000$ s

P = 60\ \text{units}, \quad g = 9.9\ \mathrm{N/kg}, \quad T = 400\ \mathrm{K}

[From graph: at surface

\frac{\Delta P}{\Delta t} = \frac{(60 - 55)\ \text{units}}{(4000 - 3900)\ \mathrm{s}} = .05\ \text{units/s}

[line from (3900, 55) to (4000, 60) is straight]]

From Eq. (3),

v_o = \frac{(.05\ \text{units/s})(8.3\ \mathrm{J/K\text{-}mol})(400\ \mathrm{K})}{(60\ \text{units})(44\times10^{-3}\ \mathrm{kg/mol})(9.9\ \mathrm{N/kg})} \cong 6.4\ \mathrm{m/s}

(c) Probe was 15 km above surface

\Delta t = \frac{15\ \mathrm{km}}{6.4\ \mathrm{m/s}} \cong 2344\ \mathrm{s} \text{ before touchdown; ie., at}

t = (4000 - 2344)\ \mathrm{s} = 1656\ \mathrm{s}

at $t \approx 1650$ s, $P = 11$ units

slope at $t = 1650$ s is about the same as slope of line from (1500 s, 9 units) to (2000 s, 15 units)

\text{slope} \cong \frac{(15 - 9)\ \text{units}}{(2000 - 1500)\ \mathrm{s}} = .012\ \text{units/s}

$\approx .01$ to two sig. figures (like slope found above)

From Eq. (3),

T = \frac{PMg v_o}{R\left(\dfrac{\Delta P}{\Delta t}\right)}

\Rightarrow\ \frac{T_2}{T_1} = \frac{P_2}{P_1}\,\frac{(\Delta P/\Delta t)_1}{(\Delta P/\Delta t)_2} \qquad \begin{matrix}\text{Point 1: surface} \\ \text{Point 2: 15 km above surface}\end{matrix}

T_2 = \frac{11}{60}\left(\frac{.05}{.01}\right)400\ \mathrm{K} \cong 367\ \mathrm{K}

(d)

\begin{aligned} g(h) &= \frac{GM}{(R+h)^2} \\ &= \frac{GM}{R^2}\left(1 + \frac{h}{R}\right)^{-2} \\ &\cong g_o\left(1 - \frac{2h}{R}\right) \\ &\equiv g_o - \Delta g \end{aligned}

g_o = \frac{GM}{R^2}

where

\Delta g = \frac{2h}{R}g_o

\frac{\Delta g}{g_o} = \frac{2h}{R} = \frac{2(15\times10^{3}\ \mathrm{m})}{5\times10^{6}\ \mathrm{m}} = 6\times10^{-3}

In part (c), the estimate of $P$ and $\Delta P/\Delta t$ at 15 km above the surface was much cruder than 6 parts out of a thousand.

Hand sketch for (a), Forces: a thin slab with P(y+Δy)A pushing down on the top, P(y)A pushing up on the bottom, and weight mg downward.

Hand sketch for (a), geometry: a slab of area A and thickness Δy at height y above the ground, with a stick figure standing on the ground.

Hand sketch for (d): a planet of mass M and radius R, with height h above the surface marked; below it g0 = GM/R².

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