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Условие Решение

USAPhO 2006, Четвъртфинал — Задача 4

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Всички материали от USAPhO 2006

Preliminary Exam — Open Response Questions · 25 т.

Указания

DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN

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

#1 – P. 1/3

A hand-held calculator may be used. Its memory must be cleared of data and programs. You may use only the basic functions found on a simple scientific calculator. Calculators may not be shared. Cell phones may not be used during the exam or while the exam papers are present. You may not use any tables, books, or collections of formulas.
Each of the four questions is worth 25 points. The questions are not necessarily of the same difficulty. Good luck!
In order to maintain exam security, do not communicate any information about the problems or the solutions on this exam (or on the multiple-choice exam) until after February 13th.

Possibly Useful Information

Quantity	Value
Gravitational field at the Earth's surface	$g=9.8$ N/kg
Newton's gravitational constant	$G=6.67\times10^{-11}$ N·m $^2$ /kg $^2$
Average Earth-Sun Distance	1 Astronomical Unit (A.U.) $=1.5\times10^{11}$ m
Binomial expansion	$(1+x)^n\simeq1+nx$ for $\lvert x\rvert\ll1$

Moment of Inertia about Center of Mass - Uniform Object

(will not be provided on the second screening exam)

Object	Moment of Inertia
Disk	$\frac12MR^2$
Sphere	$\frac25MR^2$
Rod	$\frac1{12}ML^2$

Solutions to Problems

Each question is worth 25 points. Any correct solution should be awarded equivalent points. Suggested partial-credit points are presented in square brackets in the right margin after the equation number. You may further break down the listed points into one point increments. If it is clear they have done an intermediate step, they should get credit for it even if they have not presented it. Students should not be penalized in a subsequent part for using the wrong answer to a previous part. (No double jeopardy.)

Условие

Consider a spherical asteroid of radius $R_a$ and mass $M_a$ in a circular orbit of radius $r\gg R_a$ around a spherical planet of radius $R_p$ and mass $M_p$ . The asteroid is "tide-locked;" that is, its rotation is such that the same point on its surface is always facing the planet. Let the gravitational constant be $G$ . If the asteroid is too close to the planet, it may be torn apart by tidal forces. Ignore any oblation due to tidal effects.

a) Consider a particle of mass $m$ sitting on the surface of the asteroid closest to the planet. Find the normal (contact) force from the surface of the asteroid acting on the particle.

b) Approximate your answer to part (a) to first order in $\dfrac{R_a}r$ .

c) Assume that the asteroid and the planet have the same mass density. What is the smallest value of $r$ such that the particle remains on the surface, in terms of $R_p$ and/or $R_a$ only? This value is known as the "rigid Roche limit."

Решение

(a) (13 points)

The magnitude of the net force on an object of mass $m$ in uniform circular motion of radius $r$ is

F_{\mathrm{net}}=m\frac{v^2}r\qquad\text{(4-1)}

[1 pt]

but

v=2\pi r/T\qquad\text{(4-2)}

[1 pt]

where $T$ is the period of revolution of the object. Combining the previous two expressions yields

F_{\mathrm{net}}=m\frac{4\pi^2r}{T^2}\qquad\text{(4-3)}

[1 pt]

Both the particle and the asteroid have the same period of revolution. In the case of the asteroid the net force is equal to the gravitational attraction between the planet and the asteroid, so

F_{\mathrm{net}}=M_a\frac{4\pi^2r}{T^2}=\frac{GM_aM_p}{r^2}\qquad\text{(4-4)}

T=\sqrt{4\pi^2\frac{r^3}{GM_p}}\qquad\text{(4-5)}

[2 pt]

The distance from the particle of mass $m$ to the center of rotation is

r_m=r-R_a\qquad\text{(4-6)}

The magnitude of the net force on the particle of mass $m$ is

F_{\mathrm{net}}=m\frac{4\pi^2r_m}{T^2}\qquad\text{(4-7)}

[1 pt]

which can be combined with Eq. 4-5 for the period of the asteroid to yield

F_{\mathrm{net}}=GmM_p\frac{r_m}{r^3}\qquad\text{(4-8)}

[2 pt]

The magnitude of the force of gravity from the planet on the particle of mass $m$ is

F_p=\frac{GmM_p}{r_m^2}\qquad\text{(4-9)}

[1 pt]

The magnitude of the force of gravity from the asteroid on the particle of mass $m$ is

F_a=\frac{GmM_a}{R_a^2}\qquad\text{(4-10)}

[1 pt]

The magnitude of the net force on the particle can also be written as

F_{\mathrm{net}}=F_p-F_N-F_a\qquad\text{(4-11)}

[1 pt]

where $F_N$ is the normal force, so

F_N=F_{\mathrm{net}}+F_a-F_p\qquad\text{(4-12)}

=Gm\left(M_p\frac{r_m}{r^3}+M_a\frac1{R_a^2}-M_p\frac1{r_m^2}\right)\qquad\text{(4-13)}

=Gm\left(M_p\frac{r-R_a}{r^3}+M_a\frac1{R_a^2}-M_p\frac1{(r-R_a)^2}\right)\qquad\text{(4-14)}

[2 pt]

(b) (7 points)

Rewrite the terms, where possible, so that

r-R_a=r(1-x)\qquad\text{(4-15)}

where

x=R_a/r\qquad\text{(4-16)}

is a small quantity. Then Eq. 4-14 can be written as

F_N=Gm\left(M_p\frac{1-x}{r^2}+M_a\frac1{R_a^2}-M_p\frac1{r^2(1-x)^2}\right)\qquad\text{(4-17)}

[2 pt]

Apply a binomial expansion to the last term,

(1-x)^{-2}\simeq1+2x\qquad\text{(4-18)}

[2 pt]

then Eq. 4-14 will be approximated by

F_N\approx Gm\left(M_p\frac{1-x}{r^2}+M_a\frac1{R_a^2}-M_p\frac1{r^2}(1+2x)\right)\qquad\text{(4-19)}

=Gm\left(M_a\frac1{R_a^2}-3M_p\frac1{r^2}x\right)\qquad\text{(4-20)}

=Gm\left(M_a\frac1{R_a^2}-3M_p\frac{R_a}{r^3}\right)\qquad\text{(4-21)}

[3 pt]

(c) (5 points)

Set the normal force equal to zero,

F_N=0\qquad\text{(4-22)}

[1 pt]

and solve Eq. 4-21 for $r$ ,

r=\sqrt[3]{3\frac{M_p}{M_a}}R_a\qquad\text{(4-23)}

[1 pt]

The mass of the planet is equal to the density $\rho$ times the volume, so

M_p=\frac43\pi\rho R_p^3\qquad\text{(4-24)}

[1 pt]

A similar expression holds true for the asteroid, so the mass ratio is

\frac{M_p}{M_a}=\frac{R_p^3}{R_a^3}\qquad\text{(4-25)}

[1 pt]

Combining Eq. 4-23 and Eq. 4-25,

r=\sqrt[3]3\,R_p\qquad\text{(4-26)}

[1 pt]

Оригинал в Архива: Quarterfinal 2006.pdf · официални решения: Quarterfinal 2006 Solutions.pdf