Cuộc thi Vật lí quốc tế cho học sinh trung học từ năm 2011

Như chúng ta đã biết, Olympic Vật lý Quốc tế (IPhO) là một cuộc thi Vật lí quốc tế cho học sinh trung học, lần đầu tiên được tổ chức tại Warsaw (Ba Lan) năm 1967. Năm nay, có thêm một cuộc thi có qui mô trên toàn thế giới như IphO nhưng hình thức và tên gọi thì có một chút khác biệt, cuộc thi mang tên World Physics Olympiad (WoPhO) – Olympic Vật lí Thế giới là một cuộc thi mới dành cho học sinh trung học trên toàn thế giới và Indonesia là nước đăng cai WoPhO lần đầu tiên.

Cuộc thi WoPhO được tiến hành trong một năm, bao gồm ba vòng:

1. Vòng tuyển lựa (từ ngày 05 tháng 01 đến ngày 30 tháng 6 năm 2011) tạo cơ hội cho đông đảo học sinh trung học toàn thế giới tham gia cuộc thi thông qua Internet. Có mười bài toán lí thuyết, không có bài thí nghiệm, đề bài được viết bằng tiếng Anh và tiếng của nước đăng cai. Đề có thể tải tại website WoPhO.Org (hoặc tại đây). Bài giải cần được gửi về Ban Tổ chức trước ngày 30 tháng 6. Ngôn ngữ gửi bài cho vòng thi này là một trong số tiếng Anh, Indonesia, Nga, hay Trung Quốc .

Có 10 câu hỏi, mỗi câu hỏi tối đa 10 điểm. Những người tham gia đạt trên 78% số điểm và qua được buổi phỏng vấn sẽ có thể được tham dự Vòng chung kết, với tổng số tối đa là 100 người.

Nếu học sinh sử dụng một chương trình máy tính để giải quyết bất kỳ vấn đề, mã của chương trình sau đó phải được gắn liền với các cách giải và kèm theo mô tả (tức là ngôn ngữ máy tính được sử dụng, hướng dẫn sử dụng chương trình, các thông số và ký hiệu được sử dụng, và cách giải thích đầu ra số liệu và đồ thị, vv.)

Một Hội đồng Giám khảo Quốc tế, được lựa chọn từ những giáo sư, giảng viên, các lãnh đạo đoàn APhO, IPhO và những người đạt huy chương vàng APhO, IPhO, tham gia lựa chọn đề và chấm điểm cho bài thi Vòng tuyển lựa, phỏng vấn thí sinh đạt điểm trên 78%, giới thiệu 100 thí sinh tham dự vòng chung kết.

2. Vòng thảo luận trên internet (từ 1 tháng 6 đến 27 tháng 12, bằng tiếng Anh) là dịp để mọi người tham gia thảo luận về lời giải của các bài toán trong Vòng tuyển lựa cùng với các bài toán khác ở trình độ Olympic vật lí. Đây cũng là diễn đàn của các nhà vật lí tương lai.

Giải thưởng sẽ được trao cho người tham dự tích cực nhất hoặc/và có lời giải sáng tạo nhất. 3. Vòng chung kết sẽ được tiến hành trên hòn đảo du lịch nổi tiếng Bali, từ 28/12/2011 đến 5/1/2012), đây là thử thách cuối cùng với những người tham gia, một dịp để thi tài với các nhà vô địch về vật lí, những người đạt huy chương vàng APhO và IPhO.

Vòng chung kết gồm vòng thi lí thuyết với ba bài thi, được làm trong năm giờ, và vòng thi thí nghiệm với hai bài thi, được làm trong năm giờ.

Tổng số điểm tối đa cho vòng thi lí thuyết là 30 điểm, cho vòng thi thí nghiệm là 20 điểm. Điểm số cao nhất được lấy làm 100%. Giới hạn dưới về điểm cho huy chương vàng là 90% , cho huy chương bạc là 78% và cho huy chương đồng là 65%. Không có giải khuyến khích. Thí sinh nào đạt điểm cao nhất và đạt điểm cao hơn mọi thí sinh huy chương vàng APhO và IPhO sẽ nhận học bổng trị giá 20 000 USD và danh hiệu “Người chiến thắng của Olympic Vật lí”. Nếu điểm cao nhất thuộc về thí sính huy chương vàng APhO và IPhO, thì thí sính này sẽ nhận 15 000 USD và danh hiệu trên. Ngoài ra, sẽ có giải thưởng cho thí sính đạt điểm cao nhất riêng về lí thuyết và về thí nghiệm, cũng như một số giải đặc biệt.

Mong rằng có nhiều học sinh của Trường Chuyên Hùng Vương Gia Lai nói riêng và của Việt Nam nói chung tham gia cuộc thi này ngay từ vòng tuyển chọn.

Olympic Vật lý quốc tế lần thứ 42 năm 2011 (IPhO 42) sẽ được tổ chức tại Đại học Chulalongkorn ở thủ đô Bangkok Thái Lan ngày 10-18 tháng 7 năm 2011.

Trước đó, IPhO 41 được tổ chức tại Croatia, đề bài và lời giải các học sinh có thể tải về:

Theoretical exam:

• Cover sheet • Problem 1: text, answer sheet, solution • Problem 2: text, answer sheet, solution • Problem 3: text, answer sheet, solution

Experimental exam:

• Cover sheet • Problem 1: text, answer sheet, solution • Problem 2: text, answer sheet, solution

Nguồn: http://wopho.org

SAU ĐÂY LÀ 10 CÂU HỎI CỦA CUỘC THI WOPHO VÒNG 1 NĂM 2011

Charged Discs

Two thin metal discs of radius 5 cm each are suspended by electrically insulating threadssuch that the discs are parallel (see Fig. 1a) and close to each other (for example theirdistance could be 2 mm).

1. Calculate the force between the two discs if they are charged with small charges +qand −q respectively. As q is small, the displacement of the discs and the possibility ofelectric discharge can be neglected.

2. Now consider only one disc; calculate the surface charge distribution on a metal disc ofradius R having total charge +q. (This charge distribution might be useful to answerthe next question.)

After this, the two original discs are each charged +q. A third metal disc of radius R∗ > 5cm is carefully inserted between the two discs; the third disc is neutral and is suspended byan electrically insulating thread. The three discs are all parallel to each other and theircenters lie along the same horizontal line (so that when viewed head-on the discs areconcentric circles). The resulting set-up is shown in Fig. 1c.

3. Find the radius R∗ of the third disc such that the net electrostatic force acting on eachcharged disc is zero. (The fringing effect is neglected in this problem).

Figure 1: Charged discs set-up

Cylinder Collision

A hollow cylinder with mass M and radius R is at rest on a horizontal plane. In the interiorof this cylinder, there is a solid disk with mass m and radius r. Initially, the center of thedisk is at a distance l from the center of the cylinder and moves with velocity v y as shownin Fig. 1. Unless otherwise specified, all collisions are elastic and frictions can be ignored.

Figure 1:

1. Determine the velocity (the x and the y components of the velocity) of the disk andthe cylinder immediately after the first collision. Write your answer in terms of m, M ,v and θ.

2. Determine the velocity (the x and the y components of the velocity) of the disk andthe cylinder immediately after the second collision. Write your answer in terms of m,M , v and θ.

3. If initially the disk is placed at l = (R− r)/2, determine the velocity of the disk andthe cylinder immediately after the n-th collision.

4. What is the condition for l such that immediately after the n-th collision m moves withvelocity v y and M is at rest? Determine the distance between two successive positionsof the center of M when it is at rest.

5. For this part, the friction between the the disk and the cylinder cannot be ignored. Asin part (a), initially the cylinder is at rest, while the center of the disk is at a distancel < (R− r) from the center of the cylinder and moves with velocity v y as shown inFig. 1. If during the collision process the point of contact does not slide, determine theangular velocity of the disk and the cylinder immediately after the first collision.

Dielectric SlabWaveguide

1 Total Internal Reflection

The electric field of a polarized monochromatic plane wave can be generally represented asE(r, t) = E exp i(k.r − ωt), where E is the amplitude of the wave, k the wavenumber, andω the frequency. Suppose that a monochromatic plane wave with frequency ω travels inthe medium of refractive index n1, and is incident on the boundary of another medium ofrefractive index n2. The incoming wave forms an angle θi with respect to the normal ofthe boundary. Throughout this problem, we only consider transverse electric (TE) polarizedwave where the electric field is perpendicular to the plane of incidence and all media arenon-magnetic.

1. In the case of n1 > n2, there exists a critical angle θc where the incoming wave will betotally reflected for θi > θc (total internal reflection or TIR). The phase of the reflectedwave lags by δ with respect to the incident wave. Derive δ and state it in terms of n1,n2, and θi.

2. Using the necessary boundary conditions, derive the reflectance R for the case of TIR.Show that the wave is perfectly reflected for all θi > θc.

2 Constructive Phase Matching

The most simple dielectric waveguide is a planar slab with thickness d and refractive indexn1 located in a homogeneous background medium with refractive index n2 (n2 < n1). In thecase of TIR, the slab can be used to guide waves without loss, with the additional conditionthat the waves interfere constructively. In other words, the wavefronts should be preservedas the waves travel inside the waveguide. The wavenumbers for the vacuum, medium n1,and medium n2 are taken to be k0, k1, and k2, respectively.

1. Find the necessary condition for the constructive phase matching.

2. The wave can only be guided without loss for certain values of θ. Show that in thesecases, θ must satisfy the equations:

k1d cos θ − δ = mπ; m = 0, 1, 2, 3, .... (1)

Verify that the equations above can also be written as:

√u2 + v2 =

k0d

2

√n21 − n2

2, (2)

u tanu = v or − u cotu = v, (3)

with u = k1d2

cos θ and v = d2

√k21 sin2 θ − k22.

3 Maxwell’s Equations

The Maxwell wave equation for the electric field in a dielectric medium of relative permittivityε is (

∂2

∂x2+

∂2

∂y2+

∂2

∂z2

)E(r, t) = µ0εε0

∂2E(r, t)

∂t2. (4)

In the case of the slab waveguide shown in the figure above, ε = n21 for 0 < z < d, and

ε = n22 for z < 0 or z > d. Taking the system coordinates such that the wave travels in the

xz-plane, the electric field can be generally written as

E(r, t) = E(x, z, t) = E(z) exp i(βx− ωt), (5)

where β is the effective propagation constant along the waveguide due to the translationalsymmetry of the structure in the x-direction. In the case of waveguiding the TE polarizedwave (E(z) = E(z)y), E(r, t) should be simple harmonic inside the slab and decayexponentially outside.

1. What is the relation of β to k1 and θ?

2. From the boundary conditions at z = 0 and z = d, derive from the Maxwell equationsthe condition for waveguiding as found in Part 2.

Page 2

4 Mode Solutions

The waveguide mode solutions are solutions of θ where waveguiding occurs inside the slab.The solution for m = 0 (see Part 2) is commonly called the fundamental mode (the lowestmode or the first mode), the m = 1 mode is called as the second mode, and so on.

1. Sketch curves in (u, v) coordinates that represent Eqs. (2)-(3). Determine thenecessary condition for only one mode solution to exist.

2. Show that the maximum number of modes supported by the dielectric slab is

M =

⌈k0d

π

√n21 − n2

2

⌉, (6)

where the de symbol denotes the ceiling function for which the expression inside isincreased to the nearest integer.

3. Verify that the number of mode solutions is incremented by one for every increase offrequency:

∆ω =πc

d√n21 − n2

2

. (7)

4. From eq.1, show that the group velocity (dω/dβ) of each supported mode solution is

vg =d tan θ + ∂δ

∂β

n1dc cos θ

− ∂δ∂ω

. (8)

5. Show that the maximum time disparity for different modes in the dielectric slabwaveguide to travel a distance L is

τ =L

c(n1 − n2). (9)

6. For n1 = 1.7, n2 = 1.5, λ = 800 nm (in vacuum), and d = 1 µm, find all the modesolutions for θ (with θ > θc). Plot the electric field E(z) for these solutions.

Page 3

Magnetic Dipole

Oscillation

A magnetic dipole with magnetic moment m1 is placed at the coordinate origin parallel to thex-axis.

1. Determine the resulting magnetic field in all space.

2. Another dipole is placed at a distance r from the origin at an angle θ to the x-axis. Themagnetic moment of the second dipole, m2, forms an angle α to the x-axis. The whole set-upcan be seen in Fig. 1. Determine the torque on the second dipole.

3. Determine the interaction energy between the two dipoles.

4. Determine the force on the second dipole.

5. The second dipole is tied to the first dipole via a massless string such that the distancebetween the two is fixed at r. While the orientation of the first dipole at the coordinate originis fixed, the orientation of the second dipole may change. It is also allowed to move freely inthe xy-plane around the first dipole. Write down the equation of motion of the second dipole.The mass and moment of inertia of the second dipole are taken to be m and I respectively.

6. Initially the second dipole is at rest on the x-axis, with the magnetic moment forming anangle α0 to the x-axis (α0 << 1). At t = 0, the second dipole is released and allowed to movefreely. Write down the equation of motion of the second dipole assuming θ and α are small.Take I = mr2/5.

7. The system undergoes simple harmonic oscillation. You are asked to determine the normalmode frequencies of oscillation. The system is in a normal mode when the oscillating variablesare in phase and can be written as follows: θ = θ0 cos (ωt+ φ) and α = α0 cos (ωt+ φ). Thereare two possible values of ω (denoted by ω1 and ω2). Determine ω1 and ω2.

8. For each normal mode, determine the ratio of the amplitude of α to θ, c1 = α1/θ1 andc2 = α2/θ2.

9. The system can be described by the following equations:

θ = θ1 cos (ω1t+ φ1) + θ2 cos (ω2t+ φ2) ,

α = c1θ1 cos (ω1t+ φ1) + c2θ2 cos (ω2t+ φ2) .

Using the initial conditions, determine the values of θ1, φ1, θ2, and φ2.

Figure 1:

Helical Rope

A device is built by placing two metal strips, each of mass m, on a large, frictionlesscylinder. Two identical massless elastic ropes each with spring constant k and obeyingHooke’s Law are used to connect the metal strips such that the two ropes are initially attheir natural length x0 and parallel to each other. The contact points of each rope on thesame strip are diametrically opposite to each other, and the whole device can be seen onFigure 1. Strip A is bolted to the cylinder, while strip B is free to move along and rotateabout the cylinder’s axis.

Figure 1: The device in its initial configuration. The screw can be used to prevent rotationof strip B.

1. The cylinder is now oriented such that its axis is vertical under a constantgravitational field g, and strip A is above strip B. Strip B is now rotated N times whileits distance from strip A is maintained at x0. After this, strip B is prevented fromrotating by a screw, as shown in Figure 1.

(a) Find an equation that, given numerical values for the initial conditions, wouldallow you to calculate x1, the new equilibrium position.

(b) Under certain conditions the metal strip will undergo simple harmonic oscillation.Calculate the frequency of oscillation for small ∆x, in terms of k, r, N , x0, and x1.

2. The cylinder is now oriented horizontally, the ropes are returned to their initialposition, and strip B is prevented from rotating by means of the screw.

(a) A horizontal stretching force F is now applied to strip B. If the force is increasedvery gradually, the ropes break when the force reaches a value F0. What, then, isthe minimum amount of constant force required to break the ropes?

(b) If strip B is rotated N times before the screw is put in place keeping the totallength of the ropes x0, calculate the minimum horizontal force required to breakthe ropes if said force is:

i. increased very gradually.

ii. kept constant.

3. The system is returned to its initial horizontal configuration. All constraints on strip Bare now removed, and the strip is rotated through an angle θ0 while keeping thedistance between the two metal strip x0, and then released (initially x(0) = 0, andθ(0) = 0).

(a) Find the equation of motion of strip B!

(b) Solve the equation of motion for x (t) and θ (t)!

(c) Find the maximum velocity and maximum angular velocity, and also the time Trequired for strip B to reach strip A!

Page 2

Liquid Air

A mixture of oxygen and nitrogen gas is stored in a closed container equipped with apiston on one end at a temperature of T = 77.4 K. The total amount of the gas mixtureis 1.1 mole and its initial pressure is 0.5 atm. With the help of the piston the gas mixtureis slowly compressed at constant temperature.

Using plausible assumptions, plot the pressure of the system as a function of its volumeuntil one tenth of the initial volume, if the ratio of the number of moles of oxygen to thenumber of moles of nitrogen is

a)nO2

nN2

=1

9.

b)nO2

nN2

=2

9.

c)nO2

nN2

=1

4.

Find the pressure and volume at distinctive points of these isothermal curves.

You can use the following data:

• Boiling point of liquid nitrogen at 1 atmosphere: 77.4 K

• Boiling point of liquid oxygen at 1 atmosphere: 90.2 K

• Heat of vaporization of oxygen: 213 J/g.

Sliding Block

A rectangular block of width 2b, length 2a, and mass M rests on a rough surface which hasa coefficient of kinetic friction µ. At some time, the block is given a sharp kick, such that itsuddenly attains a horizontal velocity v0. Under certain circumstances the rear end of the

Figure 1: The block, after given its initial velocity.

block will begin to lift and the block will subsequently rotate about its front lower edge,which will remain in contact with the surface.

1. Derive the equation of rotational motion of the block in terms of θ, a, b, µ, and g.

2. Find the physical condition, namely the range of µ, that allows this to happen.

The next question assume this condition is fulfilled, and concerns the subsequent motion ofthe block.

3. Consider a final state in which the block is at rest in the position shown in Fig. 2,where its center of mass has undergone a total horizontal displacement x. Is such a

Figure 2: A presumed final position of the sliding block.

position possible? If yes, calculate the initial velocity required to achieve it for thefollowing values: a = 0.8 m, b = 1.0 m, µ = 0.9, x = 1.65 m, θmax = 1.27 s−1.Note: Knowing a, b and µ the initial velocity can be solved numerically.

Spring and MassProblem

1. A mass M moves toward a semi infinite spring with initial velocity v0, as shown inFig. 1. The spring has mass per unit length µ and spring constant times the springlength K ≡ kL. The mass and the spring collide at x = 0 and t = 0. Write down thevelocity of the mass M after the collision as a function of time, and also write down thevelocity of the mass M as a function of position.

M

0v

Figure 1:

2. For this part another mass m is placed at the other end of the spring. After the initialwave front from the collision of mass M with the spring reaches this mass, how longwould it take for this mass to leave the spring? Also calculate the velocity of mass mwhen it leaves the spring. Assume the waves in the spring travel faster than the initialvelocity of mass M but the spring is long enough so that when the mass m leaves thespring, the reflected waves from m have not yet returned to M .

Page 1 of 1

A Model for Collisions BetweenTwo Solid Objects

One way mechanical energy is lost during collision between two solid objects is by way ofacoustic waves propagating inside both objects. Even though the real situation is quitecomplicated, in this problem it is modeled in a simple way. First treat each solid rod as aspring with unstretched length Ll and Lr. Spring constant times the length of the springare then Kl and Kr, respectively. Mass density (mass per unit length) of the spring aregiven by ρl and ρr. Indexes l and r represent the left and right springs.

The left spring is moving with velocity +v0/2, while the right spring is moving in theopposite direction with velocity −v0/2. Each spring is initially relaxed. At t = 0, thesprings collide at x = 0. The displacement of each point on the springs are described by afunction y(x, t) so that x+ y(x, t) represents the position of said point at time t; this pointwas at x when t = 0.

1. Derive the wave equation on the springs and write down the speed of the wave.

The general solution to the wave equation is given by y(x, t) = ψ(ct− x) + φ(ct+ x), wherec is the speed of the wave propagation. The form of ψ and φ are determined from theboundary conditions.

2. Write down the boundary conditions at x = 0, x = −Ll and x = Lr.

3. Write down the function of y(x, t) before the collision (t ≤ 0), i.e. y0,l(x, t) andy0,r(x, t).

At t = 0, an acoustic wave starts to propagate in both springs away from the collision pointx = 0. The dynamics of the system is analyzed using the space-time diagram as shown inFig. 1. The horizontal axis represents time, and the vertical axis represents the position ofpoints on the spring. Each line in the diagram represents an acoustic wave front thatemerges each time a wave front arrives at the border.

For example, the line AB represents the position of the wave front emerging from thecollision at point A (x = 0) as a function of time. Let functions fl(clt+ x) and fr(crt− x)describe the waves emerging from the collision that propagates in the left and right springs,where cl and cr are the speed of the wave propagation in left and right spring, respectively.The space-time diagram indicates that Ll/cl > Lr/cr in this problem. As the wave front offr(crt− x) arrives at point B, a new reflected wave, gr(crt+ x), emerges. The same eventalso occurs in the left spring at point C.

Now back in the right spring, when the wave front of gr(crt+ x) arrives at the end of thespring (x = 0, at point D in the diagram), a new reflected wave hr(crt− x) and a newtransmitted wave hl(clt+ x) are generated. These phenomena always occur when a wavefront arrives at the border; a new reflected wave or new reflected and transmitted wavesare generated.

4. Write down the wave function y(x, t) in the region I, II, III, IV, V, VI and VII in termsof y0, fr, fl, gr, hr and hl.

Page 1 of 2

x

Figure 1: Space-time diagram

5. Using the boundary condition(s), determine the form of fr(crt− x) and fl(clt+ x) interms of the springs’ properties and initial velocity.

6. Determine the velocity of the contact point (x = 0) immediately after the initialcontact.

7. Using the boundary condition(s), determine the form of gr(crt+ x) in terms of thesprings’ properties and initial velocities.

Now consider a case where both springs are identical except in its length. In this case,ρl = ρr = ρ, Kl = Kr = K. Take Lr < Ll.

8. Determine y(x, t) in region III and IV. Draw a graph for y(x) at t = 0.4Lc. For drawing

the graph, you may use Lr = 0.6L, Ll = L and v0 = 0.5c.

9. Determine y(x, t) in region V. Draw a graph for y(x) at t = 0.8Lc, use the same Lr, Ll

and v0 as in the previous question.

10. When will the two springs separate? Draw a graph for y(x), use the same Lr, Ll and v0as in the previous question.

11. Calculate the coefficient of restitution e between the springs.

12. Calculate the ratio of the translational kinetic energy of the springs after the collisionto the kinetic energy before the collision.

Page 2 of 2

Lagrange PointsStability

In a system that rotates with the Earth around the Sun, there are five equilibrium points(where the sum of the forces is zero). These 5 points are known as Lagrange Points (namedafter Joseph Lagrange, the first person to study this three-body system). Exact analysis ofthis system is very complicated and chaotic. In the following problem, the mass of the twobodies (M1 and M2) are taken to be much larger than that of the third body (m). Thedistance between M1 and M2 is taken to be R.

m

1M

2M

r

1r 2r

R

1mr 2mr

1. Basic equations of the system

(a) Write down the vector of the total gravitational forces Fg on m.

(b) By assuming M1,M2 >> m, determine the angular velocity of the M1 and M2

system (Ω).

(c) In a frame that rotates with the system, there are fictitious forces on m. Writedown the vector of the total forces on this mass (FΩ) in this frame.

(d) Choose a coordinate system where the three masses are in the xy-plane and theangular velocity Ω is in the positive z-axis. The center of the coordinate is set atthe center of mass of M1 and M2 on the x-axis. Write the position of m asr = x(t)i+ y(t)j. In this rotating frame, write down the total forces on m in thex- and y-axis using parameter α = M2

M1+M2and β = M1

M1+M2when the velocity of m

is zero.

2. Identifying Lagrange PointsThere are 5 points with zero net forces in this rotating system. Three of them (callthem L1, L2 and L3) lie on the line connecting M1 and M2 (the x-axis) and the othertwo (call them L4 and L5) lie on the xy-plane on symmetric positions above and belowthe x-axis; that is, y4 = −y5.

(a) First consider the case of finding the position of L1, L2 and L3. Use x = (ν − α)R,with ν the distance of m from M1 in units of R. Write down the equation of forcethat must be satisfied to identify these points. Express this equation in terms of νand α.

(b) The equation above gives rise to three cases (each for L1, L2 and L3) to consider,ν < a, a < ν < b and b < ν. Determine the values of a and b.

From here on, we will also assume that α is small (in the Earth-Sun system, α is3.0 × 10−6). Use only the lowest order non-zero term in α, ignore all higher order termsin α. The following three questions will help you determine the three Lagrange pointson the x-axis.

(c) For the first case, ν < a, write ν = −1 + δ1 with δ1 a small positive number thatdepends on α. This value of ν will determine the position of the first Lagrangepoint at x = −R(1 + ξ1). Determine ξ1 as a function of α.

(d) For the second case, a < ν < b, write ν = 1 − δ2 with δ2 a small positive numberthat depends on α. This value of ν will determine the position of the secondLagrange point at x = R(1 − ξ2). Determine ξ2 as a function of α.

(e) For the third case, b < ν, write ν = 1 + δ3 with δ3 a small positive number thatdepends on α. This value of ν will determine the position of the third Lagrangepoint at x = R(1 + ξ3). Determine ξ3 as a function of α.

Determining the fourth and fifth Lagrange points requires a more complicated method.First decompose the gravitational force on m into components parallel andperpendicular to the vector r.

(f) Find the unit vector parallel to the vector r, e‖. Find also the unit vectorperpendicular to the vector r on the xy-plane, e⊥.

(g) Find the component of the force on m parallel to the vector r, F‖Ω, and find the

component perpendicular to the vector r, F⊥Ω .

(h) Specify the condition that must be satisfied by the force component perpendicularto the vector r in order that mass m be in equilibrium. With this condition,determine the relation between rm1 and rm2.

(i) Specify the condition that must be satisfied by the force component parallel to thevector r in order that mass m be in equilibrium. With this equation, determinethe relation between rm1 and R.

(j) Now determine the position of the fourth Lagrange point (x4, y4) and the fifthLagrange point (x5, y5).

3. Lagrange Point StabilityTo test the stability of these Lagrange points, small perturbation are given to the massm around its equilibrium points. Because the forces in this system depend on the

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position (x, y) and the velocity (vx, vy) of the mass m, the restoring forces must becalculated for variations in position and velocity. Expand the total force as follows:

Fx(x0 + δx, y0 + δy, vx,0 + δvx, vy,0 + δvy) =∂Fx∂x

δx+∂Fx∂y

δy +∂Fx∂vx

δvx +∂Fx∂vy

δvy

Fy(x0 + δx, y0 + δy, vx,0 + δvx, vy,0 + δvy) =∂Fy∂x

δx+∂Fy∂y

δy +∂Fy∂vx

δvx +∂Fy∂vy

δvy.

This force has taken into account the contribution of the velocity of the mass m. Allthe partial derivatives are evaluated at the point (x0, y0, vx,0, vy,0).

(a) Write down the general form for 1m∂Fx

∂x, 1m∂Fx

∂y, 1m

∂Fy

∂x, 1m

∂Fy

∂y. Show that ∂Fy

∂x= ∂Fx

∂y.

(b) Calculate 1m∂Fx

∂vx, 1m∂Fx

∂vy, 1m

∂Fy

∂vx, 1m

∂Fy

∂vy.

These eight coefficients should act as a restoring constant (analog to the springconstant). Now we are ready to check the stability of the five Lagrange points.Consider only the lowest order term in α, ignore all higher order terms.

(c) The first Lagrange Point

i. Show that 1m∂Fx

∂x= c1Ω2. Determine c1.

ii. Show that ∂Fy

∂x= ∂Fx

∂y= 0.

iii. Show that 1m

∂Fy

∂y= c2αΩ2. Determine c2.

iv. By substituting δx = Aeλt and δy = Beλt, with A and B nonzero, determine λas a function of α and Ω only.

v. There are four solutions to λ. Write down the condition that these solutionsmust satisfy in order that the first Lagrange point is stable and thendetermine the stability of this point.

vi. For the Earth-Sun system α is 3.0 × 10−6 and Ω is 2π/year. If this point isstable, determine its period of oscillation (in days), if not, determine its timeconstant 1/λ (in days also).

(d) The second Lagrange Point

i. Show that 1m∂Fx

∂x= c3Ω2. Determine c3.

ii. Show that ∂Fy

∂x= ∂Fx

∂y= 0.

iii. Show that 1m

∂Fy

∂y= c4Ω2. Determine c4.

iv. By substituting δx = Aeλt and δy = Beλt, with A and B nonzero, determine λas a function of α and Ω only.

v. There are four solutions to λ. Write down the condition that these solutionsmust satisfy in order that the second Lagrange point is stable and thendetermine the stability of this point.

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vi. For the Earth-Sun system: if this point is stable, determine its period ofoscillation (in days), if not, determine its time constant 1/λ (in days also).

The third Lagrange point is similar to the second Lagrange point hence it need not beconsidered.

(e) The fourth Lagrange Point

i. Show that 1m∂Fx

∂x= c5Ω2. Determine c5.

ii. Show that 1m

∂Fy

∂x= 1

m∂Fx

∂y= (c6 + c7α)Ω2. Determine c6 and c7.

iii. Show that 1m

∂Fy

∂y= c8Ω2. Determine c8.

iv. By substituting δx = Aeλt and δy = Beλt, with A and B nonzero, determine λas a function of α and Ω only.

v. Define M1/M2 = ξ. Find the range of value of ξ for the fourth Lagrange pointto be stable.

The fifth Lagrange point has the same behavior as the fourth Lagrange point, hence itneed not be considered.

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