Examination Theory Solution IOAA2007

Theoretical Competition

1st IOAA

December 5, 2007

-----------------------------------------------------------------------------------------------------------SOLUTION FOR QUESTION 1. (30 points for 15 short questions) 1.1 Z

42.5o

α= 23.5

o

W S

E

•

42.5o

x

N

•

A •

AB ⊥ BE

42.5o

B

Equator

E = East point BE = 23.5o = ⏐declination of the Sun⏐ AB BE = o sin 42.5 sin 90o − 42.5o

(

)

sin 42.5o AB = BE = 23.5o tan 42.5o o cos 42.5

local time = (23.5 tan 42.5o / 15) hrs after 6:00 = 7:26 am. The official time at 75o W should be 16 min. less. Ans. 7:10 am.

1

Theoretical Competition

1st IOAA

December 5, 2007

-----------------------------------------------------------------------------------------------------------Alternative solution (for 1.1)

Z′

Z

Latitude λ = 42.5o

N

α= 23.5o DG ⊥ BC

E

Bx G x C

φ

Y′

Ax xD

Sun’s rays

α x O

R

Y

Longitude 71o X&X ′

The Earth’s position relative to the Sun is shown in the figure. Note that

OB = R sin λ AB = OB tan α BC = BD = BE = R cos λ

DG BA BO tan α = = = tan α tan λ BD BD R cos λ = tan 23.50 tan 42.5o = 0.39843 = sin(23.48o )

sin φ =

(

)

) Hence, the Sun will rise at (71 + 23.48 ) × 4 min . − 5 hours. = 77.92 min. o

(

o

after 6 a.m.. This is at 7:18 a.m.

2

Theoretical Competition

1st IOAA

December 5, 2007

-----------------------------------------------------------------------------------------------------------1.2 The angular separation is maximum when Venus

R 90o Sun

46o

Sun, Venus and Earth form a right-angled triangle as shown.

Earth

Here

R⊕

R = R⊕ sin 46o

(

)

= 1A.U. sin 46o = 0.72 A.U.

1.3

If the same face of the Earth were to face the Sun all the time then the Earth would make one complete turn relative to fixed stars in one solar year (365.25 solar days). This implies that in 365.25 solar days our actual Earth makes (365.25+1) complete turns relative to fixed stars. Hence 365.25 solar days are the same time interval as 366.25 sidereal days; and

183 solar days

183 × 366.25 sidereal days 365.25 = 183.50 sidereal days

≡

OR 1 solar day = 24 x 60 x 60 s = 86400 s 1 sidereal day = 23 x 3600 + 56 x 60 + 4.1 = 86164.1 s 183 solar days =

1.4

183.50 sidereal days

During a full Moon we see the whole face of the Moon. Moon's diameter = angle in radians distance to Moon 0.46 × π = 180 180 Distance to the Moon = (Moon's diameter) × 0.46 × π 180 = 2 × 1.7374 × 106 m × 0.46 × π 8 = 4.328 × 10 m = 4.3 × 105 km

Hence

(

)

3

Theoretical Competition

1st IOAA

December 5, 2007

-----------------------------------------------------------------------------------------------------------1.5 Diameter of Earth’s orbit around the Sun 2a

α

D

star

a = 1 A.U. = 1.496 × 1011 m 1 pc = 3.0856 × 1016 m D = 100 × 3.0856 × 1016 m

α =

2a 2 × 1.496 × 1011 = radian D 100 × 3.0856 × 1016 = 0.96966..... × 10−7 radian = 5.555779... × 10−6 degree = 0.02 arc second

1.6

According to Kepler’s third law we have

(period)

2

= (constant)(semi-major axis)

3

T 2 = (constant)a 3

(year) This constant is 1 (A.U.)

2 3

when T is measured in years and a in A.U.’s.

For this comet we have a =

31.5 + 0.5 = 16.0 A.U. 2

T 2 = (16) = 16 × 4 × 4 × 16 = (64) 3

T = 64

1.7

2

years

According to Kepler’s second law we have; The area is swept out at constant rate throughout the orbital motion. Now, for this comet the area swept out in one orbital period is πab where

a is the semi-major axis and b the semi-minor axis of the orbit. a = 16.0 A.U. b 2 = a 2 − (distance from a focus to centre of ellipse )

2

4

Theoretical Competition

1st IOAA

December 5, 2007

-----------------------------------------------------------------------------------------------------------= (16.0) − (16.0 − 0.5) 2

2

b = 3.968 A.U.

πab = 199.5 (A.U.)

2

Hence, the area swept out per year is 199.5 (A.U.)

2

64 years

1.8

= 3.1

(A.U.)

2

/ year

From the Wien’s displacement law λmaxT = 2.8977 × 10−3 m.K

2.8977 × 10−3 m = 7.244 × 10−7 m 4000 = 724 nanometers

λmax =

1.9

From the Stefan-Boltzmann law, the total power emitted is

(

L = 4πR

2

2 

4

4 ⎛ R ⎞⎛ ⎞⎟ 2 ⎛ 7500 ⎞ T ⎟ ⎜ ⎜ ⎟ σT ⎜⎜ ⎟⎟ ⎜⎜ ⎟⎟ = L (2.5) ⎜⎜⎜ ⎟⎟⎟ ⎟ ⎜T ⎠⎟ 5800 ⎝ ⎠ ⎝⎜ R ⎠⎝ 

)(σT ) = (4πR )( 4

4 

)

L = (6.25)(2.796) = 17.47 = 17.5 L L = 17.5L

1.10

flux density =

distance = =

luminosity 4π × (distance)

2

0.4 × 3.826 × 1026 = 1.398 × 1019 m −14 4π × 6.23 × 10 1.398 × 1019 pc = 453 pc 3.0856 × 1016

5

Theoretical Competition

1st IOAA

December 5, 2007

-----------------------------------------------------------------------------------------------------------1.11 Lsupernova = 1010 L

Let D be the distance to that supernova. Then the flux intensity on Earth would be I = This must be the same as 1010 L 4πD ∴

2

1010 L 4πD 2

L 4π (D⊕− )

2

L

=

(

)

4π 1 A.U.

D = 10

10 2

2

A.U. = 105 A.U.

1.496 × 1016 = pc = 0.485 pc 3.0856 × 1016 = 0.485 × 3.2615 ly = 1.58 ly

1.12

The difference in frequencies is due to the relativistic Doppler shift. Since the observed frequency of emission from the gas cloud is higher than the laboratory frequency ν 0 , the gas cloud must be approaching the observer. Hence,

ν = ν0

c +v c −v 2

⎛ν ⎞ ⎜⎜⎜ ⎟⎟⎟ − 1 ⎜⎝ ν 0 ⎠⎟

v = 2 c ⎛ ν ⎞⎟ ⎜⎜ ⎟ + 1 ⎜⎜ ν ⎟⎟ ⎝ 0⎠ =

0.001752379 = 0.000875422 2.001752379

(

v = (0.000875422) 2.99792458 × 108 m.s−1

)

= 0.00262445 × 108 m.s−1 = 262.445 km.s−1

6

Theoretical Competition

1st IOAA

December 5, 2007

-----------------------------------------------------------------------------------------------------------Circular aperture of diameter D 1.13 (representing the eye)

B

A′ B′

θ

A

A and B are two distant objects. A ′ and B′ are the central maxima of their diffracted images.

The angular position φ1 of the first minimum relative to central maximum λ . D According to Lord Rayleigh the minimum angle of resolution is φ1 .

of each diffraction pattern is given by φ1 = 1.22

λ . θ We may take distance AB to be the diameter of the crater. Hence the

Hence the images of A and B will be resolved if θ > φ1, D > 1.22

diameter D of the eye’s aperture must be, at least, Dmin = 1.22

λ . θ

80 km

θ =

distance from Earth to Moon 80 × 103 m = = 2.081 × 10−4 radian 8 3.844 × 10 m

λ = 500 × 10−9 m

∴

Dmin

1.22 × 500 × 10−9 = m = 2.93 × 10−3 m −4 2.081 × 10 = 2.9 mm

Hence, it is possible to resolve the 80 km-diameter crater with naked eye. 1.14

The spherical boundary at which the escape velocity becomes equal to the speed of light is of radius R = R =

2GM 

c2 2 × 6.672 × 10−11 × 1.989 × 1030

(2.99792458 × 10 ) 8

2

m

2 × 6.672 × 1.989 × 103 m 2.998 × 2.998 = 2.95 km

=

7

Theoretical Competition

1st IOAA

December 5, 2007

------------------------------------------------------------------------------------------------------------

1.15

The flux ratio versus magnitude difference implies ⎛ f ⎞⎟ m1 − m2 = − 2.5 log ⎜⎜⎜ 1 ⎟⎟ ⎝⎜ f2 ⎠⎟ f1 f2

(m2 −m1 )

= 10

2.5

So for a magnitude difference of (−1.5) − 6 = − 7.5 we find a flux ratio of fmin fmax

−7.5

= 10 2.5 = 10−3

And thus the visible stars range only over a factor of 1000 in brightness.

8

Theoretical Competition

1st IOAA

December 5, 2007

-----------------------------------------------------------------------------------------------------------QUESTION 2

A PLANET & ITS SURFACE TEMPERATURE

SOLUTION a.) Intensity

b.) Absorption rate

L 4πD 2 A = (1 − α ) πR 2I I =

= (1 − α )

(1 point)

LR2 4D 2

(1 point)

c.) Light energy reflected by the planet per unit time is

αLR2 απR I = 4D 2 (1 point) 2

Hence the planet’s luminosity is

αLR 2 4D 2 d.) Here, we will neglect the planet’s internal source of energy. Lplanet =

(1 point)

Let T be the black-body temperature of the planet’s surface in kelvins. Since the planet is rotating fast, we may assume that its surface is being heated up uniformly to approximately the same temperature T . The total amount of black-body radiation emitted by the planet’s surface is from Stefan-Boltzmann law: 4πR 2 ⋅ σT 4 , σ being Stefan-Boltzmann constant. At equilibrium, that is when the temperature remains steady, this emission rate must be equal to the absorption rate in b.).

(1 point)

Hence 4πR2 ⋅ σT 4 = (1 − α )

LR2 4D 2 1

⎡ L ⎤⎥ 4 (1 point) T = ⎢(1 − α ) ⎢ 16πσD 2 ⎥⎦ ⎣ e.) In this case the emitted black-body radiation is mostly from the planet’s surface facing the star. The emitting surface area is now only 2πR2 and not 4πR2 . Hence the surface temperature is given by T ′ , where

2πR 2 ⋅ σ (T ′) = (1 − α ) 4

LR 2 4D 2

(1 point)

Theoretical Competition

1st IOAA

December 5, 2007

-----------------------------------------------------------------------------------------------------------1

1 ⎡ L ⎤⎥ 4 4 T ′ = ⎢(1 − α ) = 2 ⋅ T ≈ (1.19)T ( ) 2⎥ ⎢ 8 πσ D ⎣ ⎦

(1 point)

1

f.)

⎡ L ⎥⎤ 4 T = ⎢(1 − α ) ⎢ 16πσD 2 ⎥⎦ ⎣ 1

⎡ ⎤4 26 ⎢ ⎥ 3.826 10 × ⎥ T = ⎢(1 − 0.25) × 2⎥ ⎢ −8 11 16π × 5.67 × 10 × 1.523 × 1.496 × 10 ⎥ ⎢ ⎣ ⎦ o (2 points) = 209.8  210 K = -63 C

(

)

Theoretical Competition

1st IOAA

December 5, 2007

----------------------------------------------------------------------------------------QUESTION 3

BINARY SYSTEM

Solution a) The total angular momentum of the system is (M 1r12 + M 2r22 )ω

L = Iω =

(0.5 points)

We have also M 1r1 = M 2r2 and D = r1 + r2 which yield

L =

M 1M 2 M1 + M 2

D 2ω

……………..(1)

(0.5 points)

The kinetic energy of the system is 2 2 1 1 1 M 1 (r1ω ) + M 2 (r2ω ) = M 1r12 + M 2r22 ω 2 2 2 2 1 M 1M 2 …….……..(2) = D 2ω 2 2 (M 1 + M 2 )

(

K .E . =

)

(0.5 points) (0.5 points)

b) From Newton’s laws of motion we have M 1ω 2r1 = M 2 ω 2r2 =

GM 1M 2

(1 point)

D2 These equations together with those in a) yield ω2 =

G (M 1 + M 2 ) D3

…….…………(3)

(1 point)

c) In order to find the quantity Δω , we must also realize that M 1 + M 2 = constant

….…………(4)

(0.5 points)

And that, since there is no external torque acting on the system, the total angular momentum must be conserved. L =

that is,

M 1M 2 M1 + M 2

D 2 ω = constant

M 1M 2D 2 ω = constant

Now, after the mass transfer, ω M1 M2 D

→ ω + Δω → M 1 + ΔM 1 → M 2 − ΔM 1 → D + ΔD

(0.5 points)

Theoretical Competition

1st IOAA

December 5, 2007

-----------------------------------------------------------------------------------------

(

Hence, M 1M 2D 2 ω = (M 1 + ΔM 1 )(M 2 − ΔM 1 )(D + ΔD ) ω + Δω 2

)

After using the approximation (1 + x ) ∼ 1 + nx and rearranging, we get n

⎛ M − M ⎞⎟ Δω ΔD 2⎟ +2 = ⎜⎜⎜ 1 ⎟ ΔM 1 ⎜⎝ M 1M 2 ⎠⎟ ω D

………..(5)

(1 point)

From equation (3), ω 2D 3 is also constant. That is, ω 2D 3 = (ω + Δω ) (D + ΔD ) 2

3

This gives, ΔD 2 Δω ………………(6) = − D 3 ω 3 (M 1 − M 2 ) Δω = − ω ΔM 1 ………..(7) M 1M 2

Hence

(0.5 points) (0.5 points)

d) Given that M1 = 2.9 M☼ , M2 = 1.4 M☼ orbital period, T = 2.49 days and that T has increased by 20 s in the past 100 years, we have

ω=

2π and Δω T

=

−

ω ΔT …………(8) T

1 ⎛ M 1M 2 ⎞⎟⎟ ΔT ΔM 1 = + ⎜⎜⎜ ⎟ 3 ⎜⎝ M 1 − M 2 ⎠⎟ T ⎞⎟ ⎛ ⎞ ⎞⎟ ⎛⎜ ΔM 1 1⎛ 1.4 20 ⎟⎟ ⎜⎜ 1 ⎟⎟ ⎟⎟ ⎜⎜ = ⎜⎜ ⎟ M 1Δt 3 ⎜⎝(2.9 − 1.4)⎠⎟ ⎜⎜⎝ 2.49 × 24 × 3600 ⎠⎟ ⎝⎜100 ⎠⎟⎟

e) f)

(0.5 points)

(0.5 points)

= 2.89 × 10−7 per year

(0.5 points)

Mass is flowing from M 2 to M 1 .

(0.5 points)

From equations (6) and (8): ΔD 2 Δω 2 ΔT =− =+ = 6.20 × 10−7 per year (1.0 points) D Δt 3 ω 3 T

Theoretical Competition

1st IOAA

December 5, 2007

----------------------------------------------------------------------------------------QUESTION 4 Solution

GRAVITATIONAL LENSING S2

S

P

S1

DS - DL L

DS

ξ

φ

β

DL

θ

Figure 1

O a). From figure 1, for small angles, tan θ ≈ θ , hence

PS1 = PS+ SS1

θDS = βDS + (DS − DL ) φ θ−β =

4GM (DS − DL ) DS ξc 2

………………….(1)

ξ also, hence, DL ⎛ 4GM ⎞ ⎛ D − DL ⎞⎟ ⎟ θ 2 − βθ = ⎜⎜ 2 ⎟⎟⎟⎜⎜⎜ S ⎜⎝ c ⎠⎟ ⎝⎜ D D ⎠⎟⎟

( 1 point)

Note that θ =

L

………………….(2)

S

For a perfect alignment in which β = 0 , we have θ = ±θE , where θE =

⎛ ⎞⎟ ⎛⎜ DS − DL ⎞⎟ ⎜⎜ 4GM ⎟⎜ ⎟ ………………….(3) ⎜⎝ c 2 ⎠⎟⎟ ⎝⎜⎜ DLDS ⎠⎟⎟

( 1 point)

b). From equation (3), for a solar-mass lens with DS = 50 kpc,

DL = 50 − 10 = 40 kpc θE =

⎛ 4GM ⎞⎟ ⎜⎛ D − D ⎞⎟ −9 L⎟ ⎜ ⎟⎜ S radian ⎜⎝⎜ c 2 ⎠⎟⎟ ⎜⎝⎜ D D ⎠⎟⎟ = 0.956 × 10 L

−4

= 1.97 × 10

S

arc second

( 1 point)

1

Theoretical Competition

1st IOAA

December 5, 2007

----------------------------------------------------------------------------------------c.) The resolution of the Hubble space telescope having diameter of 2.4 m is, θHubble

5 × 10−7 m λ = 1.22 = 1.22 × = 2.54 × 10−7 radian for light of 2.4 m D

wavelength 500 nm.

( 1 point)

Hence the Hubble telescope could not resolve this Einstein ring. ( 1 point) d). The quadratic equation (2) has two distinct roots, namely,

where θE =

2

β θ1 = + 2

⎛ β ⎞⎟ ⎜⎜ ⎟ + θ 2 E ⎜⎝ 2 ⎠⎟⎟

β θ2 = − 2

⎛ β ⎞⎟ ⎜⎜ ⎟ + θ 2 E ⎜⎝ 2 ⎠⎟⎟

( 1 point)

2

( 1 point)

⎛ 4GM ⎞⎟ ⎛⎜ D − D ⎞⎟ L⎟ ⎜⎜ ⎟⎜ S ⎜⎝ c 2 ⎠⎟⎟ ⎜⎜⎝ D D ⎠⎟⎟ L

S

This implies that there are two images for a single isolated source.

θ1,2

e).

β

=

1 = ± 2

f).

β ± 2

2

⎛ ⎞ ⎜⎜ β ⎟⎟ + θ 2 E ⎜⎝ 2 ⎠⎟⎟ β

2 2 ⎡ ⎤ 2 ⎛ 1 ⎞⎟ ⎛ 1 ⎞⎟ ⎜⎜ ⎟ + ⎜⎜ ⎟ = 1 ⎢1 ± η + 4 ⎥ ⎢ ⎥ ⎜⎝ 2 ⎠⎟⎟ ⎝⎜ η ⎠⎟⎟ 2⎢ η ⎥ ⎣ ⎦

(1 point)

From equation (2) θ 2 − βθ − θE2 = 0

(θ + Δθ ) − (β + Δβ )(θ + Δθ ) − θ 2

2 E

=0

Δθ θ = Δβ 2θ − β

⎛ Δθ ⎞⎟ ⎟ ⎜⎜ ⎜⎝ Δβ ⎠⎟⎟

θ =θ1,2

1 η2 η ± 1+ ⎡ ⎤ θ1,2 η ⎥ 4 = 1 ⎢1 ± = = 2 ⎢ ⎥ 2 2 2θ1,2 − β 2⎢ ⎥ 4 η η + ⎣ ⎦ ±2 1 + 4

(1 point)

(1 point)

2