physics examination problems solutions and hints...

PHYSICS EXAMINATION PROBLEMS SOLUTIONS AND HINTS FOR STUDENT SELF-STUDY

Module Code PHY1022

Name of module Introduction to Astrophysics

Date of examination January 2014

1. (i) The first part of this question is bookwork.

The leptons are tau neutron and electron. The hardrons are kaons and sigma+.

The baryons are neutrons and protons. The mesons are pions and kaons.

uud is a proton. udd is a neutron.

(ii) The ground state wavelength is 20 nm.

The ground state energy is 6 ×10−22 J.

The first excited state energy is 24 ×10−22 J.

P(0 < x < L / 4) = Ψ2 (x)dx = 2L

sin2 π xL

⎛⎝⎜

⎞⎠⎟0

L /4

∫ dx0

L /4

∫

which can be solved by a suitable trig identity to get P = 0.091 .

The second part can be solved by integration, but it is easier to note the symmetry of the wavefunction, so that P(L / 4 < x < 3L / 4) = 1− 2 × 0.091 = 0.818 .

2. The first part is bookwork.

Two photons of opposite momentum must be produced to conserve momentum. 2Eγ = 2mec

2 to give an energy of 8.2 ×10−14 J and a wavelength of 2.42 ×10−11m.

The first part is bookwork.

R1 =dN1dt

= λN1 = λN0e−λt1 = R0e

−λt1

R1R0

= e−λt1 so ln R1R0

⎛⎝⎜

⎞⎠⎟= −λt1 and hence λ = t1

−1 ln R0R1

⎛⎝⎜

⎞⎠⎟

.

Half life is 4332s, and number in sample is 7.2 ×105 nuclei.

3. First part is bookwork. Momentum uncertainty is 3.5 ×10−20 kgms-1 .

vg =

dωdk

=dωdk

=d(hf )d(h / λ)

=dEdp

vg =ddp

p2

2m⎛⎝⎜

⎞⎠⎟=2p2m

=pm

= u .

4. Use m − M = 5 logd − 5 to give M = −7.4

Next part is bookwork.

Use M* − M = −2.5 log L*

L

⎛⎝⎜

⎞⎠⎟

to give M = −17.67 and then m − M = 5 logd − 5 to give m = −5.7 .

This would be the brightest object in the sky apart from the Sun and Moon.

5. First part is bookwork.

ML

⎛⎝⎜

⎞⎠⎟ cluster

=1.2 ×1014 /1000

5 ×109= 24

M

L

⎛⎝⎜

⎞⎠⎟

For nearby stars the mass to light ratio is going to be greater than one (since there are more, less massive stars than

Sun, which are less luminous). I would accept a mass-to-light ratio from 2 to 10.

The cluster contains a substantial fraction of dark matter.

5. a = 0.049AU

v = 134kms−1

Δλλ

=vc

so Δλ = 9 ×10−5nm .

The planet’s mass is 1×1027 kg

The luminosities of the giant branch star come from L = 4πR2σT 4 and are 2154 and 19386 in solar

luminosities. The radius of the star should look like R = 400R + 200R sin 2πt / P( ) and differentiate this to get the

radial velocity v = dR

dt= 200R

2πPcos(2πt / P) . This has a maximum when the cosine is unity, so

v = 33.7kms−1 . This occurs when the star is 400 solar radii in radius, so the escape velocity is

2GMR

= 27.7kms−1 . So the star will shed mass through the pulsations as vmax > vesc .

6. The first part is bookwork.

The redshift is 7.1

The recession speed is 0.97c.

The wavelength of 738.7 nm in the rest frame of the galaxy is 91.2 nm.

Photons below this wavelength have an energy capable of ionizing hydrogen, so will be absorbed.


Module Code PHY1025

Name of module Mathematics Skills

Date of examination 2014

1 (i) 𝑥 = 2

(ii) 𝑥 = 𝜋 3⁄ (iii) 𝑓(𝑥) = 5𝑥 − 3𝑥 + 7

2 (i) A = −−5 −5 52 3 −48 7 −11

(ii) (𝑥, 𝑦, 𝑧) = 4,− ,− (iii) 𝜆 = −1

3 (i) (a) 𝑢(𝜌) = 2𝑎ln𝜌

(b) 0

(ii) =

(iii) = , = 1

= ( ) , = 2

4 (i) (a) ∫ √

√𝑑𝑥 = 1 + √𝑥 + 𝐶 = 𝑥 + 2√𝑥 + 𝐶

[Note that it is not necessary to use the substitution provided in the hint; the hint was provided in case of difficulties with the integration process.]

(b) ∫ 𝑑𝑥 = (ln ) + 𝐶

(ii)

(iii) 𝑚 (𝑅 − 𝑅 )

5 (b) √1 + tan 𝑥 ≈ 1 + (c)

(d) The function has periodicity (due to the cos function). The function is even, since ln(cos 𝑥) = ln(cos(−𝑥)).


Module Code PHY2021

Name of module Electromagnetism I


1. (i) Coursework for Gauss's law.

Charge per unit length is λ = πkr4

2

Electrostatic field inside:

€

E =

kr3

4εˆ r and outside is:

€

E =

kR4

4ε0rˆ r

The arbitrary constant, V, cancels giving a potential difference of 0.01655 V. 2. (i) Coursework. Coursework. Coursework. Coursework. The important thing here is to consider the z-component of the dipole moment,

€

pz = zρrdrdθdz∫ in which there is a z^2 term. Also be careful with the integration

limits. Direction is

€

+ˆ z

The potential is 2.485x

€

10−9 V. Note that

€

cos θ =0.30.4

. Also note that there is a

typographical error in the exam paper in that the units of λ0 should be Cm −3 rather than Cm −1 .

3. (i) Coursework. Coursework. General solution is

€

Φ = c1 ln r + c2 and then you need to apply boundary conditions.

Electric field is

€

E =

Vln b / a( )r

ˆ r

Solution given. Solution given. The capacitance per unit length is 5.9x

€

10−11 F/m.

4. (i) Coursework. Coursework. There is a mistake in this question - as written, the proton will not reach z=0. The magnetic

field should be 0.002 T, rather than 0.02 T in strength. Then, to solve this, first work out the radius of gyration (the particle moves in a circle). Then you can work out the time taken for a full circle (circumference of the circle divided by the speed). Then finally, work out the fraction of the circle that it goes through before reaching z=0. The radius of gyration is 0.1565 m. The time taken to travel a full circle (2*pi radians) is 32.78 microseconds. The angle it

moves through is given by

€

sin θ =(0.1565 − 0.1)

0.1565 so that the time taken to reach z=0

is 6.27 microseconds. (ii) Coursework 6A, 15 V, 37 V, 22/3

€

Ω 5. (i) The flux is a cosine wave, while the emf is a sine wave. Make sure you label the time axis with

appropriate points. The flux is

€

Φ = BL2 cos(ωt) The emf is

€

ε = BL2ω sin(ωt) You can either look at the force on the bits of the loop using F=BIL and then work out the

torque. Or you can use

€

τ =

m × B .

The torque is 0.0107 N.m. 6. (i) Coursework. Coursework. Solution given. The direction is in the x-direction. Coursework.

Vector potential is

€

A =

µ0Jd y2

ˆ z

PHYSICS EXAMINATION PROBLEMSSOLUTIONS AND HINTS FOR STUDENT SELF-STUDY

Module Code PHY2022Name of module Quantum Mechanics 1Date of examination January 2014

1.! ! To normalize let Ψnorm = CΨ , then C 2 = 1

Ψ 2 dVall space∫

. This makes Ψ 2 equal to, not

just proportional to, the probability density for a position measurement.! !

!

SOLUTION TO DEGREE EXAMINATION QUESTIONfor Coordinator

Name of setter Usher Paper Question

Name of module Quantum Mechanics 1

Year of examination 2014 PHY2022 1Initials of checker

[recall]

[recall]

[application]

[recall/application]

[synthesis]

! ! Reasonable estimate of uncertainty in position for this wave packet is L (or 2L).Hence

Δp = !

2L.

! ! px2 = !

2

2L2. Square root of this is close to (but not equal to) estimate above.

2.! ! First 10 marks - see lecture notes/exercises.

!





[application]! !! ! ! ( i i )!

E1n = u0n* V x( )u0n dV

all space∫ (or 1D equivalent).

! ! Construct this integral when V(x) is a delta-function, and remember what delta func-

tions do inside integrals. Change in energy is 2V0. For 1st excited state u 2 = 0 at cen-

tre of well so perturbation has no effect.

3.! ! First 12 marks, see lecture notes.

! ! N = 13

,

! !





! ! S = − 1

2

4.! (a)

!





[synthesis]

[synthesis]

[application]

! (b)! First excited state n = 1 (note counting convention for harmonic oscillator), because it has one node.

! (c)! To find classically allowed/forbidden region, set energy (in this case

32!ω ) equal to the

potential energy 12mω 2x2 . Answer, forbidden region is

x > 3!

mω⎛⎝⎜

⎞⎠⎟1 2

.

! (d)! Forbidden regions are smaller, because of shape of parabolic potential - more energetic

particles can access larger region of x.

! (e)! x2 = 3!

2mω

! ! E = 3

2!ω

5.! ! First 14 marks - see lecture notes.

! ! ℓ = 2 so m = −2, −1, 0,1 or 2 so possible outcomes are Lz = −2!, − !, 0, ! or 2!

!




Year of examination 2014 PHY2022 5 (cont)Initials of checker

6.! ! First 4 marks - see lecture notes.

! ! V r( ) = − e2

4πε0r





(4)

(3)

ionization energy is −E n = 1( ) .

! ! Ionization energy is proportional to Z2 (Z the charge of the nucleus). Therefore ioniza-tion for Helium+ (which is a nucleus of two protons with one orbiting electron) is 4 times that of hydrogen: 54.4 eV.

! ! Relations between quantum numbers and generation of radial functions: see lecture notes.

1


Module Code PHY3051/3054

Name of module Electromagnetism II


1. Maxwell’s equations for an empty vacuum were given in the lectures. Remember to define

both the physical quantities and the vector operator Del. Obtain the wave equation for B by taking the curl of both sides of the 4th Maxwell equation and

then substituting the 3rd Maxwell equation into the right had side of this expression. The form of the solutions of the wave equation were discussed within the lectures. The forms

presented in the question are for the case of a linearly polarized electromagnetic wave, with the unit vectors nE and nB indicating the direction of polarization of the respective field quantities.

Expressions for E0 in terms of B0 and nE in terms of nB can be found by substituting the given

solutions into the 3rd Maxwell equation and then comparing the magnitude and orientation of the vector quantities on each side of the equation. This should yield

nE = − k× nB( ) and E0 =ωkB0

Using the expressions just derived

B = kωE0

16x+ y− 2z( )

2. The component of electric field tangential to the conducting surface must vanish, so Er = -Ei. It is first necessary to explain why a hollow waveguide without a conducting core must have a

cut-off condition, namely that the boundary conditions cannot be satisfied at all points on the inner surface of the hollow guide without quantizing the wave in some manner. When a core is introduced a quantization condition is no longer required to satisfy the boundary conditions.

General hint: a cylindrical guide with a core is similar to the case of the rectangular waveguide when the rectangular waveguide is deformed so that two of its sides are joined together. We can then think of a quantization along the radial direction due to the need to satisfy the metallic boundary conditions, and a periodic boundary condition in the azimuthal direction.

If the central conductor is removed then the waveguide will indeed have a lower cut-off frequency. This is because quantization orthogonal to the axis of the guide is required to satisfy

2

the boundary conditions, so there must be a finite component of the wavevector orthogonal to the guide.

Hint: the inner and outer diameters of the guide a very similar, so, to a good approximation, one can use the radius at the middle of the empty space within the guide to evaluate the quantization condition. Then

ω 2 = c2k2 = c2 kr2 + kφ

2!" #$ .

So,

f1 =c2πa

and f3 =3c2πa

and finally k2 = 8( ) / a .

3. (i) Derivation of the Clausius-Mossotti expression is well described in many text books and as

assigned as self-study. The atomic polarisability increases as the material is cooled, so consider what might happen to

the denominator in the C-M expression. Assuming that the volume and hence the number of atoms per unit volume scales linearly with

temperature near the phase transition, the C-M expression can be manipulated to show that

ε∝1

T −TC( ).

(ii) The moment lies perpendicular to the plane of the triangle with

m =34a2I

Now the dipole moment lies parallel to the z-axis and we may use the expression for a dipole field

B = µ04πr3

3 r.m( ) r−m"# $%

Hence,

Bx =µ04πr3

34a2I 3xz

r, By =

µ04πr3

34a2I 3yz

r, Bz =

µ04πr3

34a2I ×3 z2

r−1

#

$%

&

'(

4. The expressions for the retarded potentials and the derivation of the electric field generated by

the Hertzian dipole were described within the lectures. The question leads you through this derivation and the final expression can be found within the lecture notes.

3

5. The definition of s and p polarization states and the form of the respective Fresnel coefficients were given in the lecture notes.

In making sketches of the dependence of the Fresnel reflection coefficients upon the angle of incidence you should pay particular attention to the values of the intercepts, which you should evaluate and label. Note that since incidence upon a less dense medium is being considered, one expects to observe total internal reflection beyond a critical angle (48.75°). The reflection coefficient should also vanish at the Brewster angle for p-polarised light.

The derivation of the Brewster angle was described within the lectures. The possible latitudes are +/- 53.06°.


Module Code PHY3062

Name of module Methods of theoretical physics


1. (a) 21

dd

1)2exp()exp(

dcosh

1

02

0

S

�

� ³³³

ff

f�

�f

tt

xxx

xx

, where )exp(xt is used.

(b) > @

1d

d1)2exp(

)2exp(2d

cosh1

122

02

�

³³³ff

f�

�f

tt

xxx

xx

, where )2exp(1 xt � is used.

(c) > @ � �

� �,41

d4d

1)2exp()exp()2exp(

4dcosh

121

032

2

30

3II

t

ttx

xxx

xx

I � �

�

³³³ff

f�

�f

where )exp(xt is used.

Both 1I and 1I can be evaluated using contour integration calculating the residues at iz .

� �³

f

�

0

221 41

d S

t

tI and

� �³f

�

0

322 163

1

d S

t

tI . Thus,

4S

I .

2. (i) ³�

�

S ST

T2

02 1

2cos

d

aa. Substitute )exp( Tiz and use contour integration around the unit circle.

(ii) Consider a harmonic function BA � 4M , which is an imaginary part of .)log()( BizAzf �

Using the boundary conditions yields ¸¹

·¨©

§� ¸

¹·

¨©§ � �

yx100

0 tan2

21

SMM

4S

MM .

3. (i) WKB� for highly excited states > @ ¸¹·

¨©§ � �³ 2

1d)(2 nxxVEm

b

a

!S , where a and b are the

classical turning points (Bohr-Sommerfeld quantisation condition).

� � 3231223231223/2

21

405.121

2¸¹·

¨©§ �¸

¹

·¨©

§|¸

¹·

¨©§ �¸

¹

·¨©

§ n

mn

mE

!! DDJS, where .

32

d11

0³ � ttJ

(ii) Three nearest neighbours > @ 232 3)1(1)( xxxxP |�� (for 1��x ).

4. J 2A ; m

T2

22J! ;

JD2

V ; 3

1223

1223

1

min945.0

3227

¸¹

·¨©

§|¸

¹

·¨©

§¸¹·

¨©§

mmE

DD !!.

5. Rewrite as 322

2

<J<[<

� dd

, where xmE 2

1

2

2¸¹·

¨©§�

![ ,

EDJ � 2 and 0�E . Using < as a

new variable and[<dd

f as an unknown function yields a first order differential equation

with separable variables. The final answer is

� �»»¼

º

««¬

ª�¸

¹·

¨©§�

�

0

21

2

2cosh

12

xxmE

E

!

D< .


Module Code PHY3063

Name of module Stars


1. The unphysical nature of the n=5 solutions is shown by considering the outer boundary condition where the density

tends to zero, so θ must also tend to zero. Consider what value of ξ in the two solutions discussed will yield this.

For the next part use this (unphysical) value of ξ in the integral for the mass (which need not be performed) to see the

problem.

2. When explaining the Gamow peak first explain clearly what is meant by dN/dE. For the final part recall that dN/dE

must then be multiplied by a nuclear cross section.!

3. For the core bounce to occur we need the buoyancy term to dominate over the gravitational one at small radii, and

hence have a higher power of the radius in the denominator.

4. The material is either in the lectures (first half) or the self-study packs (second half).

5. The derivation of the radial temperature profile of an accretion disc is given in the lecture notes – make sure it is done

by considering an annulus of the disc. The final part is solved by considering an optically thick element of gas where

the energy received from the star is balanced by the element’s own black body emission.

PHYM002: Quantum Mechanics II

Hints and tips for the 2013/14 exam

1. To identify the first excited state, one has to bear in mind that the energylevels in this case depend upon n and m. For the g.s., n + |m| = 0, whilefor the first excited state n + |m| = 1, which has three solutions: n = 1,m = 0, and n = 0, m = ±1. By symmetry, the matrix elements vanish forall transitions except m − m′ = ±1. (This is similar to particle moving ona ring, see, e.g., Lecture 06.) Thus, there only four non-zero matrix elementscorresponding to transitions (1, 0) ↔ (1,±1). The radial integrals for thesetransitions are identical, and the angular integrals are all equal to π. The scaleof the perturbation is given by eaE , as usual. This gives Stark effect patternsimilar to that of 3d hydrogen atom, except that the zero energy state is nownon-degenerate (rather than doubly degenerate).

2. Spin-related multiplicity is 2S + 1 = 3, and orbital multiplicity is 2L+ 1 = 3,so there are nine states. The level is split into components with a given totalangular momentum J = 0, 1, 2. The corresponding multiplicities are 2J+1, i.e.,1, 3, 5. The energy levels are given by the formula A

2[J(J+1)−L(L+1)−S(S+

1)], which gives −10, −5, and 5meV, respectively. Magnetic field splits each ofthese fine-structure levels into the components differing by angular momentumprojection MJ onto the direction of magnetic field. Thus, the upper level issplit into five sublevels as ±3µBB, ±3

2µBB, and 0; the middle level is split into

three components as ±3

2µBB, and 0, the lower level is non-degenerate and thus

is not split. The crossover criterion is µBB ∼ A, which gives B ∼ 100T .

3. The symmetries of the two factor in the wave functions must be opposite, dueto Pauli principle. The spin wave funciton is symmetric for S = 1 and antisym-metric for S = 0, so that the coordinate wave function is symmetric for S = 0.(There was an ambiguity in the question; all solutions which clearly demon-strated the proper symmetry of the spin and coordinate wave functions weretreated as correct.) The three wavefunctions for S = 1 are given by |↑↑⟩, |↓↓⟩,and 1/

√2 [|↑↓⟩+ |↓↑⟩], while the S = 0 wave function is given by antisymmetric

combination of |↑↓⟩ and |↓↑⟩.The ground-state configurations for He and Si are 1s2 and 3p2, which givesS = 0, and S = 1, respectively. In the case of He, this configuration fol-lows from the theorem about two-particle spin configuration, while for Si itis not applicable, due to electrons occupying lower shells. (Alternatively, onemight explain the difference by noticing the degeneracy of the 3p shell.) Themultiplicity for Si is 3x3 = 9 (see, e.g., the preceding question). Thus, theground-state configurations are 1S0 and 3P . (One can also determine the

1

ground-state total angular momentum J by noticing that the shell is less-than-half filled, so that J = |L − S| = 0. This gives 3P0.) As the orbitalmultiplicity is 3, there are three coordinate wave functions. One of themis Ψm=1 = 1/

√2 [ψ1(r1)ψ0(r2)− ψ1(r2)ψ0(r1)], the other two can be obtained

by replacing the pair (ψ1, ψ0) with either (ψ−1, ψ0) or (ψ1, ψ−1). (The wavefunctions of helium were not required in this question.)

4. The tricky issue here is the normalisation of the wave function: it is assumedto be a plane wave of unit amplitude. The initial and final states are givenby exp(ikx) and exp(−ikx), with E = h2k2/2m. The density of final statesis m/2πhp, note that neither spin degeneracy nor p → −p doubling is included:we consider the scattering which conserves the spin, and which only occurs intothe states with negative momenta. The particle flux is J = p/m, as usual, andthe reflection coefficient can be found as R = Γi→f/J . At large energies, thereflection coefficient vanishes, as it should.

5. It is important to notice that the tunneling gives rise to the off-diagonal matrixelements in the left-right basis, while the electric field gives diagonal terms.Indeed, quantum tunneling takes the particle from the left well into the rightwell. In a strong field, the asymmetry between the wells is more importantthan tunneling, so that the particle is sitting in the right well, thus minimisingits energy. As the field is switched off, the particle oscillates between the wells.This process is described by equations ihcL = ∆cR, etc. The equations can beeasily solved: cR(t) = cos∆t/h, cL(t) = i sin∆t/h.

2

PHYSICS HINTS AND TIPS

Module Code PHYM006

Name of module Relativity and Cosmology

Date of examination June 2014

1. (i) The first part of this question is bookwork. To show that the acceleration and velocity four vectors are orthogonal we note that in the rest frame [Aµ ] = (0,a) and [U µ ] = (c,0,0,0) so that UµA

µ = ηµνUνAµ = 0 . For the

rocket we see that on the ship’s clock the time is simply L / c . For the Earth bound clock the expression for the

distance travelled is ct = L(1− (3 / 4)2 )1/2 + 34ct so t = 7L / c .

(ii) This part is all bookwork.

2. The first part of this question is bookwork. Then we use the norm of the four velocity gµνU

µUν = c2 to get

(1− 2GM / c2r)U 0U 0 = c2

therefore (U 0 )2 = c2

(1− 2GM / c2r)

So [U µ ] = c(1− 2GM / c2r)1/2

,0,0,0⎛⎝⎜

⎞⎠⎟

. Remember the astronaut is stationary in r,θ,φ .

Then use the equation for acceleration given (covariant derivative) to get

A0 = dU 0

dτ+ Γ00

0 U 0U 0 + ... = 0

A1 = dU1

dτ+ Γ00

1 U 0U 0 =GM (1− 2GM / c2r)

r2c2c2

(1− 2GM / c2r)=GMr2

A2 = A3 = 0 So [Aµ ] = (0,GM / r2 ,0,0)

Finally a2 = gµνAµAν =

1(1− 2GM / c2r)

G2M 2

r4

And a = 1(1− 2GM / c2r)1/2

GMr2

which tends to a = GMr2

in the Newtonian limit.

3. S = ds =dx2 + dy2

y∫ dy.∫ For points along y = 0 the integral is divergent.

Now we work out the Christoffel symbols and find

Γ000 = Γ11

0 = Γ100 = Γ01

0 = 0 and Γ010 = Γ10

0 = Γ111 =

−1y

and Γ001 =

1y!

! ! So!geodesic!equations!are! d2xdτ 2

−2ydxdτ

dydτ

= 0 !and! d2ydτ 2

+1y

dxdτ

⎛⎝⎜

⎞⎠⎟2

−1y

dydτ

⎛⎝⎜

⎞⎠⎟2

= 0 !

!! ! The!two!components!of!the!Ricci!tensor!are! R00 =

1y2!and! R11 =

−1y2.!

! ! The!Ricci!scalar!is!the!contraction!of!the!Ricci!tensor!! ! R = gµνRµν = g

00R00 + g11R11 = −y2 1

y2− y2 1

y2= −2 !which!is!a!constant!–ve!curvature.!

! 4.!! The!first!part!of!this!question!is!bookwork.!The!Friedmann!equations!are!

! ! 1RdRdt

⎡⎣⎢

⎤⎦⎥

2

=8πG3

ρm,0R0R

⎡⎣⎢

⎤⎦⎥

3

+ ρr ,0R0R

⎡⎣⎢

⎤⎦⎥

4

+ ρΛ

⎡

⎣⎢⎢

⎤

⎦⎥⎥−kc2

R2!

1Rd 2Rdt 2

=−4πG3

ρm,0RR0

⎡

⎣⎢

⎤

⎦⎥

3

+ 2ρr ,0RR0

⎡

⎣⎢

⎤

⎦⎥

4

− 2ρΛ

⎡

⎣⎢⎢

⎤

⎦⎥⎥

For a static solution we need R = R = 0 so

−4πG3

ρm,0 − 2ρΛ⎡⎣ ⎤⎦ = 0 and ρΛ =ρm,0

2.

We also need R = 0 and for a closed universe k = +1 so

0 = ρm,0 + ρΛ,0 −3c2

8πGR02

0 = ρm,0 +ρm,0

2−

3c2

8πGR02

ρm,0 =c2

4πGR02

PHYSICS EXAMINATION PROBLEMSSOLUTIONS AND HINTS FOR STUDENT SELF-STUDY

Module Code PHYM435Name of module Solar and extra-solar planets and their

atmospheresDate of examination Jan 2014

1. (i) From classwork. The key points here are that there should be a lower crater density on planets undergoing resurfacing, and there should be a size cut-off on planets with an atmosphere. (ii) a) The minimum impact velocity ~ escape velocity from Earth ~ 11 km/s.Maximum impact velocity (for plausible impacting objects) set by the maximum velocity of something on a bound orbit at Earth's distance, or about 70 km/s. A "typical" velocity might be, say, Earth's velocity around the Sun, ~30 km/s.b. From E~(1/2)mv^2 ~ energy of explosion, estimate of order 10^11 kg.c. Will be appreciably slowed if it passes through a column of atmosphere with mass equal to its own mass, so M_atm ~ 10^11 kg.d. Surface pressure set by weight of the atmosphere, so P~ M_atm*g/A ~ 10^7 Pae. Earth's surface pressure is 10^5 Pa, Venus is about 90 times greater, so this is approximately the surface pressure on Venus. f. from hydrostatic equilibrium plus definition of opacity (NOTE: we used different definition of alpha, kappa at times). If you took alpha to be the quantity s.t. tau = alpha *int(rho dz), then tau = P*alpha/g

2. a. descriptions from class/readingb. likewise c. mass ratio ~ 10^-3, so the reflex velocity of the star is roughly 10^-3 times the orbital velocity of the planet. From the planet's semi-major axis, deduce planet's v~42.4 km/s, so the stellar reflex motion is about ~42 m/s (this is what you’d need to detect)d. apparent astrometric wobble ~ 0.05 milliarcsecond (not detectable), and would need to wait only about a month (say 0.25 times orbital period of 0.35 years). e. dF/F = (Rp/R)^2, so Rp = 2.7x10^5 kmf. This is about 4x Jupiter’s radius. Cue discussion of max radius for a degenerate body (~Rj). g. “see” to about optical depth unity. Can use this to estimate nh. T ~ 9 hours.

3. a. dT/dz ~ Q(D-z)/K (if you assume same geometry as me; other factors if not)b.integrate to find the quoted equationc. Bottom temperature of the ice layer is just set by T at which water undergoes phase transition, not be details of Q (i.e., Q just changes the depth of the layer, not its temperature contrast)d. D ~ sqrt(2 (delta T)* K/Q)e. from classwork (buoyancy driving over damping by viscous or thermal diffusion)

Charles Williams

PHYM012

Charles Williams

PHYM012

PHYM012

PHYM012

f. Because delta T independent of Q, find Ra ~ Q^(-3/2), so for Q> Q_crit, Ra < Ra_crit for convection (i.e., if heating is too big, no convection!). Which is weird.g. straightforward application of Rayleigh number stuff. I got about 13 km.

4. a) bookworkb) rho’ = epsilon *rho_0 * (1 - (z/2H) - 1/2 (z^2/H^2))c) multiple ways of proceeding here, but all involve just showing that this solution can’t satisfy equations 1 and 3 simultaneouslyd) Fact that box is square + continuity equation implies u ~w and d/dx ~ d/dz ~ 1/H, giving quoted resulte) Re = UH/kf) balance of (1 and 2) gives V = Omega *R* epsilong) balance of (2 and 3) gives U ~ H^2 Omega^2 R epsilon / kh) in eqn 2, only possible linearised balance is U ~ kV/Omega*H^2 = k*R*epsilon/H^2i) For U1 to hold need Omega*v term << k*del^2 u term, since otherwise principal balance would partly involve this term. But this is just the condition for the Ekman number … hence need Ek> 1.j) find this condition nowhere near satisfied (i.e., for any plausible rotation rate, Ekman number is NOT greater than one, 2*Omega*v term is NOT ignorable, and hence the dynamical balance expressed by U1 does not hold).

5. (i) is all from lecture notes. (unsaturated emission line, unsaturated absorption line, saturated

absorption line in a,b,c respectively)(ii) a) use definition of pressure scale height, form a ration w/r/t to the solar scale height, and use g=GM/r^2 to find ratio. (it’s a really tiny number.) b) by Wien’s law, lambda ~ 0.29 cm*K/T, T=10^6 means peak is in X-raysc) L = 4piR^2 sigma T^4, so 18 L_sund) from lecture notes. e) bulk modulus essentially an energy density; here E ~ cp is energy per relativistic electron. use p = h/lambda to get U ~ h*c*n_e^(4/3)f) Follow argument used in class, but with this crazy bulk modulus, etc, to derive height (it should be a very small number).

Charles Williams

PHYM013

PHYM013

PHYM013

PHYM013

PHYM013

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