Problem T1. Main sequence stars (11 points)

Transcription

1 Proble T1. Main sequence stars 11 points Part. Lifetie of Sun points i..7 pts Since the Sun behaves as a perfectly black body it s total radiation power can be expressed fro the Stefan- Boltzann law as P = πr σt = W. Forula.5, nuber.1, units.1 pts. ii..5 pts Fro the energy conservation law. pts. Then p c = He c + e c + W W = p c He c e c = MeV. Forula.1, nuber.1, units.1 pts. iii..5 pts The fusion of four protons creates two positrons which in turn annihilate with two electrons eaning that an additional energy of W 1 = e c =. MeV is released. Then the total energy released is W = W +W 1 = 6 MeV. Noticing that particles annihilate per one He ato., forula.1, nuber.1, units.1 pts. iv. 1. pts Over the course of Sun s lifetie the central part of the Sun will undergo fusion and release energy. The total nuber of reactions that will take place is N = M 1 p. pts. nd thus, the total energy released is W E = NW = 1 M = J p. pts. The total lifetie of the Sun can be approxiated as τ = E P = y. Forula., nuber.1, units.1 pts. The current age of the sun τ = y is approxiately two ties saller than the calculated theoretical age Part B. Mass-luinosity relationship of stars.5 points i.. pts Since all of the star s ass is below the point Q, the gravitational acceleration is the sae as that of a point ass with a ass of M. pts. Then. pts. a Q = GM GM R = R ii.. pts By applying Gauss s law for gravity for a sphere surrounding the stellar core R π a P = πg M. pts;. pts. a P = GM R iii.. pts Since the gravitational acceleration decreases linearly along the thickness of the spherical layer, the average acceleration experienced by the spherical layer is a avg = a P +a Q = 9GM R Furtherore, a piece of the sall spherical layer with an area has a ass of = π R 7M = 7 M π R Fro the Newton s second law. pts. F = a avg = 6 GM π R iv.. pts The previously calculated force acting on the sall piece of the narrow spherical layer can also be expressed as. pts. Then F = p c = 6 GM π R p c = 6 GM π R v. 1 pt Fro the ideal gas law π R p c = nr g T c where n is the nuber of oles of protons and electrons inside the stellar core.6 pts;. if electrons are forgetten. Since the ass of an electron is negligible copared to the ass of a proton, n = and M pn a = M pn a p c πr 6 = MR gt c p N a. pts. Then p c = Mk B T c π R p = Mk BT c p vi.. pts Cobing both expressions for p c, one gets. pts.. pts. page 1 of 5 6 GM π R = Mk B T c π R p R = 1 GM p 16 k B T c

2 vii. 1.5 pts Writing out the energy balance for a spherical shell with a radius of x and thickness dx concentric to the star πx dt dx κ = P. pts and rearranging the ters, one gets πκdt = P dx x. pts. Integrating fro x = R to x = R yields. pts. Then T R R dx πκ dt = P R T c x 1 πκt R T c = P R R πκt c = P R P = πκt c R. pts. When siilar expression is obtained without integration leading to a wrong factor, only. for integration is lost. Substituting κ = ftc ρ c, ρ c = M πr ultiately end up with P = 1. pts. Thus γ = and R = 1 16 G p π Tc ft c M k B Part C. Proton-proton fusion chain.5 points GM p k B T c, we i. 1.5 pts First, we ust convert the units to base units: [c] = /s, [G] = kg 1 s, [k B ] = kg s K 1.1 pts, [N ] = ol 1, [ ] = kg/s.1 pts, [e] = C, [k e ] = kg C s Let α = [c] β [G] γ [k B ] δ [N ] ε [ ] µ [e] φ [k e ] ω. Then we can create an equation for each unit: : β + γ + δ + µ + ω = s: β γ δ µ ω = kg: γ + δ + µ + ω = K: δ = ol: ε = C: φ ω =..1 pts for each equation. fter solving the syste of equations and setting ω = 1, we get β = 1, γ =, δ =, ε =, µ = 1, φ =, and ω = 1 apart fro δ and ε,.1 pts for each value. Thus α = k ee c = pts for the nuerical value. ii. 1 pt Let the distance to the centre of ass for both protons be x. Then the force acting on one of the protons is F x = kee x and thus the potential energy is Π = x F xdx = k ee x dx x = k ee x.. pts out which.1 goes for correctly treating the distance to the centre of ass and distance between the protons. By applying the energy conservation law at x = rp get. pts. Furtherore k e e = pv r p p v. pts. T can be expressed as = k BT T = k ee k B r p = K.1 pts for forula. This is around T T c and x =, we = 6 ties larger than the actual teperature of the stellar core iii. 1 pt The total energy of a proton oving at speed v is W = pv and the potential energy, as expressed in the last subtask, is Πr = kee r = αc r. The oent at which the proton dives into the tunnel happens when W = Πr = Πr = αc r. pts. Thus r = αc pv Then the probability of the tunnelling taking place is [ r 1 p exp 1 p αc r 1 ] dr = r. pts = exp 1 p αc π r = exp παc v. pts. page of 5

3 Proble T. Water tube points i..5 pts There is a water colun of height H between points P and Q creating an additional pressure of p P p Q = ρgh = Pa. Forula. pts, value.1 pts, units.1 pts. ii. 1.5 pts The external forces acting on the syste are sketched on the figure to the right. F = p = 1 N forula.1 pts, value with units.1 pts and F 1 = 1.1p = 11 N forula.1 pts, value with units.1 pts is the atospheric pressure acting on the pistons. F = M +g =.1 N forula.1 pts, value with units.1 pts is the gravitational force acting on the water-piston syste, where M = H +.1hρ =.1 kg forula.1 pts, value with units.1 pts is the ass of the water colun. N is the total noral force exerted by the etal cylinder. There is no horizontal coponent for the noral force since it cancels out due to syetry. N can be expressed fro the Newton s nd law applied on the vertical axis N + F + F F 1 = N N + p + M + g 1.1p = N =.1p M+g =.1p H+.1hρ+g = 5.9 N. Forula. pts, value with units.1 pts. shown force in the sketch:.1 pts. pts overall. F F F 1 N Each correctly iii. 1. pts Notice that N =.1p 1.6 pts, where p 1 = 1N/ = 59 kpa forula. pts is the pressure at the joint of the two tubes. Therefore, p Q = p 1 ρgh h = 57 kpa. Forula.1 pts, value with units.1 pts. and p P = p 1 + ρgh = 6 kpa. Forula.1 pts, value with units.1 pts. lternatively, applying the Newton s law on the vertical axis for the piston, one gets. pts,. pts, p p Q + g + 1.1p P 1.1p =.1p + g p Q + 1.1p Q + ρgh = p Q = p 11ρgH 1 g = 57 kpa forula. pts, value with units.1 pts. p P = p Q + ρgh = p 1ρgH 1 g forula. pts, value with units.1 pts. = 6 kpa iv.. pts Newton s nd law on the vertical axis for the top piston can be written out as g + p p Q + T =. pts. T = p Q p g = 11ρgH + 1 g g = = 11ρgH 1 g =.5 N The negative sign of the tension force eans that the steel bar is being copressed, not stretched. Forula. pts, value with units.1 pts, sign or direction of T.1 pts. v. 1 pt During the ipact, the etallic tube coes to rest but the two pistons keep oving downwards because the pistons and the tube aren t strongly connected. pts. s a result, the volue between the two pistons increases since the area of the botto piston is larger than the top piston and vacuu is created. pts. This causes the atospheric pressure to try to reverse the change and push the pistons upwards. pts. Because no energy is lost in the water-piston syste for siplicity we assue the friction between the tube and water / pistons to be negligible, after the pistons have returned to their initial position, their speed will be of equal agnitude and of opposite sign, pointing upwards, which in turn akes the tube jup. pts. vi. pts Neglecting the pressure of water vapors and of the water colun of c, the pressure between the pistons is zero, hence the net force acting on the syste water + pistons is F =.1p + + Mg = 5.9 N 1. pt. Because the force is constant throughout the whole process, the change of oentu for the water-piston syste can be expressed as 1. pt page of 5 M + v M + v = F τ v =.1p M + g τ M + v τ = p 1 + Mg where v = gl is the speed of the tube when it reaches the ground. Thus M + τ = gl =.1 s. p 1 + Mg Forula. pts, value.1 pts, units T.1 pts.

4 Proble T. ccelerating shock wave 11 points i. 1 pt In the reference frae of the shock wave, the electron s initial velocity is v 1 = v x w, v y, v z. pts. fter deflecting against the shock wave, the horizontal coponent of the velocity gets flipped. pts. Thus, the electron s velocity in the oving frae of reference, after deflecting against the shock wave, is v = w v x, v y, v z Moving back into the laboratory frae of reference, the final velocity is v = w v x, v y, v z. pts for x coponent,.1 both for x and y coponents. ii. 1 pt fter being hit by the shock wave, the electron starts oving with speed v = w. Due to the agnetic field, it oves along a circular trajectory, and at the initial oent of tie, the trajectory is perpendicular to the front. dditionally, the electron periodically undergoes collisions against the shock wave, and the x-coordinates of the collision points grow in tie. This is enough to draw an approxiate sketch of the electron s trajectory. Grading: trajectory is ade fro circular segents. pts which are connected at the reflection points so that instantaneous change of direction is clearly seen. pts.. pts if the trajectory starts parallel to the x-axis,.1 pts if the direction of otion is shown by arrow or described in another way;. pts if the reflection points advance in the sae direction as the shock wave. iii..5 pts The Lorentz force acting on the electron acts as a centripetal force evb = v R. pts. Thus R = v eb = w eb iv. 1 pt Before the first collision, the electron s x-coordinate is x 1 t = R sin π t T. pts and the shock wave s x-coordinate is x t = wt y The second ipact happens when x 1 t = x t. pts. Thus w B e sin eb t = wt B e sin t = 1 B e t Substituting u = Be t, one gets sinu = u. pts. This equation can be solved nuerically to get u = 1.95 Thus t = 1.95 B e v..5 pts Every tie a collision happens, the electron and the front are at the sae place, with the sae value of the x x-coordinate. This eans that the electron s and shock wave s average velocities in the direction of the x-axis are the sae. In other words, v x = w.5 pts. vi. 1.5 pts It is easier to find the value of k by taking a derivative fro both sides of the equation as it gets rid of the constant. Then v y + kv x =. pts. The only forces acting on the electron are the Lorentz force and the repulsion forces between the electron and the shock wave. pts. Since the shock wave affects the electron only in the horizontal direction. pts, the acceleration s vertical coponent coes purely through the Lorentz s force. pts. This eans that ÿ = ev x B holds throughout the electron s otion. pts. In other words, v y = Be v x. Plugging this to the conservation law, we get Be v x + kv x =. pts. Thus k = Be.1 pts. vii. 1 pt By taking a derivative fro the conservation law v y + Be x = const, we get a y + Be v x =.5 pts. Over the long run, the average x-directional oving speed of the electron is the sae as that of the shock wave s. Thus a y + Be w =. pts and a y = Be w viii. 1 pt Over the course of one period, there is a constant acceleration a x acting on the electron in the x-direction, both in the lab frae, and in the shock wave s frae; the behaviour is the sae what would be if there were a free fall acceleration g = a. If we let x be the relative distance between the electron and the shock wave, and the initial x-directional oentu at x = be p x, then the quantity E = p x = p x + a xx is conserved over the course of one period energy conservation law. pts. Thus p x = p x a x x On the phase diagra, this corresponds to a parabola who s axis of syetry is at p x =. pts. Furtherore, at x =, the oentu of the electron gets flipped due to the collision against the shock wave, eaning that there is a straight line fro, p x to, p x. pts. This gives enough inforation to draw the phase diagra correctly drawn figure.1 pts, arrow shown.1 pts. page of 5

5 p x p x -p x ix. 1.5 pts The area under the phase diagra can be found by integrating p x xdx fro x = to x 1 = p x a x and ultiplying the result by two since the phase diagra is syetrical about p x =. Thus x1 x1 S = p x a x xdx = p x x 1 x 1 x x 1 dx = = p xx 1 = p x = v a x a x. pts. Due to the conservation of this quantity we use its initial value taken fro the proble text, v a x = 1.6w B e vb e w a x = 1.. pts, where v is the electron s speed at x =. The electron will fall behind the shock wave when v > ev or ev v > = w ε. pts. The horizontal acceleration coes fro Lorentz force a x = Bevy. pts. Thus 1 B e. w w ε = B ev y 1. wε = vy. pts. Since v y v x, W f v y. pts. W f = ε w = ε.. ev x. pts In the reference frae of the shock wave, initially, the electron s x-directional and y-directional oenta are p x and p y respectively. In the liiting case, the electron s final x-directional oentu is. Since the shock wave acts only in the x direction, p y will stay sae throughout the otion. The Lorentz invariant of the -oentu, initially and after the electron has coe to rest, can be written out as.6 pts, E = p xc + p yc + c E ev = p yc + c.6 pts. Subtracting one equation fro the other, we get EeV e V = p xc = rel c w = E w c. Thus. pts, E w c EeV + e V = E = ev c w 1 ± 1 w c. pts with inus sign we would obtain p yc = E ev c < E ev e V c /w < which is not acceptable. Thus, we need to take the plus sign. pts: E = ev c w w c for w c we can approxiate E = evc w So, the electron will fall behind the shock wave if its relativistic energy E evc w ; ; page 5 of 5