The solar wind. or rather: The energy budget of the corona-solar wind system. Øystein Lie-Svendsen

Transcription

1 Introduction Facts The solar wind or rather: The energy budget of the corona-solar wind system Øystein Lie-Svendsen Norwegian Defence Research Establishment and Institute of theoretical astrophysics, University of Oslo, Norway Kodai Winter School, December 2006

2 Introduction Facts Why study the solar wind? The solar wind represents loss of mass and energy (tiny!) from the Sun. The solar wind affects us ( space weather ): disturbances at high latitudes ( magnetic storms ) affecting communications, power transmission, oil installations causes the aurora may damage orbiting satellites harmful to humans in space The solar wind is an exciting physical system in its own right. We still do not understand what drives the solar wind there is work to do!

3 Introduction Facts The solar wind at solar minimum from Ulysses < 1.4 AU 5.4 AU (McComas et al., 2000)

4 Introduction Facts Comparison of minimum and maximum from Ulysses (McComas et al., 2003)

5 Introduction Facts Ulysses mass flux density at solar minimum

6 Introduction Facts Summary of observed solar wind properties Parameter Slow Fast Speed (km/s) km/s km/s Density m m 3 Flux m 2 s m 2 s 1 Helium abundance n α /n p 4% 5% T e K K T p K K Origin streamer boundary? (polar) coronal holes protons at 1 AU are supersonic, c p kt p /m p 30 km/s u wind (hence protons cannot carry information inward) while electrons remain subsonic, kte /m e 1200 km/s u wind (hence electrons may, through heat conduction, carry information inward)

7 Introduction Facts The white light corona T e 10 6 K, slightly less in open regions, slightly more in closed regions.

8 Introduction Facts The corona in the extreme ultraviolet SOHO/EIT, Fe XII 195 Å

9 Introduction Facts Polar coronal holes as seen by SOHO/UVCS Temperatures from line broadening Flow speeds from Doppler dimming (Zangrilli et al., ApJ, 574, 477, 2002) (Cranmer et al., ApJ, 511, 481, 1999)

10 Introduction Facts Escape from the gravitational potential Hence u wind v esc mv esc 2 mm S = G N R S 2GN M S v esc = 620 km/s. R s

11 Introduction Facts The fast wind from coronal holes What do these observations tell us? Protons and ions are much hotter than electrons, indicating strong (?) ion heating. Rapid acceleration of the solar wind, complete within a few solar radii.

12 Introduction Facts An enormous range of scales...

13 Introduction Facts And a transition to supersonic flow

14 Introduction Facts Questions Why is the corona hot, of order 10 6 K? Why do we even have a corona?? Is it a coincidence that the solar wind speed is of order the gravitational escape speed? Is there a relation between the coronal temperature and the wind speed? What does the fast acceleration of the wind say about driving mechanism?

15 Introduction Solar wind energy loss Heat conduction Radiation Energy balance of the corona The fact that the corona is hot either means that it is strongly heated or it cannot easily lose energy unless it becomes hot Hence we must understand not only the sources of energy but equally important the loss mechanisms.

16 Introduction Solar wind energy loss Heat conduction Radiation Sources of energy From the colder chromosphere below? Anomalous heat conduction up the temperature gradient has been proposed ( velocity filtration ), but does not seem to work. Radiative heating? Should rather equilibrate the upper atmosphere temperatures. And what can absorb photons in a fully ionized plasma? Waves from the photosphere? A likely candidate! Sound waves may not carry the required amount of energy, but Alfvén waves may work (Poynting flux dissipation). Release of magnetic energy in the corona (reconnection)? Yes, also possible through nanoflares, and lots of energy available. But no conclusive evidence. Some form of external energy source must exist we still do not know what.

17 Introduction Solar wind energy loss Heat conduction Radiation Losses of energy The possible loss mechanisms are The solar wind carries energy as it flows out. Heat conduction down towards the chromosphere. Radiation either from the corona or from layers below Sources of energy are uncertain, but losses are not uncertain!

18 Introduction Solar wind energy loss Heat conduction Radiation Calculating the energy flux As each particle flows out from the Sun it carries kinetic energy 1 2 m svs 2 gravitational and electric potential energy E pot = G N M S m s ( 1 r 1 R S ) + Q s V E (r) where M S and R S are the mass and radii of the Sun, r the heliocentric distance, Q s the particle charge and V E the electric potential (unspecified yet, but not negligible). ionization energy (13.6 ev per electron-proton pair), which we neglect. (Why?)

19 Introduction Solar wind energy loss Heat conduction Radiation The energy flux density is ( ) 1 F Es = d 3 v v r f s (r, v) 2 m sv 2 + E pot where f s is the velocity distribution function (VDF) and v r the radial component of v. Assuming a drifting Maxwellian VDF, f s = ( ms 2πkT s ) 3/2 exp ( m s(v u s ) 2 ) 2kT s we find F Es = n s u s ( 1 2 m su 2 s kt s + m s V G + Q s V E ) where V G is the gravitational potential.

20 Introduction Solar wind energy loss Heat conduction Radiation Electron-proton flow Adding possible heat conduction, the total electron/proton energy flux is ( 1 F E = n p u p 2 m pup 2 + m p V G + ev E ) 2 kt p + q p +n e u e ( ev E + 5 ) 2 kt e + q e neglecting terms m e. Neglect q p, assume charge neutrality and no current, and T e = T p = T, [ ( 1 1 F E = nu 2 m pup 2 + G N M S m p 1 ) ] + 5kT + q e R S r

21 Introduction Solar wind energy loss Heat conduction Radiation Assume all energy supplied to the wind at r = r 1. Then total energy flux is conserved through any surface beyond r 1. Since also particle flux is conserved, the conservation of F E yields 5kT 1 = 5kT m p(v 2 esc + u 2 2) + q e2 n 2 u 2 q e1 n 1 u 1 where v esc = 2G N /r 1 = 620 km/s at the surface. Neglecting also heat flux and T 2 we get T 1 = 1 ( 10 m pv esc (R S ) 2 RS u 2 ) 2 + r 1 v esc (R S ) ( 2 ( ) ) = R S u 2 2 K + r km/s

22 Introduction Solar wind energy loss Heat conduction Radiation Hence wind requires million degree corona For the fast wind (700 km/s) we then have ( ) T RS K + 1 r 1 With r 1 = 2 R S, T K A high-speed wind driven by thermal heating near the Sun requires a million-degree corona. Even a very slow wind, u 2 0, gives, for r 1 = 2 R S, T K.

23 Introduction Solar wind energy loss Heat conduction Radiation Heavy ions Neglecting the electric potential, the same argument gives the coronal ion temperature or m i T i = 1 m i ( v 2 5 k esc + u2) 2 min(t i ) = 1 m p 5 m p k v esc(r S ) 2 R S r K m i m p R S r 1 Choosing again r 1 = 2 R S, and m i /m p = 16 (oxygen) we get T 1 = K SOHO/UVCS finds that indeed T (oxygen) 10 8 K.

24 Introduction Solar wind energy loss Heat conduction Radiation How much energy does the solar wind need? Scaling the observed solar wind energy flux density at Earth back to the Sun, we find ( ) 1 AU 2 1 F E (R S ) = (nu) E 2 m ( p v 2 esc + ue 2 ) R S With u E = 700 km/s and (nu) E = m 2 s 1, we get F E (R S ) 70 W/m 2. Hence not much compared to the radiative energy flux of (1 AU/R S ) W/m 2 50 MW/m 2...

25 Introduction Solar wind energy loss Heat conduction Radiation Estimating the wind speed Near the Sun each particle (electron-proton pair to be correct) receives energy (1 + a) 1 2 m pv 2 esc, which is converted to potential and kinetic energy at Earth (plus a small amount of thermal energy), (1 + a) 1 2 m pv 2 esc = 1 2 m p(v 2 esc + u 2 E ) where a specifies the excess energy provided. Hence u E v esc = a. Choosing the escape speed at the surface, v esc 620 km/s, we find a = 0.1 u E = av esc 200 km/s. u E = 10 km/s a

26 Introduction Solar wind energy loss Heat conduction Radiation Wind speed (contd.) We should therefore expect the speed of a stellar wind to at least be of order the gravitational escape speed in the region of the atmosphere where most of the energy needed by the wind is supplied. In this simplified picture gravity sets both coronal temperature mpv 2 esc k and the wind speed v esc

27 Introduction Solar wind energy loss Heat conduction Radiation Heat conduction In a collision-dominated fully ionized plasma (such as the low corona) the amount of heat conduction is determined by Coulomb collisions between particles. Cross section σ v v 4 collision frequency ν T 3/2 mean free path T 2 From this you can show that the heat flux in a given temperature gradient is proportional to T 5/2 T. Elaborate calculations give q e = κ e ln Λ T 5/2 T where κ e W m 1 K 7/2 (e.g., Braginskii, 1965) and ln Λ 20 in the corona.

28 Introduction Solar wind energy loss Heat conduction Radiation Heat conduction causes the transition region When the corona is heated, energy flows down towards the chromosphere in the form of heat. With no loss of energy (we discuss radiation later) 5/2 dt q e = κt dz = constant. or dt dz T 5/2. Hence the temperature gradient must steepen as the temperature is reduced towards the chromosphere.

29 Introduction Solar wind energy loss Heat conduction Radiation Energy balance between coronal heating and heat conduction Assume that an energy flux density q 0 (measured at r = R S ) is deposited at the temperature maximum in the corona at a distance r 1 from the Sun centre, and that all of this energy is conducted downwards (no solar wind loss). Also assume radially expanding flow (spherically symmetric Sun). Assuming a constant heat flux and integrating, you find T 1 [ ( 7 q 0 R S 1 R )] 2/7 S 2 κ r 1 as the maximum temperature of the corona.

30 Introduction Solar wind energy loss Heat conduction Radiation Heat conduction as a coronal thermostat Choosing r 1 = 2 R S we find q 0 = 0.1 W/m 2 T 1 = K q 0 = 10 W/m 2 T 1 = K q 0 = 100 W/m 2 T 1 = K q 0 = 50 MW/m 2 (!) T 1 = K Heat conduction works like a thermostat we get a corona of order 10 6 K with almost any heat input!

31 Introduction Solar wind energy loss Heat conduction Radiation Competition between wind and heat conduction We found that if the solar wind were the only energy loss, T 1 (solar wind) 1 G N M S m p R S 5 kr S r K for r 1 = 2 R S. At the same time heat conduction predicts T 1 (heat conduction) [ ( 7 q 0 R S 1 R )] 2/7 S 10 6 K 2 κ r 1 In an open corona, which of these two processes will dominate? Answer: We have to resort to numerical modelling, showing that the solar wind energy loss dominates (Hansteen & Leer, 1995), carrying of order 90% of the energy deposited in the corona.

32 Introduction Solar wind energy loss Heat conduction Radiation Consequences of a dominant solar wind energy loss The energy flux measured at 1 AU should be approximately equal to the actual energy flux deposited in the corona. The wind is not just the evaporating tail of the corona (as is the escape of gas from the Earth s atmosphere), but must have a large impact on the corona. Both the open corona and closed magnetic loops should have temperatures of order 10 6 K. Since T 1 (solar wind) M S /R S, stars with much lower M S /R S should have negligible heat conduction and therefore no transition region and corona.

33 Introduction Solar wind energy loss Heat conduction Radiation Radiation Radiative loss is the conversion of (electron) kinetic energy into photon energy by collisional excitation of an atom or ion. Radiative loss from a fully ionized plasma with no bound electrons is difficult. Maybe the corona doesn t radiate well? If the thermal energy of free electrons is much smaller than the lowest possible excitation energy, the radiative loss should be small too. The most important line, Lyα, caused by the n = 1 to n = 2 excitation of hydrogen, has E = 10.2 ev, corresponding to T E/k = K. But hydrogen is already fully ionized at 10 5 K... At low temperatures, kt E, only electrons in the tail of the Maxwellian can excite the atom, so the rate exp( E/kT ) is extremely sensitive to T.

34 Introduction Solar wind energy loss Heat conduction Radiation The radiative loss rate The loss from excitation of hydrogen can be written L RH = n H n e L H (T e) n 2 el H (T e ) where the last relation can be obtained by assuming balance between ionization and recombination of hydrogen. Similarly for the loss from minor ion excitation, L Ri = n i n e L i(t e ) n 2 el i (T e ). The total radiative loss rate may then to a good approximation be written (T e = T ) L R = n 2 el(t ) = P2 4k 2 T 2 L(T ) where P = 2nkT is the total electron + proton pressure and L(T ) is a function of temperature only.

35 Introduction Solar wind energy loss Heat conduction Radiation Temperature dependence of the loss rate L R (From the model of Fontenla et al., ApJ, 355, 700, 1990.)

36 Introduction Solar wind energy loss Heat conduction Radiation Consequences of the form of the loss rate L R n 2 e small radiative loss in the (outer) corona. L R has a maximum at some T R K, max(l R ) = P 2 L(T R) 4k 2 T 2 R so a heating rate larger than max(l R ) cannot be balanced by radiative loss. L R decreases with T for T > T R system is unstable beyond T R leading to runaway heating. The corona, at T 10 6 K T R, is in the unstable temperature region. Radiation does not matter for the energy balance of the corona.

37 Introduction Solar wind energy loss Heat conduction Radiation Maximum radiative cooling at T R : The chromosphere-corona dividing line If heating is such that T < T R we have a cool chromosphere. If heating is such that T > T R we form a corona. For T > T R runaway heating occurs until either heat conduction (inefficient at T 10 5 K) or the solar wind energy loss kicks in, both requiring T 10 6 K. Once a corona is formed, its very low radiative loss means that even if the heating rate is strongly reduced again, the corona will persist. If heating decreases less rapidly than n 2 e, then beyond some distance r TR the heating will be larger than the maximum radiative cooling rate P 2 L(T R )/(4k 2 T 2 R ).

38 Introduction Solar wind energy loss Heat conduction Radiation When will stars have coronae? In order to have a wind energy loss (without heat conduction) we found that the coronal temperature had to reach the gravitational escape temperature given by 5kT esc = G NM S m p. r Equating this temperature with the maximum temperature, T R, supported by radiative loss, we have or T R = G NM S m p 5kr R r R = G NM S m p 5kT R.

39 Introduction Solar wind energy loss Heat conduction Radiation When will stars have coronae (contd.)? A corona cannot be formed beyond r = r R = G n M S m p /(5kT R ). If the heating rate is so low, or density so high, that r TR > r R, a corona is not formed the chromosphere is converted directly into a slow, cool wind. With T R = K, we find for our Sun that r R 0.7 AU! So the Sun must have a corona. However, for, say, a red giant star with M = 16 M S, R = 400 R S, we get r R 6 R, so it may not have a corona.

40 Introduction Solar wind energy loss Heat conduction Radiation Balance between heat conduction and radiation Energy supplied beyond r TR is not converted (directly) to radiation. If it is converted into heat conduction, in a steady state the radiative loss in the transition region must balance the heat flux divergence, 2 L(T (r)) AP 4kT (r) 2 = d(aq e). dr where A(r) = A 0 (r/r S ) 2 is the flow tube area. Using q e = κt 5/2 dt dr, using T instead of r as the independent variable (since the temperature is monotonically increasing), and assuming constant pressure P in the transition region, this energy balance may be rewritten d((aq e ) 2 ) dt = κ(ap)2 2k 2 T L(T ).

41 Introduction Solar wind energy loss Heat conduction Radiation Radiative energy balance (contd.) Assuming a thin transition region, so we may treat A as a constant, this equation may formally be integrated between T 1 and T 2. Expressing the result in terms of P we get 2k P = 2 (q2 2 q2 1 ) κk(t 1, T 2 ) where K(T 1, T 2 ) T 2 T 1 T L(T ) dt. Assuming that T 1 is so low that all heat has been absorbed, q 1 = 0, we get 2 P κk(t 2 ) k q 2 The transition region pressure is directly proportional to the downward heat flux q 2.

42 Introduction Solar wind energy loss Heat conduction Radiation Radiative energy balance (contd.) Pressure of the transition region, and thus the coronal density, is not an independent parameter, but is set by the energy balance! When P P bal 2 κk(t 2 ) k q 2 the transition region will move until P = P bal is restored. Hence the transition region is dynamic (and difficult to model... ). The solar wind mass flux directly affected by the energy balance of the transition region. Hence we must use so-called radiative energy balance models of the solar wind (Hammer 1982, Withbroe 1988). Even if 90% of the coronal heating goes to the wind, the remaining 10% is important since it sets the mass flux.

43 Introduction Solar wind energy loss Heat conduction Radiation Radiative energy balance (contd.) 2 P = κk(t 2 ) k q s m 1 q 2 (e.g., Rosner et al., 1978, Hansteen & Leer, 1995). With a coronal temperature of 10 6 K and a coronal hole density, inferred from observations, of n m 3, we find q 2 37 W/m 2.

44 Introduction Solar wind energy loss Heat conduction Radiation Radiative energy balance (contd.) Since we have shown that the wind needs about 70 W/m 2, this implies that more than 1/3 of the energy needs to be conducted down. Difficult to obtain in solar wind models which typically get only 10% conducted down. Hence solar wind models get a lower coronal density than observations indicate. Are the models wrong or the density observations incorrect?

45 Conservation equations Isothermal solar wind Interaction with the interstellar medium Basic conservation equation In a fluid flow with particle density n(r) and mean flow velocity u(r), assume that each particle carries an intrinsic quantity b, which can be charge (b = q), mass (b = m), number of particles (b = 1), momentum (b = mu i ) etc. Considering flow through an arbitrary, infinitesimally small volume in space, you can show that the following equation must be satisfied: (nb) + (nub) = δb t δt where the right-hand side expresses the rate of change of b per unit volume and time due to external processes.

46 Conservation equations Isothermal solar wind Interaction with the interstellar medium For b = 1 (number of particles), the right-hand side expresses the production (or loss) rate of particles due to chemical and ionization processes. In the case that the total number of particles is conserved, we thus get the continuity equation n t + (nu) = 0. In a steady state, spherically symmetric flow this simplifies further to d dr (r 2 nu) = 0 where u is the radial flow speed.

47 Conservation equations Isothermal solar wind Interaction with the interstellar medium Momentum conservation Setting b = mu x (momentum in x-direction) t (mnu x) + (mnu x u) = n(mg x + ee x ) P x + where we have included gravity, the electric force and the pressure gradient force on the right-hand side. With no production or loss of particles, and a steady state, spherically expanding flow you can simplify this to mnu du dr + nmg nee + dp dr = 0 (where gravity is assumed to point inwards).

48 Conservation equations Isothermal solar wind Interaction with the interstellar medium Setting Energy conservation b = 1 2 mu kt into the general conservation equation, we have t (1 2 mnu2 + 3 ( 1 2 kt ) + 2 mnu2 u + 3 ) 2 ktnu = +mnu g + neu E P u u P q where the work done by gravity, the electric force, the pressure has been included on the right-hand side, as well as the heat flux divergence. With the same simplifying assumptions as for the continuity and momentum equations, we end up with. 3 2 u dt dr + T d(r 2 u) r d(r 2 q) dr r 2 = 0 nk dr

49 Conservation equations Isothermal solar wind Interaction with the interstellar medium Isothermal solar wind We consider an electron-proton wind, assuming charge neutrality n e = n p = n and no current u e = u p = u. The steady state electron and proton momentum equations may then be combined, and, using the continuity equation, written as u du dr ) (1 c2 u 2 = G NM S r 2 + 4kT m p r where c 2kT /m p is the proton thermal speed. Both sides must be zero when u = c, the critical point. The critical point is located at r c = G NM S 2c 2 = With T = 10 6 K, r c 6 R S. ( vesc ) 2 RS. 2c

50 Conservation equations Isothermal solar wind Interaction with the interstellar medium Critical point density Requiring a subsonic (u c) solution near the Sun and a supersonic solution (u c) at infinity, the solution must go through the critical point. With n 0 the density at r R S, you can show that the critical point density is [ 3 n c = n 0 exp ( vesc which becomes very small when c v esc. c ) 2 ].

51 Conservation equations Isothermal solar wind Interaction with the interstellar medium The isothermal wind mass flux We now know the flow speed and density at the critical point, hence we have the solar wind mass flux without solving the equation! Scaled to Earth, we get (nu) E = where r e = 1 AU. ( RS r E ) e3/2 n o c ( vesc c ) ( 4 exp v esc 2 ) 2c 2

52 Conservation equations Isothermal solar wind Interaction with the interstellar medium The isothermal flux as a function of coronal temperature (nu) E [m -2 s -1 ] (nu) E [m -2 s -1 ] r c =R S T [K] r c =R S c/v esc The flux at Earth for n 0 = m 3. Horizontal dashed line = observed solar wind flux.

53 Conservation equations Isothermal solar wind Interaction with the interstellar medium A physically sensible solution requires r c > R S, or v esc > 2c or T < m p 8k v 2 esc K.

54 Conservation equations Isothermal solar wind Interaction with the interstellar medium Comparison with adiabatic wind The isothermal wind produces a reasonable mass flux with c 0.2v esc, or T K. Our energy flux argument, which neglected heat conduction entirely, predicted a coronal temperature necessary for a wind, (with v esc at r = R S ). T 1 m p 10 k v esc K Thus the isothermal case, with infinitely efficient heat conduction, produces a wind at a lower temperature. But even the isothermal wind requires c/v esc not much less than unity to have a realistic mass flux. Even in this extreme case a million-degree corona is required, or both mass and energy fluxes become extremely small.

55 Conservation equations Isothermal solar wind Interaction with the interstellar medium The isothermal result The mass flux problem nu n 0 c ( vesc c ) ( 4 exp v esc 2 ) 2c 2 indicates that the mass flux is extremely sensitive to the coronal temperature when c v esc. Why is the (fast) solar wind mass flux then so constant? Answer: The mass flux is limited by the energy supply to the corona: F E nu 1 2 m p(vesc 2 + ue 2 ) F E 1 2 m pvesc 2 As long as the total coronal heating rate remains constant, and most of the energy is lost in the solar wind, the mass flux cannot change by a large amount.

56 Conservation equations Isothermal solar wind Interaction with the interstellar medium Interaction of the wind with the interstellar medium As the wind expands, its density decreases as r 2. At some distance the dynamic pressure of the wind becomes smaller than the pressure of the interstellar medium, and a termination shock will form. Here the supersonic, tenuous solar wind is abruptly slowed down, converted into the slower flow of the interstellar gas. This point marks the end of the solar system.

57 Conservation equations Isothermal solar wind Interaction with the interstellar medium Fluid shock relations Flux conservation We assume the shock is at rest, so time derivatives may be neglected. The continuity equation d dr (r 2 nu) = 0 must be satisfied everywhere. Denoting upstream (solar wind side) quantities by subscript 1 and downstream values by subscript 2, the continuity equation may be integrated across the shock to read r 2 1 n 1 u 1 = r 2 2 n 2 u 2 or, if the shock is so thin that r 1 r 2, n 1 u 1 = n 2 u 2.

58 Conservation equations Isothermal solar wind Interaction with the interstellar medium The momentum equation Shock relations (contd.) Pressure balance m p nu du dr + d (nkt ) = 0. dr With the aid of the continuity equation this may be written d m p dr (nu2 ) + 2 r m pnu 2 + d (nkt ) = 0. dr Integrating across the shock, assuming a very thin shock and that all variables (but not their derivatives) remain finite within the shock, we obtain m p n 1 u n 1 kt 1 = m p n 2 u n 2 kt 2 expressing pressure balance at the shock.

59 Conservation equations Isothermal solar wind Interaction with the interstellar medium Shock relations (contd.) Energy conservation From the steady state energy conservation equation in spherical geometry [ ( 1 d 1 r 2 r 2 nu dr 2 m pu )] 2 kt = 1 d(r 2 q) r 2 dr (neglecting gravity) we obtain the energy conservation requirement across the shock ( 1 n 1 u 1 2 m pu ) ( 1 2 kt 1 + q 1 = n 2 u 2 2 m pu ) 2 kt 2 + q 2. If heat conduction can be neglected outside the shock (it may not be negligible inside) we get the final result: 1 2 m pu kt 1 = 1 2 m pu kt 2.

60 Conservation equations Isothermal solar wind Interaction with the interstellar medium Summary of shock relations n 1 u 1 = n 2 u 2 m p n 1 u1 2 + n 1 kt 1 = m p n 2 u2 2 + n 2 kt m pu kt 1 = 1 2 m pu kt 2 Since we have 3 equations and 6 unknowns, three of these need to be specified in order to have a unique solution, for instance the solar wind speed u 1 and the interstellar gas density n 2 and pressure n 2 kt 2. In the pressure balance, the magnetic as well as cosmic ray pressure of the interstellar gas may have to be included.

61 Conservation equations Isothermal solar wind Interaction with the interstellar medium Location of the termination shock From the pressure balance condition, neglecting the thermal solar wind pressure, the heliospheric distance r T to the termination shock is m p (nu) E u E r T = 1 AU m p n 2 u2 2 + n 2kT 2 Choosing n 2 = 10 5 m 3, T 2 = 7000 K, and u 2 = 25 km/s on the upwind side of the Sun (Axford & Suess) we find m p n 2 u Pa n 2 kt Pa m p (nu) E u E Pa

62 Conservation equations Isothermal solar wind Interaction with the interstellar medium Location of termination shock (contd.) On the upwind side the dynamic pressure dominates over the thermal pressure of the interstellar gas At 1 AU the solar wind dynamic ( ram ) pressure is 10 4 times larger than the interstellar gas pressure, so the termination shock must be far from us! With these parameters we find r T 140 AU.