I only need the answer to Q. 1 d), Q 3 e). It’s mostly math involved.  Question 1 (a) In…

Question 1

(a) In the region of the Sun where energy is transported by radiation, the flux of radiant energy at a

radial distance r from the centre is F (r )=-(16(boltz. const)T^3)/3kpdT/dr

 , where  k and p are the mean opacity and density of the solar material at r . Show that this relation can be re-written in terms of the temperature gradient and the solar luminosity L as follows,

               dT/dr=-(3kpL)/(64pi(boltz const)T^3r^2)

(b) At radius r=0.4RS , the temperature, density and opacity of the solar material are

T  =5 *106 K, p= 5 *103  kg m 3 ,  k=0.5 m2  kg^- 1 . Use this data to determine the radial temperature gradient at this distance from the centre.

(c) Compare your value in part (b) with the linear temperature gradient (between centre and surface) in the Sun.

(d) The speed of sound waves propagating through the Sun can be estimated from the following relation, c= 10^4 sqrt(T/10^4) m/s. Use this and the value of linear temperature gradient dT/dr

to estimate the time it takes a sound wave to travel from the centre to the surface of the Sun.

Solar System Physics Coursework: Planetary Physics

Surviving beneath a planetary surface

  Qestion 2  The method of dimensions

a) We have looked in the lectures at steady heat conduction through an ice layer to find out how quickly a lake might freeze over. The equation governing unsteady heat conduction (in the vertical direction z ) is

pCdT/dt=kd^T/dz^2  , X is the thermal conductivity, C specific heat capacity ,r density. We can get a ‘cheap and cheerful’ estimate of the rate at which heat diffuses through the ground by

neglecting the differentiation in the equation above so that     pCT/t ~ kT/z^2

As a result of a forest fire, the ground temperature is raised significantly. Using the simplified equation above, estimate how long it takes this heating to affect a mole’s tunnel 2 metres under the surface.  The thermal conductivity of soil is 1.1Wm-1K-1 , the specific heat capacity is 1000Jkg-1K-1  [this is a rough value, the true value will depend on the moisture content of the soil, from 800 for dry soil to 1500 for a wet soil], and the density of a (clay-rich) soil is 1.25´103 kgm-3 . Does your estimate depend on the magnitude of the surface temperature rise caused by the fire?

Question 3 Exact solutions of the heat conduction equation

The simple method of question  (2) is useful but it is helpful to have a more accurate method of determining the effect of the surface temperature change. For one thing, it gives no sense of whether the mole will experience the same temperature rise as the surface or something smaller. A little thought suggests this will depend on how long the surface is heated for. A related problem is periodic heating e.g. the warming of the ground in the day and the cooling at night; or, on longer timescales, the seasonal heating of the surface; or, on even longer timescales, the effect of ice ages. To investigate these effects, we consider harmonic solutions of the heat conduction equation i.e. solutions in the form of sines and cosines.

(a)  We write the heat conduction equation in the form:  dT/dt = X d^2T/dz^2 , where X is the thermal diffusivity. Assume that a solution exists of the form:

T (z, t) = Tmean  + DT Re{exp (iwt – ikz )} 

Re denotes the ‘real part’

In  the  above  equation,  we  assume  that  the  surface  is  at z = 0 ,  and  that   z  increases  vertically downwards. Substitute this solution in the equation and show that the oscillation parameters k and w

are related by (the dispersion relation):

iw = -c k 2 .

(b)  Using  sqrt -i = 1/sqrt 2 (1- i)  show that possible solution is therefore

T(z,t) = Tmean + DT exp (-b z ) cos (wt – b z )   b =  sqrt (omega/(2X))

(c)  State expressions, in terms of T (mean) and delta T, for the maximum and minimum temperature values on the surface.

 (d) We’ll study the meaning of this equation by considering the diffusion of heat through the surface rock layer with thermal conductivity equal to density 2.3´103 kgm-3 ,thermal conductivity 2.5Wm-1K-1 ,and specific heat capacity is 103 Jkg-1K-1 

The skin depth is defined as the depth below the surface where the variation in temperature has

dropped to 1 e of its surface value. Show that this depth is given by  z(skin) = sqrt 2k/ (p C w) 

 (e) Show that for daily (so called ‘diurnal’) variation w = 7.27 ´10-5 s-1 . Hence deduce, the skin depth of rock for diurnal variations in temperature. How deep, would you have to be below the rock surface to