When meteors enter the atmosphere, they are typically traveling at around 15km/s

Question

When meteors enter the atmosphere, they are typically traveling at around 15km/s. At such

speeds, heat does not have time to be conducted away and even the air has no time to be pushed out of the way. This results in adiabatic compression of the air ahead of the meteor. The density of the atmosphere in kg/m3 is given by

rho=1.22e^(h/8100)

where h is the altitude in meters.

a) For a spherical meteor of radius 5m entering the atmosphere on a vertical trajectory, As a result of compressing all the air in its way into a cylinder of radius 5.5m and height 0.3m, there

will be a very high pressure and temperature in front of the meteor. Calculate the density of air in this compression zone when the meteor has reached an altitude of 4050m by finding the kg of air compressed and the volume of the compression zone. Hint: dm = ?Adh

b) Calculate the original volume of the gas before it was compressed. You may use h=40,000m as the top of the atmosphere.

c) Calculate the average pressure using   where h = altitude in meters. Calculate the average pressure as Pavg = [?Pdh]/?h

d) Now find the adiabatic constant using the average pressure from (b) and the original volume from (a). Take Cp/Cv = 1.35 for air.

e) Show that the ideal gas law can be written as   where M for air is about

25.6x10-3 kg/mol. f) Use the results from (e) (d) and (a) to get the pressure of the compressed gas at h=4050

meters. Take Cp/Cv = 1.35 for air. Compare the pressure to the ultimate strengths for rock(107Pa) and iron(108Pa) to see if the meteor explodes. g) Using the results from (e) and (f), calculate the temperature in the compression zone at h=4050 meters. h) At the temperature calculated in (g), determine the wavelength of the maximum intensity and

the total power emitted by the hot gas (using a surface area of 2? 5.52 for the flattened spheroid of gas).

a) For a spherical meteor of radius 5m entering the atmosphere on a vertical trajectory, As a result of compressing all the air in its way into a cylinder of radius 5.5m and height 0.3m, there will be a very high pressure and temperature in front of the meteor. Calculate the density of air in this compression zone when the meteor has reached an altitude of 4050m by finding the kg of air compressed and the volume of the compression zone. Hint: dm = pAdh b) Calculate the original volume of the gas before it was compressed. You may use h=40,000m as the top of the atmosphere. c) Calculate the average pressure using P(h) =1[atm]e^-h/7000 where h = altitude in meters. Calculate the average pressure as Pavg = [integrate Pdh] delta h d) Now find the adiabatic constant using the average pressure from (b) and the original volume from (a). Take Cp/Cv = 1.35 for air. e) Show that the ideal gas law can be written as P=PRT/M where M for air is about 25.6x10^-3 kg/mol. f) Use the results from (e) (d) and (a) to get the pressure of the compressed gas at h=4050 meters. Take Cp/Cv = 1.35 for air. Compare the pressure to the ultimate strengths for rock(10^7Pa) and iron(10^8Pa) to see if the meteor explodes. g) Using the results from (e) and (f), calculate the temperature in the compression zone at h=4050 meters. h) At the temperature calculated in (g), determine the wavelength of the maximum intensity and the total power emitted by the hot gas (using a surface area of 2pi 5.5^2 for the flattened spheroid of gas).

Explanation / Answer

a)

dm = ro*A*dh

Area A =pi*r^2 =pi*5.5^2 =95 m^2

ro =1.22*exp(-h/8100)

m =1.22*95*integral (from 4050 to infinity) exp(-h/8100) = 1.22*95*8100*exp(-4050/8100) =5.694*10^5 kg

Volume of the compression zone is

V =pi*R^2*H =pi*5.5^2*0.5=47.517 m^3

ro = 5.694*10^5/47.517 =11983 kg/m^3

b)

original volume of gas was

V0 = pi*R^2*(h-h0) =pi*5.5^2*(40000-4050) =3.41*10^6 m^3

c)

P(h) =1*exp(-h/7000)

Pavg =1/(40000-4050)*integral(from 4050 to 40000) exp(-h/7000) =

=1/35950 *7000*[exp(-4050/7000) -exp(-40000/7000)] =0.1085 atm

d) This question is clearly saying gamma= Cp/Cv =1.35

C = Pavg*V0^gamma = 0.1085*10^5*(3.41*10^6)^1.35 =7.16*10^12

(because 1 atm =10^5 Pa)

at point f) you will compute the pressure P in the compressed zone using gamma =1.35

and the equation

Pavg*V0^gamma =P*V^gamma

(where V and V0 are the volumes computed above)

e) Law of perfect gases (ideal gas law) is (N is the number of gas moles)

P*V= N*R*T

P*V = m/M*R*T

P = (m/V)*(1/M)*R*T

P =(ro/M) *R*T

f)

see point d) above.

P = Pavg* (V0/V)^gamma = 0.1085*(3.41*10^6/47.517)^1.35 =3.9*10^5 atm =

=3.9*10^5*10^5 Pa =3.9*10^10 Pa

The pressure is 390 times bigger than the maximum strength of iron.

The pressure is 3900 times bigger than the maximum strength of rock,

The meteor will explode.

g)P =ro*R*T/M

ro =11983 kg/m^3

R =8310 J/Kmol/K

M =25.6 kg/kmol

T = 25.6*(3.9*10^10)/11983/8310 =10026 K

h)

see http://en.wikipedia.org/wiki/Black-body_radiation#Wien.27s_displacement_law

Wien law states:

lambda = b/T = 2.898*10^-3/10026 =2.89*10^-7 m =289 nm

Total radiated power is (Stefan Boltzmann law) for emissivity =1 (black body)

(http://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law)

P =A*sigma*T^4

A= 2*pi*5.5^2 =380 m^2

T =10026 K

sigma =5.67*10^-8 J/s/m^2/K^4

P =2.18*10^11 W

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