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/m 3 is given by

p=1.22*e^(-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 0.5m, 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 = [integral(Pdh)]/dh

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(10^7 Pa) and iron(10^8 Pa) 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*pi*5.5^2 for the flattened spheroid of gas).

note that p = rho (aka density), and the carrot sign (^) represents an exponent.

Explanation / Answer

a)

At h = 4050 m, density p = 1.22*e^(-4050 / 8100) = 0.74 kg/m^3

Area A = pi *r^2 = 3.14 * 5.5^2 = 94.985 m^2

dm = p*A*dh

dm = 0.74*94.985*0.5 = 35.1444 kg

New volume = A*dh - pi*h^2 /3 *(3*r_sphere - h)

= 94.985*0.5 - 3.14 *0.5 ^2 /3 *(3*5 - 0.5)

= 43.698 m^3

New density = 35.1444 / 43.698 = 0.804 kg/m^3

b)

V = pi *r^2 *h

V = 3.14 *5.5^2 *40000 = 3799400 m^3

c)

P = 1*e^(-h / 7000)

Integral P dh = Integral e^(-h / 7000) dh

= -7000 *e^(-h / 7000)

Integrating from h = 0 to 40000 m,

P_avg = -7000*[e^(-40000 / 7000) - 1] / 40000

P_avg = 0.1744 atm

d)

Adiabatic constant = 0.1744*10^5 *3799400 = 6.627*10^10

e)

Using ideal gas eqn,

PV = m (R/M) T

But m/V = p = density

Hence, P = p*(R/M)*T

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