Question 15.9: Meteorite Entry A meteorite enters the earth’s atmosphere at...

Use the isentropic and normal shock tables to determine the total properties after the shock.
Assumption
Close to the meteorite, the shock can be assumed to be normal to the direction of the flow. Flow is isentropic before and after the shock, and air can be assumed to be a perfect gas.
Analysis
Obtain the stagnation properties at State-1 before the shock by using the table panel of the gas dynamics TESTcalc and entering M1 = 20 (roll the pointer over a variable to view a more accurate value on the message panel):

\begin{gathered}p_{t 1}=\left[\left(\frac{p}{p_{t}}\right)_{@ M_{1}}\right]^{-1} p_{1}=\left(\frac{1}{2.0788 \times 10^{-7}}\right)\left(\frac{1}{1000}\right)=4,810 MPa \\T_{t 1}=\left[\left(\frac{T}{T_{t}}\right)_{@ M_{1}}\right]^{-1} T_{1}=\left(\frac{1}{0.012355}\right)(200)=16,188 K\end{gathered}

The total temperature remains unchanged after the shock, but the total pressure decreases as dictated by the shock strength:

p_{t 2}=\frac{p_{t 2}}{p_{t 1}} p_{t 1}=\left(\frac{p_{t e}}{p_{t i}}\right)_{@ M_{3}} p_{t 1}=\left(1.0714 \times 10^{-4}\right)(4,810)=0.515 MPa

The flow after the shock is isentropic; therefore, the stagnation pressure must be the same as the total pressure:

p_{s 2}=p_{t 2}=0.515 MPa ; \quad T_{s 2}=T_{t 2}=T_{t 1}=16,188 K

TEST Analysis
Launch the gas dynamics TESTcalc and select air as the working fluid. Evaluate the states, as described in the TEST-code (see TEST > TEST-codes), to verify the manual results.

Discussion
The high values of stagnation pressure and temperature explain why most meteorites disintegrate upon entering the earth’s atmosphere. At such extreme conditions, dissociation of oxygen and nitrogen is very likely, and would alter the gas properties, so the results presented in this example should be considered approximate.