# 2- The earth’s atmospheric pressure p is often modeled by assuming that the rate dp/dh at Which p changes with the altitude h above sea level is proportional to p. Suppose that the pressure at sea level is 1013 millibars (about 14.7 pounds per square inch) and that the pressure at an altitude of 20km is 90 millibars.

a) Solve the initial value problem

Differential equation: dp/dh = kp
Initial Condition: p = p0 and k from the given altitude-pressure data.

b) What is the atmospheric pressure at h = 50km?

c) At what altitude does the pressure equal 900 millibars?

Is that a typo? Would p be a function of h?
dp/dh= kh?

2- The earth’s atmospheric pressure p is often modeled by assuming that the rate dp/dh at Which p changes with the altitude h above sea level is proportional to p. Suppose that the pressure at sea level is 1013 millibars (about 14.7 pounds per square inch) and that the pressure at an altitude of 20km is 90 millibars.

a) Solve the initial value problem

Differential equation: dp/dh = kp
Initial Condition: p = p0 when h = 0,
to express p in terms of h. Determine the values of p0 and k from the given altitude-pressure data.

b) What is the atmospheric pressure at h = 50km?

c) At what altitude does the pressure equal 900 millibars?

If you believe the problem is right, then thre is no relation of p to height stated in the model, and the problem is unsolvable. I think the rate dp/dh should be proportional to h. That will yield a solution.

a) p = 1013e^(-0.121h)
b) 2.383 millibars
c) 0.977 km

any ideas on how to get those?

dp/dh = kp --->

P = P(0) Exp[k h]

For h = 0 the pressure is p(0) which is given as 1013 mb. Next solve k by demanding that for h = 20 km you get the pressure at that height.

1. k = -ln(90/1013)/20

Now you have the equation for p as a function of h.

b) p(50) = 1013 Exp[-0.121*50] = 2.383 mb

c) Solve for h when p = 900 mb

900 = 1013 Exp[-0.121 h]

h = 0.977 km

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