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.

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

To solve for k, we can use the given data that at an altitude of 20 km, the pressure is 90 millibars. Plugging this into the equation:

90 = 1013 * e^(20k)

Dividing both sides by 1013:

e^(20k) = 90/1013

Now take the natural logarithm of both sides:

20k = ln(90/1013)

Solving for k:

k = ln(90/1013) / 20

Once you have the value for k, we can substitute it back into the equation:

p = 1013 * e^(kh)

To find the atmospheric pressure at h = 50 km, we plug this value into the equation:

p(50) = 1013 * e^(k * 50)

To find the altitude at which the pressure equals 900 millibars, we set p = 900 and solve for h:

900 = 1013 * e^(kh)

Taking the natural logarithm of both sides:

ln(900/1013) = kh

Solving for h:

h = ln(900/1013) / k

Substituting the value of k we found earlier, we can calculate the value of h.

To find the value of k, we can use the given data point at an altitude of 20km where the pressure is 90 millibars. Plugging in the values into the equation, we get:

90 = 1013e^(k * 20)

To solve for k, divide both sides by 1013:

e^(k * 20) = 90/1013

Take the natural logarithm of both sides:

k * 20 = ln(90/1013)

Finally, solve for k:

k = ln(90/1013) / 20

Now that we have found the value of k, we can substitute it back into the equation to solve for p in terms of h. The equation is:

p = 1013e^(k * h)

To find the atmospheric pressure at h = 50km, we simply substitute h = 50 into the equation:

p = 1013e^(k * 50)

Calculate the value of p using the given values of p0 and k, and plug in h = 50.

To find the altitude at which the pressure equals 900 millibars, we can set p = 900 and solve for h in the equation:

900 = 1013e^(k * h)

Take the natural logarithm of both sides and solve for h.