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Homework Help: Barometric formula to calculate ground pressure

  1. Jan 16, 2012 #1
    1. The problem statement, all variables and given/known data
    A spacecraft approaches the surface of an unfamiliar planet and finds that the pressure of the atmosphere 50 km above the surface is 2.0 × 10^5 Pa. As the spacecraft descends, it is found that the pressure of the atmosphere 40 km above the surface increases to 5.0 × 10^5 Pa. Estimate the pressure at the surface of the planet and hence decide whether it is safe for the spacecraft to land, assuming that the atmosphere is thin and isothermal, and that the spacecraft can withstand an external pressure of 1.0 × 10^7 Pa.

    2. Relevant equations
    P(z) = P(0)e^(-z/y)

    3. The attempt at a solution
    I know that pressure decreases exponentially with the increase in height

    Any pointers on where to start would be appreciated as always!
    Last edited: Jan 16, 2012
  2. jcsd
  3. Jan 16, 2012 #2


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    Okay, so at an altitude of 40 km, that equation becomes ________?
    And at 50 km, it becomes _______?
  4. Jan 16, 2012 #3
    So at 40km:

    (5x10^5 Pa) = P(0) e^(-40^3/γ)

    and at 50km:

    (2x10^5 Pa) = P(0) e^(-50x10^3/γ)

    .....????? do I need to find the scale height (γ)? If so wouldn't be different every time?
  5. Jan 16, 2012 #4


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    Okay, good, you now have two equations in two unknowns -- P(0) and γ. The scale height γ is a constant.

    What methods do you know about solving two equations in two unknowns?
  6. Jan 17, 2012 #5
    Substitution? But I don't know any of the values to calculate y?
  7. Jan 17, 2012 #6


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    The "substitution" method of solving a system of two equations means you solve one equation for P(0), then substitute that into the other equation in place of its P(0). You end up with one equation with one unknown. Another approach would be to divide one equation by the other - the P(0) will cancel out. Then you could solve for the γ.

    Your P(z) = P(0)e^(-z/y) is a mathematical model for the pressure. It has two parameters, P(0) and y, that can be found by using the pressure measurements. Once you have them you can use the model to find the pressure at any height.
  8. Jan 18, 2012 #7
    Just got your message will have a look now thank you very much for your time on this both of you :smile: :!!)
  9. Jan 18, 2012 #8
    ok so it works out that the scale height (y) = 25000m and hence pressure at height 0m is P(0) = 2.5x10^6Pa... so the spaceship survives

    If anyone else has done the math are my answers spot on??? They look plausible...
  10. Jan 18, 2012 #9
    knowing that pressure increases expotentially you can check your answer. in 1 drop of 10 km pressure increases by what? ( i would recomend pg 166, qu 4.3 of your course book) as for calculating for the symbol y I would suggest looking at part 'a' of the assignment question.
  11. Jan 18, 2012 #10
  12. Jan 18, 2012 #11


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    I don't get those values for P(0) and y. Maybe show your work so we can sort it out.
    I checked my model with its different values and it does give 200 kPa at 50 km. Does yours?
  13. Jan 19, 2012 #12
    Thanks for checking - yeah my figures just don't add up :-(

    Ok here is my working...

    To work out y: where P(z2) = 2.0x10^5Pa, z2 = 50x10^3m, P(z1) = 5.0x10^5Pa and z1 = 40x10^5


    Divide P(z2) by P(z1) and rearrange to get y:

    P(z2)/P(z1) = P(0)e^(-z2/y) / P(0)e^(-z1/y)

    P(z2)/P(z1) = e^(-z2/y) / e^(-z1/y) ****** P(0) cancels

    P(z2)/P(z1) = e^((-z2/y)-(-z1/y))

    P(z2)/P(z1) = e^(-(z2+z1)/y)

    log (P(z2)/P(z1)) = -(z2+z1)/y

    y = -(z2+z1) / log (P(z2)/P(z1))

    substitute in the values above:

    y = -(50x10^3m + 40x10^3m) / log (2.0x10^5Pa)/(5.0x10^5Pa))

    Scale height (y) = 25x10^3 m


    Ok, so now using this as the value of y and using the equation

    P(z) = P(0)e^(-z/y) ****** rearranged for P(0)

    P(z) / e^(-z/y) = P(0)

    Using the values at z2

    2.0x10^5 Pa / e^(-(50.0x10^3 m / 25.0x10^3 Pa)) = 1.5 x 10^6 Pa (not my original)


    Please help!!!
    Last edited: Jan 19, 2012
  14. Jan 19, 2012 #13
    Found my error - used log instead of ln!!! Don't I feel stupid?!?! It all tallies up now...

    Thank you for all your help!
  15. Jan 20, 2012 #14
    Im glad youve got a different answer, cause so did I. I can stop panicing and double and treble checking my answer now.
  16. Feb 14, 2012 #15
    I thought I would have a go at this but am stuck on how you got (y) = 25x10^3 m from y = -(50x10^3m + 40x10^3m) / log (2.0x10^5Pa)/(5.0x10^5Pa)) either way using log of ln

    If anyone one could solve this cheers

  17. Feb 14, 2012 #16
    You get it if you use the natural log instead of log...

    Incidentally thank you to everyone who helped I got 92% on this assignment
  18. Feb 14, 2012 #17
    SO what did y = -(50x10^3m + 40x10^3m) / log (2.0x10^5Pa)/(5.0x10^5Pa)) change to etc

    Thanks. and do you just mean ln instead of log
  19. Feb 14, 2012 #18
    Also is it not y = -(50x10^3m - 40x10^3m) / log (2.0x10^5Pa)/(5.0x10^5Pa))

    instead of y = -(50x10^3m + 40x10^3m) / log (2.0x10^5Pa)/(5.0x10^5Pa))

    Im thinking the pressure at the surface is 1.95*10^7 but this is obtained from other sources
  20. Feb 14, 2012 #19
    Will dig out my assignment...
  21. Feb 14, 2012 #20
    There you go...


    For the final paragraph just did a little rearranging of the formula
    Last edited: Feb 14, 2012
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