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Surface temperature of a star?

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  1. Jan 17, 2016 #1
    • moved into h/w help, so template is missing
    I am trying to find the temperature of a star given its wavelength in micrometres, but I am not sure if my conversion is right therefore don't know if the answer is correct.

    Star A has a maximum emission wavelength of 1 μm and Radius 100 Rsun. What is its Effective Temperature and Luminosity?

    Where 109 is the μm → metre conversion.

    1μm → 1000nm
    So, 1000nm / 109m = 1×10-6m

    Finding the Temperature:

    T = 0.0029K m / λmax
    T = 0.0029K m / 1×10-6m

    T= 2.9×10-9 K → (how do I put this value into millions and not in scientific notation?)

    Given the result and assuming it is correct we find the luminosity with the surface area and Stephan-Boltzman Law as shown below:

    L= 4πR2 σT4 where σ is the Stephan-Boltzman constant.

    My confusion here is, how do I plug the value of the Radius since I have it in Solar Radius?
     
  2. jcsd
  3. Jan 17, 2016 #2

    Simon Bridge

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    ... you subtract 6 from the exponent, or just write out all the zeros and remove six of them.
    1 million is 1x106 ... the same as a "mega..."

    You either multiply the number of solar radiuses you have been given by the value of one solar radius in more convenient units ... or use a value of the constant given in units of "per square solar radii".
     
  4. Jan 17, 2016 #3

    SteamKing

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    You might want to check your arithmetic here. 10-9 is not capable of being expressed in "millions". When the exponent is negative, the value of the number is less than 1.
    You look up the value for the radius of the sun in meters and convert that number to the radius of your star. Did you really have to ask this question?

    BTW, interesting fact. The surface temperature of blue giant stars rarely exceeds 30,000° K. Their luminosity is due primarily to their huge size.
     
  5. Jan 17, 2016 #4

    vela

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    You didn't plug this into your calculator correctly. Use the EE key or use parentheses appropriately. Or better yet, don't use a calculator.
     
  6. Jan 17, 2016 #5

    Simon Bridge

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    Um ##10^{-9}## is ##10^{-15}## million ... though that would be a particularly cool star ;)

    Off that I thought I'd go through the rest of the post more carefully and not just answer the questions.

    Um, no: the metric exponents to prefixes go:
    -3 milli
    -6 micro
    -9 nano
    -12 pico

    So 1m = 109nm and there are 1000nm in a ##\mu##m

    You shift the decimal point 3 to the right every time you want to multiply by 1000 thus:
    0.0029 K.m = 2.9K.mm = 2900K.##\mu##m = 2900000K.nm or 2.9 million K.nm
     
  7. Jan 17, 2016 #6
    Hi Simon,

    Thank you for your answer. I noticed I somehow managed to plug the value in the wrong way and get an incorrect answer as Vela says above.
    However, I believe my conversion for micrometer to meters is wrong since the answer I get is -5.971 and that looks very weird to be a temperature value.

    I know the "λmax" value has got to be in meters but not sure though what's the right procedure.
     
  8. Jan 17, 2016 #7

    Simon Bridge

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    Your conversion from micrometers to meters is, indeed, wrong - I have pointed out the correct conversion above, in the metric prefixes list.
    ##\lambda_{max}## can be in any units of length - provided you adjust your equation appropriately.

    There is no "right procedure" - just makes sure you get the relations right.

    i.e. $$T=\frac{0.0029(\text{K.m})}{\lambda_{max}(\text{m})} = \frac{2.9(\text{K.mm})}{\lambda_{max}(\text{mm})} \cdots$$ ... it is often easier to adjust the constants in the equation than it is to put everything into base SI units.

    But you can change everything into meters, kilograms, and seconds, if you want to.
     
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