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Gravitational force and the distance from the earth

  1. Jun 18, 2004 #1
    I really need help with this question. I've been trying to figure it out for a couple of hours.

    Find the altitudes above the earth's surface where earth's gravitational field strength would be (a) five-sixths and (b) one-sixth of its value at the surface. These units have to be in km.

    thank you.
     
  2. jcsd
  3. Jun 18, 2004 #2

    jcsd

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    at the surface accelartion due to gravity is given by:

    [tex]g = \frac{GM}{r^2}[/tex]

    where r is the radius of the Earth G the universal graviational constant and M the mass of the Earth:

    therfore:

    [tex]\frac{5g}{6} = \frac{5GM}{6r^2}[/tex]

    [tex]\frac{g}{6} = \frac{GM}{6r^2}[/tex]

    re-arrange the equations, plug in the numbers and the adjust for the correct units.
     
  4. Jun 18, 2004 #3
    Substitute GM/R^2 in for g and solve for r using jcsd's equations.

    What value are you getting?

    cookiemonster
     
  5. Jun 18, 2004 #4

    jcsd

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    All you need to do is recognize th the answers to

    a)
    [tex]\sqrt{\frac{6r^2}{5}}[/tex]

    b)
    [tex]\sqrt{6r^2}[/tex]

    where r is the radisu of the Earth at the surface.
     
    Last edited: Jun 18, 2004
  6. Jun 18, 2004 #5

    jcsd

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    Sorry that's cos I missed out the indices!
     
  7. Jun 18, 2004 #6
    What do you mean by that? are the equations wrong?
     
  8. Jun 18, 2004 #7

    jcsd

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    they are correct now.
     
  9. Jun 18, 2004 #8
    they haven't changed. are they the same equations you gave me originally?
     
  10. Jun 18, 2004 #9

    jcsd

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    You should be getting answers of roughly:

    a) 6979 km

    b) 15606 km

    the equations are correct.
     
  11. Jun 18, 2004 #10
    these values are not correct either. I submitted them, and the professor said they were not correct.
     
  12. Jun 18, 2004 #11
    jcsd's solutions are correct. If you're using a different value for R, then plug it into the formulas he gave to get whatever answer your professor thinks it's looking for.

    cookiemonster
     
  13. Jun 19, 2004 #12
    I think it is just the wording of the question that is gettting the wrong answer not that the answers are wrong. NOte it says altitude, i think altitude means distance above earths surface therefore the anwers given for a and b should have the radius of the earth subtracted from them. Depending on what constants you are using I am getting a around 610km (roughly) if you need sig figs you better do it all i.e. follow the equation and just get an answer, not just subtract from the other answers and for b about 9300km real rough two sig fig calcs
     
  14. Jun 19, 2004 #13
    how are you getting 610km? When I take 6.38e3 from 6.37e3(which is the radius of the earth in km), I'm not getting that value. I'm getting the values that jcsd got, but how do i get the altitude?
     
  15. Jun 19, 2004 #14
    once i have the radius, how do i find the altitude? taking the earth's radius minus the radius i found, does not produce what jungle peanut got.
     
  16. Jun 20, 2004 #15
    Lets say you have the same answer as jcsd,
    6979km as the radius where the acceleration is 5/6 the acceleration at the surface.
    If the surface is at a radius of 6378km then I subtract 6979km from 6378km and get 601km pretty close to my rough answer.
     
  17. Jun 20, 2004 #16
    thank you all for the help! I'm terrible at working physics problems. thanks again.
     
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