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Variation of g with altitude and depth

  1. Sep 11, 2011 #1
    1. The problem statement, all variables and given/known data[/
    The value of acceleration due to gravity (g) at an altitude (h) is gh = g (1 - 2h/R).
    Similarly the value of g at a depth (d) is gd = g(1 - d/R), where R is the radius of the earth.
    2. Relevant equations

    In both the cases, my book says the value of g decreases with increase in altitude and increase in depth, by quoting these equations. I dont know how to interpret this result by mere equations. Revered members can help in this regard

    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 11, 2011 #2


    The gravitation of attraction of a spherical object is equal to the case when ALL the material "under you feet" is completely concentrated at the center. So just imagine the part of the earth which is closer to the center than you are, completely collapse to the center.

    When you are above earth's surface (at attitude), the amount of material "under your feet" do not change: it consists of the whole earth, but you are moving further away, you are moving further away from the center (where they "collapsed").

    When you are below earth's surface (at depth), the amount of material "under your feet" is only a portion of the whole earth, and the portion gets smaller as you get closer to the center. So gravity also gets weaker.
     
  4. Sep 11, 2011 #3
    Sir, i cant understand. Also i want to know how can we say the value of g is deceasing by the given mathematical equations. Thanks for the reply sir
     
  5. Sep 11, 2011 #4

    HallsofIvy

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    The "universal law of gravity" is [itex]F= -(GmM)/r^2[/itex] where G is the "universal gravitational constant" (NOT "g") , m and M are the masses of the two attracting bodies and r is the distance between the centers of the two bodies. In particular, if we take M to be the mass of the earth and R to be the radius of the earth, F= -(GM/R^2)m= -gm so that g= GM/R^2. If r= R+ h, then we have F= -GmM/(R+ h)^2.

    We can, using the "generalized binomial formula", write [itex](R+h)^{-2}= R^{-2}- 2R^{-3}h+ [/itex] higher order terms in h so if h is small compared to R, we can approximate [itex]-GmM/(R+ h)^2[/itex] by [itex]-GmM(R^{-2}- 2R^{-3}h= -GmM/R^2+ 2[GmM/R^{-2}](h/R)= -g(1- 2h/R).


    And, of course, both of the formulas have g(1 minus something) so as that "something" increases, g decreases.
     
  6. Sep 17, 2011 #5
    Thanks for the reply HallsofIvy.
     
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