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Physics Newtons Gravitation

  1. Feb 22, 2009 #1
    Suppose an object is dropped from height h, where h < R but gravity is not constant). Show that the speed with which it hits the ground, neglecting friction, is approximately given by:

    v = sqrt(2gh)(1 - (h/2r))

    where g is the acceleration due to gravity on the surface of the Earth.
    HINT: You will need to expand an expression in a MacLaurin series. Be sure to expand an expression with a small value, so higher order terms can be ignored.
     
  2. jcsd
  3. Feb 22, 2009 #2

    malawi_glenn

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    And what is your attempt?
     
  4. Feb 22, 2009 #3
    g(h)= GM/(R+h)^2 which decreases with h increases

    g(h) / g = R^2/(R+h)^2

    g(h) = g * (1+ (h/R))^-2

    now we know that for the fall if we consider energy conservation

    1/2 mV^2 = mg(h)h

    V^2 = 2*g(h)*h

    V^2 = 2*g*h * (1+ (h/R))^-2

    V = sqrt(2gh) * (1+(h/R))^-1

    now if i expand the function (1+(h/R))^-1

    Then I dont get the form which is asked for. Can you please help me to get the correct form?
     
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