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Gravitational Field Question

  1. Nov 14, 2008 #1
    1. The problem statement, all variables and given/known data

    Three objects -- two of mass m and one of mass M -- are located at three corners of a square of edge length l. Find the gravitational field g at the fourth corner due to these objects. (Express your answers in terms of the edge length l, the masses m and M, and the gravitational constant G).

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    2. Relevant equations

    g=-GM/r2

    3. The attempt at a solution

    g= ga+gb+gc
    g= Gm/l2 [tex]\hat{i}[/tex]+ (GM/(l[tex]\sqrt{2}[/tex])2)(cos([tex]\pi/4[/tex])[tex]\hat{i}[/tex]+sin([tex]\pi/4[/tex])[tex]\hat{j}[/tex])+Gm/l2 [tex]\hat{j}[/tex]

    I know you have to take the magnitude of this, but when I did that , I still get the wrong answer. Here's what I got:

    ||g||=[tex]\sqrt{2G^2/l^4(m^2+M^2)}[/tex]

    Did I start it right? Can someone help?
     
    Last edited: Nov 14, 2008
  2. jcsd
  3. Nov 14, 2008 #2

    LowlyPion

    User Avatar
    Homework Helper

    Isn't the gravitational field given by GM/r ?

    You have 3 vectors to add, but happily the 2 m's at right angles gives one lying in the direction of M

    So ... √2Gm/L + GM/(√2*L) = √2*G*(m + M/2)/L ?
     
  4. Nov 14, 2008 #3
    So do you take the magnitude of that? Im still confused, because my book says otherwise. Can you go through your process?
     
  5. Dec 2, 2008 #4
    I think what you did originally is correct, but it ask for the magnitude without the vector sign. So just put down the answer using c^2 = a^2 + b^2 and then I believe you have to indicate the degree according to the x-axis. I'm doing a similar problem. Wait, yeah you did that, never mind.
     
  6. Dec 2, 2008 #5
    I got it!
    Instead of converting M vectors into g forces of x and y, why don't you convert the other 2 m mass into direction of M which is Gm/l^2 * cos(45) * 2.
    Then add it to the g force of M
    My answer is (1.41Gm + 0.5GM)/l^2.
    Hope it helps, the post was like half a month ago, lol.
     
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