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Gravitational force?

  1. Apr 10, 2008 #1
    A starship of mass m is traveling between a pair of binary stars that are a distance R apart. the mass of star 2 is exactly twice the mass of star 1. How far from star 1 will the gravitational force on the starship be equal to zero? Express the answer in terms of R.

    When I try, I get 2/3R but the answer should be R/2.41

    GmM/R^2(x) - Gm2M/R^2(R-x) = 0
    GmM/R^2(x) - Gm2M/R^2(R) + Gm2M/R^2(x) = 0
    GmM/R^2(x) + Gm2M/R^2(x) = Gm2M/R^2(R)
    x = Gm2M/R^2(R)/[3GmM/R^2]

    x = 2/3R
  2. jcsd
  3. Apr 10, 2008 #2


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    Your whole formula is wrong! You have "R2" in the denominator of both fractions, but have multiplied by x and R-x. I have no idea why you would do that. I assume that "x" is the distance from the starship to the star 1 (it would be good practice to specifically say that) so the gravitational force from the star 1 would be GmM/x^2 and the gravitational force from star 2 would be Gm(2M)/(R-x)^2. Assuming the starship is between the two stars, the will be no (net)gravitational force on the starship when GmM/x^2= 2GmM/(R-x)^2. I would recommend you divide both sides of the equation by GmM immediately, then multiply boths sides by x^2 and (R-x)^2.
  4. Apr 10, 2008 #3
    I dunno why but I keep getting stuck at (R-x)^2= 2x^2
  5. Apr 10, 2008 #4
    You are almost there your final equation that you posted is right-- the rest is math:

    [tex](R-x)^2 = 2x^2 \Rightarrow \\
    (R-x) = \sqrt{2} x \Rightarrow \\
    R = x + \sqrt{2} x \Rightarrow \\
    x(1+\sqrt{2}) = R \Rightarrow \\
    x = R/(1+\sqrt{2}) \Rightarrow \\
    x \approx R/2.41
  6. Apr 10, 2008 #5
    And technically, the force of gravity on something is NEVER 0. Although the Net Force can make it seem so. And it could just get to a point so low that its almost irrelvant. But the gravitational force between objects can't be zero, no matter how far apart they are. So I think you should have mentioned Resultant Force somewhere in there. And are we supposed to assume that the ship is right in between the stars? Because it could be arranged like this :
    -> . · . <-
    where the left and right dots are stars and the middle dot is the ship. Since the ship is not EXACTLY in between the stars (in this case the ship is higher on the "y-axis"), the distance would probably be different than if the ship was right in between.
  7. Apr 10, 2008 #6
    but then there would be a force composant pointing downwards, hence the net force in such arrangement cant be zero and it follows that the ship must be right in between the stars.
  8. Apr 10, 2008 #7
    Very true, if gravity is the only relevant force.
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