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Newtons Theory of Gravity

  1. May 6, 2007 #1
    1. The problem statement, all variables and given/known data
    An astronaut of mass(m) is a certain distance between two planets where his net force=0. What is the ratio of the masses of the planets?


    2. Relevant equations
    [tex] F=\frac{Gm_{1}m_{2}}{r^2}\vec{r}[/tex]


    3. The attempt at a solution
    Well if one planet has more mass than the other then the astronuat will have to compinsate for the pull from the biggr planet and be closer to the smaller planet.
    I just don't know how to represent it mathematically.
     
  2. jcsd
  3. May 6, 2007 #2

    hage567

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    Find where the vector sum of the gravitational force between Planet 1 and the astronaut and the gravitational force of Planet 2 and the astronaut is zero.
     
  4. May 6, 2007 #3
    How about this:
    Lets label the distance from the bigger planet to the Astronaut [tex]\vec{r_{1}}[/tex]
    and the distance from the astronuat to the smaller planet is [tex]\vec{r_{2}}[/tex] so that [tex]\vec{r}=\vec{r_{1}}+\vec{r_{2}}[/tex]
    so:[tex] \vec{r_{1}}\ge \vec{r_{2}}[/tex]
    Big planet at Equilibruim:
    [tex]F_{planet on the astronuat}=F_{astrouat on the planet}[/tex]
    Small Planet:
    [tex]F_{small planet on the astronuat}=F_{astrouat on the small planet}[/tex]
    So could we do :
    [tex]\frac{GM_{1}m{2}}{r_{1}^2}\vec{r_{1}}=\frac{Gm_{2}m{3}}{r_{2}^2}\vec{r_{2}}[/tex]
    Or something like that?
    Where M is the mass of the big planet, [tex]m_{2}[/tex] is the astronaut,
    [tex]m_{3}[/tex] is the smaller planet
     
    Last edited: May 6, 2007
  5. May 6, 2007 #4
  6. May 6, 2007 #5

    hage567

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    What can you do to simplify that?
     
  7. May 6, 2007 #6
    So we can kick out the G's, m2
    [tex] \frac{M_{1}}{r_{1}^2}\vec{r_{1}}=\frac{m_{3}}{r_{2}^2}\vec{r_{2}}[/tex]
    then:
    [tex] \frac{M_{1}}{m_{3}}=\frac{r_{1}^2}{r_{2}^2}[/tex]
    So the ratio doesn't of course depend on the wieght of the Astronaut.
    Sound about right?
     
  8. May 6, 2007 #7

    hage567

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    Looks OK to me.
     
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