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Gravitational force

  1. Aug 4, 2009 #1
    1. The problem statement, all variables and given/known data

    When the distance separating two masses, M and m, is 1.2 x 10^10 m, the gravitational force of attraction is 5.0 N. If the mass of M becomes 3M and the separation distance becomes
    2.4 x 1010 m, what will be the force?

    2. Relevant equations

    3. The attempt at a solution
    since i have r and Fg (first part of question) i tried solving for the masses. but how do i solve for both when i need at least one to solve the other/??
  2. jcsd
  3. Aug 4, 2009 #2


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    Gold Member

    You're missing the gravitational constant in your equation. I suggest setting up a ratio between the two forces and see what happens.
  4. Aug 4, 2009 #3
    huh? im still kinda confused..
  5. Aug 4, 2009 #4


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    Gold Member

    You know : [tex]F_1 = \frac{{GM_1 m_1 }}{{r_1 ^2 }}[/tex] and that [tex]F_2 = \frac{{GM_2 m_2 }}{{r_2 ^2 }}[/tex].

    What do you get when you divide the second equation by the first, that is: [tex]\frac{{F_2 }}{{F_1 }}[/tex].

    You know:

    M_2 = 3M_1 \\
    m_1 = m_2 \\

    along with the actual initial force, [tex]F_1[/tex]. You can solve for the ratio of forces which will tell you what [tex]F_2[/tex] is.
  6. Aug 5, 2009 #5
    or, use this EASIER way :

    for the first condition, and the second condition, the gravitational constant is always the same.

    You have known what the value of F is :
    [tex]F=\frac{GMm}{r^2 }[/tex]
    and, the value of G is :
    [tex]G=\frac{Fr^2 }{Mm}[/tex]

    you can make it like this :
    [tex]\frac{Fr1^2 }{M1m1}=\frac{Fr2^2 }{M2m2}[/tex]
    substitute the variable ( like M2 to 3 Mi ) and you can eliminate the variable. At last, there will be an equation like F2 = k.F1
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