1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Astronomy Que

  1. Mar 16, 2008 #1
    Astronomy Que!!!!!

    1. The problem statement, all variables and given/known data
    Three stars lie on a line (1,2,3). The distance from star 1 to star 3 is labelled as D. If star 1 is four times that of star 3, and seven times of star 2, what is the distance between star 1 and 2 if the net force of star 2 is zero?

    2. Relevant equations

    3. The attempt at a solution
    Since the net force for star 2 is zero, I know that the force between star 1 and 2 is the same as the force between start 2 and 3


    I tried to sub in values to find the force between star 1 and 3


    but i'm not sure where to go from here...

    Thank you
  2. jcsd
  3. Mar 16, 2008 #2

    So then, the distance between star 1 and star 2 is x, and the distance between star 2 and star 3 is D-x. Can you go on from here?
  4. Mar 16, 2008 #3
    errr..I think so.
    Here's what I've 'tried' so far


    Cross multiply:
    G's cancel, so do m2


    if 4m3=7m2
    then m3=(7/4)m2


    now to find for x?



    not sure if it's the right way, or if i'm just messing around with variables, or if the force between 1 and 3 has any use (yikes!)
    Anyone willing to verify?

    but yeeeah...times like these when I wonder why I took physics as an option when I'm planning to do arts *sighs*
    thanks for your help!
    Last edited by a moderator: Mar 16, 2008
  5. Mar 16, 2008 #4
    1. Where did that G/4 come from?
    2. [tex]F \propto \frac{1}{r^2}[/tex]
  6. Mar 16, 2008 #5
    whoops, the G's cancel out too

    I got the 1/4 from dividing G7m2 by G(7/4)m2

    to get x=D/5
  7. Mar 16, 2008 #6
    Does [tex]\frac{7m_2}{m_3}=\frac{1}{4}[/tex] when [tex]m_3=\frac{7m_2}{4}[/tex]?

    You also haven't addressed my 2nd point.
  8. Mar 16, 2008 #7
    no, it is not 1/4...but 4 instead

    as for your second point: the inverse proportionality of distance to Force...
    the larger the distance, the smaller the force....
  9. Mar 16, 2008 #8
    But it's not distance, it's distance squared.
  10. Mar 16, 2008 #9
    so what you're saying is that I have to square my x and D-x values right?
  11. Mar 16, 2008 #10
    Well, if the formula is given by [tex]F=\frac{Gm_1m_2}{r^2}[/tex], should you?
  12. Mar 16, 2008 #11
    merde. i feel like I am going around in circles...

    ok, so if i squared it

    square rooting it, i get

    cross multiplying and expanding...
  13. Mar 16, 2008 #12
    Looks good to me.
  14. Mar 16, 2008 #13
    yay! thanks for your help and patience Snazzy!
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Astronomy Que
  1. Momentum Ques. (Replies: 3)

  2. Que for physics optic (Replies: 1)