Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Gravitational attraction force b/w point mass and rod

  1. Dec 29, 2011 #1
    I was wondering about a scenario where we have a unit mass M1 and a thin rod of mass M2 of length L with a distance r between them attracting each other . Is my following approach to the problem correct ??!!

    let s = r + L
    I know that F= -(GM1M2)/(s^2) where G is gravitation constant

    therefore F= -GM1*∫1/(s^2).dM2
    I know that M2 = λ*L where λ is density
    therefore dM2 = λdL . from ds = dL
    I get dM2 = λdL
    Therefore finally , F= -GM1*∫1/(s^2)*λ.ds
    Integrating from r to r+L
    I get F= -(GM1M2)/(r*(r+L))

    Equation analysis - If i put limit L-->0 , i turn the rod to a point mass, therefore i get -GM1M2/(r^2) which is actually the gravitational attraction force for two masses :D

    Thanks in advance :)
     
    Last edited: Dec 30, 2011
  2. jcsd
  3. Dec 29, 2011 #2

    mathman

    User Avatar
    Science Advisor
    Gold Member

    You should describe the geometry of the situation. How is the rod oriented with respect to the point?
     
  4. Dec 29, 2011 #3
    Point mass --> . <------distance R------> _______________ <-- Rod M2 with length L
     
  5. Dec 30, 2011 #4
    Re: Repost : Gravitational attraction force b/w point mass and rod

    The force should be
    [tex] F = G\frac{M_1M_2}{(r+\frac{L}{2})^2} [/tex]
    So something went wrong there.
     
  6. Dec 30, 2011 #5
    I knew there was something wrong, maybe take the variable or the partition set i took for integrating.May you show me how you did it ?? :) thanks
     
  7. Dec 30, 2011 #6
    Well the gravitational force between two points is [itex]F = G\frac{M_1M_2}{r^2}[/itex] where [itex]r[/itex] is the separation of the two points. Secondly in gravity the force between a point and an entire system is the same as the force between that point and the center of mass of that system (giving the center of mass the entire mass of the system), since the center of mass of a uniform rod is in the middle I added half the distance of the rod to the distance between your particle and the start of the rod to find the [itex]r[/itex] term.
     
  8. Dec 30, 2011 #7
    Oh i see , Center of mass does represent all the dm in the system so simply we can take distance as r+L/2, But if we were to proof it using the idea of F=-(GM1)/(s^2)*ΣdM2 thus F=-(GM1)/(s^2)∫1.dM2 ? How can it be done ?
     
  9. Dec 30, 2011 #8

    clem

    User Avatar
    Science Advisor

    Redoctober is right, and JHamm is wrong. You must integrate just as Red did.
    This is a standard problem in EM texts for the electric force on a charge.
    The texts all give Red's answer.
     
  10. Dec 30, 2011 #9
    Oh k . Thankyou so much for vertifying :D ! I am happy that my approach to the problem worked !!
     
  11. Dec 30, 2011 #10
    Sorry about that then, looks like I'll be eating foot tonight :s
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Gravitational attraction force b/w point mass and rod
Loading...