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!

Simple integrals for gravitational potential

  1. Jan 7, 2013 #1
    1. The problem statement, all variables and given/known data
    2. Relevant equations

    I need help solving intergral…
    [tex]\int \frac{dx}{(a+x)^2}[/tex]

    3. The attempt at a solution
    I found the integral for…
    [tex] \int \frac{dx}{(a^2+x^2)} [/tex] = 1/a arctan x/a

    But I don’t know how to apply that to the original integral which is a little different

    [tex]\int \frac{dx}{(a+x)^2} = \int \frac{dx}{(a^2+x^2+2ax)}[/tex]

    I also need to solve the following integral
    [tex]\int \frac{dx}{(a+b-x)^2}[/tex]


    It’s not homework. The reason is that I want to work through the numbers that Rybczyk gives as equation 1 for gravitational potential in his paper “Gravitational Effect on Light Propagation”

    http://www.mrelativity.net/Gravitat...ravitational Effects on Light Propagation.htm

    I have simplified his equation in my post somewhat, as I already know how to separate the two terms separated by the minus sign, and assuming the gravitational constant G and the mass of the bodies are constant, I know that they can come out in front of the integral sign.

    Also, I changed the variable names to more familiar ones. I hope the variable name substitutions helps rather than hinders. If not I'll have to rewrite this whole post using Rybczyk's exact variables..

    Thanks.
     
  2. jcsd
  3. Jan 7, 2013 #2

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    For the initial integral, use substitution, letting u = x+a .
     
  4. Jan 7, 2013 #3
    Thanks Sammy. It looks to me like du = dx in this case. Is that right?

    [tex]\int \frac{dx}{(a+x)^2} =[/tex]
    [tex]\int {(a+x)^{-2}}{dx}[/tex]

    u = a + x
    du = dx

    [tex]\int {(u)^{-2}}{du}=[/tex]
    [tex] -\frac{1}{u} + c[/tex]
    [edit] made correction (forgot the + c)

    Substituting a + x for u gives

    [tex] -\frac{1}{a+x} + c[/tex]

    Is that right?
     
    Last edited: Jan 7, 2013
  5. Jan 7, 2013 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Yes, it's right.
     
  6. Jan 7, 2013 #5
    Great, thanks Dick (quick reply!).

    Will try same for the second part of original question.

    NOTE: I editied post #3 to make a correction for the missing ( + c) for a constant.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Simple integrals for gravitational potential
  1. Simple integration (Replies: 2)

  2. Simple Integral (Replies: 3)

  3. Simple Integral (Replies: 10)

  4. Simple integral (Replies: 7)

  5. Simple Integral (Replies: 8)

Loading...