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Find acceleration due to large planar ring

  1. Nov 1, 2013 #1
    1. The problem statement, all variables and given/known data
    Firstly, my beautiful picture:

    http://imgur.com/WZTl3rr

    A ship has encountered a massive planar ring in space. The ring has radius R=1km and mass M=1.0 * 1018. Calculate the force per unit mass on the Enterprise assuming the starship is located at a point on the plane of the ring 1.1 km from the center. If necessary calculate the integral by numerical analysis.


    2. Relevant equations

    a=g(x)=-∇[itex]\Phi[/itex](x)[itex]\Phi[/itex]'

    Density=λ=[itex]\frac{M}{2πR}[/itex]
    dm=λRd

    [itex]\Phi[/itex](x)=-G∫0 [itex]\frac{λRd\Phi'}{(x-x')}[/itex]

    3. The attempt at a solution

    Because the ship is on the plane we know that θ=90°, correct?

    The system is also symmetrical so [itex]\Phi[/itex]=0.

    [itex]\Phi[/itex](r,∅)=-GλR∫0 [itex]\frac{d\Phi'}{sqrt(r^2+R^2-2rRsin(θ)cos(\Phi'))}[/itex]

    The above equation is a bit nasty but it was what my prof. used in his example of basically the same problem.

    Here is where I am having issues. My prof. did a similar example where he assumed that the distance from the center of the ring was much larger than the radius of the ring which allowed him to simplify the integral and find an approximation using a taylor series expansion. In the case I am given the distance is just outside the radius of the ring itself and such an approximation would be inaccurate. The problem also explicitly allows me the use of numerical analysis but there is a problem.

    If I tell mathematica what each value is and then calculate the integral it spits out a number for the gravitational potential. Great, but what I need is acceleration which is the -∇ of that potential. Of course, the derivative of a scalar value is zero.

    Alternatively, if I leave r as a variable and take the integral I get a super nasty conditional statement as a function of r. Pressing forward I ask for the derivative of that function with respect to r then assign my value of 1100 to r and it spits out 25.7459. That should be the acceleration and the answer is reasonable but I am wondering two things:

    1) Am I interpreting the question correctly and taking the correct steps in solving this problem?

    2) What would be a more elegant way of doing this?

    Thank you for your help!
     
  2. jcsd
  3. Nov 1, 2013 #2

    TSny

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    If you are allowed to do the integrals numerically using Mathematica's NIntegrate, then you could just set up the integral for the force directly rather than for the potential.
     
  4. Nov 2, 2013 #3

    haruspex

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    You could differentiate the integral wrt r first.
     
  5. Nov 2, 2013 #4
    I'm not sure what the integral for the force would look like. Can you change the order in which you differentiate/integrate? i.e. take the integral of the derivative


    What do you mean?

    EDIT:

    I re-did my calculation from last night having learned a bit more about the software I was using and came up with ~35 m/s2 using the same method as I described.

    I then tried again by giving it the value for r and using NIntegrate[D[f[r],r],{phi,0,2pi}] and got the same result but it was much cleaner/more condensed. So, I not only tested NIntegrate but also changing the order of my integration/differentiation. Is that what you were implying?
     
    Last edited: Nov 2, 2013
  6. Nov 2, 2013 #5

    TSny

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    Yes, you can change the order of differentiation and integration. This is what haruspex is suggesting. If you take the negative of the derivative of the integrand with respect to r, it gives the component of force per unit mass along the r direction due to the mass element ##\lambda R d\theta##. Then when you integrate you get the total force in the r direction.
     
  7. Nov 2, 2013 #6
    Great, thank you!
     
  8. Nov 2, 2013 #7
    Would this be g(r) or would the final result be g(x)? I started using r when converting to polar coordinates from the x vector.
     
  9. Nov 2, 2013 #8

    TSny

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    I think g(r) and g(x) are the same. The ship lies in the plane of the loop and I think you set up your integral such that r is the distance of the ship from the center of the loop along the x-axis.
     
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