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

How is this a good approximation?

  1. Nov 9, 2015 #1
    Hello,

    I don't understand the following.

    I have this function: [tex]V(x,y,z)=\frac{(A+B+C)r^2-3(Ax^2+By^2+Cz^2)}{r^5}[/tex] with [itex]r=\sqrt{x^2+y^2+z^2}[/itex]
    and on the textbook they say that if x,y,z are approximately equal or comparable as order of magnitude to r, and if they are all "big" enough (they are indeed tens of thousands od kilometers), one can approximate the partial derivatives of V with respect to x, y and z, with the partial derivative of the numerator alone. I can compute both derivatives but still I don't see them as approximatively equal. Should i try with spherical coordinates? However it should be clear in cartesian coordinates too, I think...

    Could someone please show me this fact? It is a pretty important point in the chapter and everybody just says this is true because we are talking about orders of magnitude...

    Thank you in advance
     
  2. jcsd
  3. Nov 9, 2015 #2

    Mark44

    Staff: Mentor

    What do you get for the partial derivative with respect to x, of the numerator, and of the function itself? Don't forget that r is a function of x, y, and z.

    If x, y, and z are approximately equal, how can you simplify ##\sqrt{x^2 + y^2 + z^2}##?
     
  4. Nov 9, 2015 #3
    So, doing this with regard to x, I get, if I'm copying from the last calculations...
    For the complete function:
    [tex]\frac{\partial V(x,y,z)}{\partial x}=-\frac{3(A+B+C)x}{2r^5}-\frac{3Ax}{r^5}+\frac{15Ax^3}{2r^7}+\frac{15(By^2+Cz^2)x}{2r^7}[/tex]
    Now:
    1) if [itex]r\approx x[/itex] then [itex]x^3/r^7\approx x/r^5[/itex]
    2) if I consider x,y,z as equal and positive as they are, then [itex]r=x\sqrt{3}[/itex]

    In case 1), by substituting the approximation into the equation above, I can sum the first 3 terms (in A) and I don't get the same as deriving the numerator alone. If then I consider y,z to be much smaller than x and r, then the fourth term is negligible and I don't move on. If instead they are comparable to x and r, I can substitute x to them and get the contribution of the third term also for B and C with the same reasoning used for the terms in A.

    In case 2) the third term and fourth term (with y and z sustituted with x) become as here: [itex]\frac{15x^3(A+B+C)}{2x^73^3\sqrt{3}}=\frac{5x^3(A+B+C)}{18x^7\sqrt{3}}[/itex] which is smaller than, e.g (not showing x for clarity), [itex]1/3\sqrt{3}[/itex]. This is a small contribution to the first two terms that have coefficients of 9/2 for A and 3/2 for B and C and is nowhere close to the expected result of the textbook that is:
    [tex]\frac{\partial V(x,y,z)}{\partial x}=\frac{3(-2A+B+C)x}{2r^5}+...[/tex]

    (notice that the coefficient for A is -3 while I get 9/2)
    (This is the derivative of the numerator alone - according to the textbook-, in my own calculations there should be a minus in front of it)

    Having said this: I hope I did not miscalculate anything, I checked several times (even if now I'm just reading my sheets) and... Should I have first substituted the approximation 2) and then computed the derivative and only in the end go back to r at the denominator? Why should this be more correct than doing the opposite?

    Thanks
     
  5. Nov 9, 2015 #4

    Mark44

    Staff: Mentor

    What you said in a previous post was "on the textbook they say that if x,y,z are approximately equal..."
    From this, I get ##r = x\sqrt{3}##, as you do in your case 2.

    I don't follow what you're doing here.
    Assuming that x, y, and z are all approximately equal, we get ##r^2 = 3x^2##.
    So the numerator of V is ##(A + B + C) \cdot 3x^2 - 3(Ax^2 + Bx^2 + Cx^2)##. What does this simplify to?
    Part of what you had in post 1 I didn't quote. The full quote is
    This is not very clear, and might be a result of translation to English from another language. I don't think this is saying that x is approximately equal to r, but only that x, y, and z are approximately equal, and therefore of the same order of magnitude of r.
     
  6. Nov 9, 2015 #5
    Uhm, I don't really understand what you mean about quotes, anyways I think they are saying that [itex]x\approx y \approx z[/itex] and that, on top of that, they are also of the same order of magnitude of r and therefore we can freely exchange x to r.
    If I simplify this I get zero, which is even worse than what I got before but it is a good answer to my last questions... Where would you go from here?
     
  7. Nov 9, 2015 #6

    Mark44

    Staff: Mentor

    I agree with ##x\approx y \approx z##, but I don't believe that the part about "order of magnitude" means that we can exchange x with r. If ##r \approx x\sqrt{3}##, then r and x are of the same order of magnitude, but they're not equal, so you can't exchange r and x.
    If x, y, and z are approximately equal, then ##V_x##, ##V_y##, and ##V_z## are all close to 0. I believe that's what they're getting at here.
     
  8. Nov 9, 2015 #7
    Yes it could be as you say that I can't substitute x to r.

    However I fear this is not where they are getting at: this is a part of a potential that is used in the description of the gravitational interaction between a spherical object and a much lighter irregular satellite, of which A, B, C are the moments of inertia along the x,y,z directions. It is called the McCullogh's Potential even though I couldn't find it anywhere. The very fact that, by deriving only the numerator, one gets [tex]\frac{\partial V(x,y,z)}{\partial x}=\frac{3(-2A+B+C)x}{2r^5}+...[/tex] (and similar for y and z) shows that for a spherical (and I add: uniformly dense) object (A=B=C) this contribution is zero and so you only have the classical gravitational potential. I agree that this correction should be small, close to zero in some cases
    and still it should depend on the structure and geometry of the object, not on the position (x,y,z) of the center of mass of the satellite in an inertial reference frame. more over,if I just change the reference frame with a rotation or two, I can get [itex]r=x[/itex] and both y and z equal to zero. (same applies for any of the given coordinates) and this is exactly my case 1)
    , but I need to put it in this exact form, don't I?
     
  9. Nov 9, 2015 #8

    Mark44

    Staff: Mentor

    I don't think so (underline added). In your first post, it says:
     
  10. Nov 9, 2015 #9
    Yes but if it is an approximation I'd like it to be the same to the degree desired. In this case, I want the numerator approximation and the complete derivative to be equal at least to the term in (x/r^5), and the differences in the higher magnitude terms... Which is not the case.. and even with the case 1) assumptions in order to transform the (x^3/r^7) terms of the complete derivative in terms of (x/r^5) in order to get the numerator ornly derivative I fail...
     
  11. Nov 9, 2015 #10

    Mark44

    Staff: Mentor

    If ##x \approx y \approx z##, then ##V \approx \frac{(A + B + C)r^2 - 3(Ax^2 + Bx^2 + Cx^2)}{r^5} = \frac{(A + B + C)r^2 - 3(A + B + C)x^2}{r^5} = \frac{(A + B + C)r^2 - 3(A + B + C)r^2/3}{r^5} = \frac 0 {r^2} = 0##
    Compare that with what you got for the approximation to ##\frac{\partial V}{\partial x}##
     
  12. Nov 9, 2015 #11
    What is it that I don't see here?
    The potential is very small, and that's also physically ok, but what's the point of comparing it with its variation due to x? This derivative is not even close to zero...
     
  13. Nov 9, 2015 #12

    Mark44

    Staff: Mentor

    What they're saying in the problem, as I read it, is that the approximation to ##\frac{\partial V}{\partial x}## (and ##\frac{\partial V}{\partial y}## and ##\frac{\partial V}{\partial z}##) is very close to the approximation of the partial of the numerator alone.

    I think you're getting hung up the exact partial derivatives, which doesn't seem to me to be what they're asking you to do.
     
  14. Nov 11, 2015 #13
    Yeah I understand now.
    I'll try to plot this in Mathematica to convice myself that at some point they are close enough.
    Thank you!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: How is this a good approximation?
  1. How Good Am I? (Replies: 271)

Loading...