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Unit Normal Vector in Acceleration

  1. Sep 3, 2013 #1
    What is the Unit Normal Vector in Acceleration.[tex]\vec{u}_n[/tex] How would I go about calculating the normal. Wiki and Wolfram Alpha were very vague about calculating the normal of an orbit. I would like to know the normal to the orbit of a mass orbiting a much larger mass not counting any other gravitating masses in proximity.

    Magnitude[tex]v = \sqrt{vx^2 + vy^2 + vz^2}[/tex]
    Unit Tangent Vector[tex]\vec{u}_v =(\frac{vx}{v} ,\frac{vy}{v} ,\frac{vz}{v})[/tex]
    Time derivative of velocity[tex]\frac{dv}{dt} =(\frac{(\frac{d(vx^2)}{dt} + \frac{d(vy^2)}{dt} + \frac{d(vz^2)}{dt})}{v})[/tex]
    Acceleration[tex]\vec{a} = \frac{dv}{dt}*\vec{u}_v + \frac{v^2}{r}*\vec{u}_n[/tex]
    How do you calculate[tex]\vec{u}_n[/tex]
     
  2. jcsd
  3. Sep 3, 2013 #2
    [itex]\vec{u_n}[/itex] is equal to [itex]\frac{d\vec{u_v}}{dt}[/itex] divided by the magnitude of [itex]\frac{d\vec{u_v}}{dt}[/itex].
     
  4. Sep 3, 2013 #3

    ZapperZ

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    If this is an orbit in a plane (which most orbits and conic section motion are), why aren't making your life easier by doing this in plane polar coordinates?

    Zz.
     
  5. Sep 3, 2013 #4
    Now to acceleration, is the equation correct. My data is not coming out right.[tex]\vec{a} = \frac{dv}{dt} * \vec{u}_v + \frac{v^2}{r} * \frac{\frac{d(\vec{v})}{dt}}{|\frac{d(\vec{v})}{dt}|}[/tex]I am interested in the centripetal acceleration!
     
    Last edited: Sep 3, 2013
  6. Sep 3, 2013 #5

    D H

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    That is not correct. You are writing velocity as a scalar times the unit vector in the direction of the velocity vector, ##\vec v = v\hat u_v##. Here ##\hat u_v \equiv \vec v / ||\vec v||##. Differentiating with respect to time yields ##\vec a = \dot v \vec u _v + v\dot{\hat u}_v##. Your second term is not the same as ##v\dot{\hat u}_v##.


    Why? It's pretty much meaningless. You want the central force acceleration, and that's given by Newton's law of gravitation. You can compute centripetal acceleration by subtracting the tangential component (##v\hat u_v##) from the gravitational acceleration, but why bother?
     
  7. Sep 4, 2013 #6
    I suggested the same thing on his/her last two or three posts, for some reason, he/she refuses to listen.

    Let me try again, maybe he/she will hear me this time.

    USE POLAR COORDINATES, THEY WERE CREATED EXACTLY FOR PROBLEMS LIKE THIS
     
  8. Sep 4, 2013 #7
    There is a much easier way of getting [itex]\vec{u_n}[/itex]. If [itex]\vec{a}[/itex] is the acceleration vector, then the component of [itex]\vec{a}[/itex] in the direction of the velocity vector is [itex]\vec{a}\centerdot \vec{u_v}[/itex]. So the vector component of the acceleration vector in the direction perpendicular to the velocity vector is [itex]\vec{a}-(\vec{a}\centerdot \vec{u_v})\vec{u_v}[/itex]. The unit normal vector [itex]\vec{u_n}[/itex] is equal to this vector divided by its own magnitude:
    [tex]\vec{u_n}=\frac{\vec{a}-(\vec{a}\centerdot \vec{u_v})\vec{u_v}}{\vert\vec{a}-(\vec{a}\centerdot \vec{u_v})\vec{u_v}\vert}[/tex]
    Now, the denominator is given by
    [tex]\vert\vec{a}-(\vec{a}\centerdot \vec{u_v})\vec{u_v}\vert=\sqrt{a^2-(\vec{a}\centerdot \vec{u_v})^2}[/tex]
    So,
    [tex]\vec{u_n}=\frac{\vec{a}-(\vec{a}\centerdot \vec{u_v})\vec{u_v}}{\sqrt{a^2-(\vec{a}\centerdot \vec{u_v})^2}}[/tex]
    Finally, since [itex]\vec{u_v}=\frac{\vec{v}}{v}[/itex]
    [tex]\vec{u_n}=\frac{v^2\vec{a}-(\vec{a}\centerdot \vec{v})\vec{v}}{v\sqrt{(av)^2-(\vec{a}\centerdot \vec{v})^2}}[/tex]
    This provides a simple expression for getting the unit normal to the trajectory even when one is using Cartesian coordinates.
     
    Last edited: Sep 4, 2013
  9. Sep 4, 2013 #8

    D H

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    Even easier, use the triple product ##\frac {\vec v \times \vec a \times \vec v}{v^2}##.

    However, this vector has zero physical meaning. So why bother?
     
  10. Sep 4, 2013 #9
    I like your thinking. Very clever and very cute. Cudos!

    There is a bit of confusion, however. Your final result is not the unit normal. The denominator is not equal to the magnitude of the vector in the numerator. I don't know whether you intended this or not. Actually, the denominator in my expression is equal to the magnitude of the vector in the numerator of your expression. That must mean that our two numerators must be mathematically equivalent. I have not tried to verify this, but it should be possible by expressing the results in component form. But, there must be a more elegant way than this.

    I agree that the unit normal does not have much physical meaning. What really is important is the curvature vector:

    [tex]\frac{d\vec{u_v}}{ds}=\frac{\vec{u_n}}{r}[/tex]
    where r is the local radius of curvature of the trajectory and ds is differential position along the trajectory: ds=vdt.
     
  11. Sep 4, 2013 #10

    D H

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    You're right. My expression is the component of acceleration that is normal to the velocity vector. Easy enough to normalize, for all that's worth.

    Even the curvature isn't all that useful for orbits. At least I've never found it to be.
     
  12. Sep 4, 2013 #11

    jhae2.718

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    For an orbit, as others have said, the easiest thing to use is polar coordinates. First construct a reference frame fixed to the large body that you'll use to approximate an inertial frame, and then construct a rotating reference frame pointing toward the orbiting satellite. Suppose the inertial frame is defined by the three orthonormal vectors [itex]\hat{\boldsymbol \imath}_1[/itex], [itex]\hat{\boldsymbol \imath}_2[/itex], and [itex]\hat{\boldsymbol \imath}_3[/itex], and that the rotating frame is defined by [itex]\hat{\boldsymbol e}_1[/itex], [itex]\hat{\boldsymbol e}_2[/itex], and [itex]\hat{\boldsymbol e}_3[/itex].

    If we say that [itex]\hat{\boldsymbol e}_1[/itex] points to the mass center of the satellite, we can define the inertial position vector as [tex]{\boldsymbol p} = r \hat{\boldsymbol e}_1.[/tex] Now, to make calculating the inertial velocity and acceleration in the [itex]{\boldsymbol e}^+[/itex] frame, we need a result called the kinematic transport theorem. Consider two reference frames [itex]{\boldsymbol a}^+[/itex] and [itex]{\boldsymbol b}^+[/itex], where the [itex]{\boldsymbol b}^+[/itex] frame is rotating with respect to the [itex]{\boldsymbol a}^+[/itex] frame with angular velocity [itex]{\boldsymbol \omega}_\text{a/b}[/itex]. The derivative of a vector [itex]{\boldsymbol p}[/itex] in the [itex]{\boldsymbol a}^+[/itex] frame, coordinatized in the [itex]{\boldsymbol b}^+[/itex] frame, comes from two components:
    1. The rate of change of [itex]{\boldsymbol p}[/itex] in the [itex]{\boldsymbol b}^+[/itex] frame
    2. The angular velocity of [itex]{\boldsymbol b}^+[/itex] relative to [itex]{\boldsymbol a}^+[/itex]
    This gives us the kinematic transport theorem, which states that [tex]\frac{{}^a\text{d}}{\text{d}t}\left({\boldsymbol p}\right) = \frac{{}^b\text{d}}{\text{d}t}\left({\boldsymbol p}\right) + {\boldsymbol \omega}_\text{a/b} \times {\boldsymbol p}.[/tex] This is an incredibly powerful result, as it lets us easily use as many reference frames as we want while still being able to find how observers in different frames see vectors changing.

    Why do we want to do this? The ultimate goal is to find the inertial acceleration. Recall that it is an axiom of Newtonian mechanics that inertial reference frames exist and that Newton's laws hold in inertial reference frames. The KTT makes it almost trivial to use non-inertial frames and still find the inertial velocity and acceleration.

    Let's apply the KTT to find the inertial velocity of our [itex]{\boldsymbol p}[/itex] vector. Let's assume we are in the plane spanned by [itex]\hat{\boldsymbol e}_1[/itex], [itex]\hat{\boldsymbol e}_2[/itex] and that [itex]\hat{\boldsymbol \imath}_3 = \hat{\boldsymbol e}_3[/itex]. (That is, we are considering planar motion, which we can do since orbits are planar.) Suppose the angular velocity vector is [itex]{\boldsymbol \omega}_\text{e/i} = \dot{\theta}\hat{\boldsymbol e}_3 [/itex]
    [tex]
    \begin{align*}
    {\boldsymbol v} = \dot{\boldsymbol p} &= \dot{r} \hat{\boldsymbol e}_1 + \dot{\theta}\hat{\boldsymbol e}_3 \times r \hat{\boldsymbol e}_1 \\
    {\boldsymbol v} &= \dot{r} \hat{\boldsymbol e}_1 + r\dot{\theta}\hat{\boldsymbol e}_2
    \end{align*}[/tex]
    If we repeat this to find the inertial acceleration, we find that
    [tex]{\boldsymbol a} = \dot{\boldsymbol v} =\ddot{\boldsymbol p} = \left(\ddot{r} - r\dot{\theta}^2\right) \hat{\boldsymbol e}_1 + \left(r\ddot{\theta} + 2\dot{r}\dot{\theta}\right)\hat{\boldsymbol e}_2[/tex](Verifying this expression is left as an exercise for the reader.)

    Now, we can apply Newton's second law, [itex]{\boldsymbol F} = m{\boldsymbol a}[/itex]. Using Newton's law of gravitation, we find that
    [tex]{\boldsymbol F} = -\frac{GMm}{r^2}\hat{\boldsymbol e}_1 = -\frac{\mu m}{r^2}\hat{\boldsymbol e}_1[/tex]where [itex]\mu=GM[/itex] is the gravitational parameter. This results in the equations of motion for an orbiting satellite[1],[tex]
    \begin{align*}
    \ddot{r} - r\dot{\theta}^2 &= -\frac{\mu}{r^2} \\
    r\ddot{\theta} + 2\dot{r}\dot{\theta} &= 0
    \end{align*}
    [/tex]

    [1] More specifically, this is the position of the satellite's mass center. For a rigid body, you can attach a third reference frame to the body and use various attitude coordinatizations to fully define the pos and attitude of the satellite.
     
    Last edited: Sep 4, 2013
  13. Sep 5, 2013 #12

    UltrafastPED

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  14. Sep 5, 2013 #13
    I have had some luck with adding a simple Centripetal Acceleration to the mix:[tex]a_{centripetal} = \frac{v^2}{r}[/tex]Making the Acceleration equations:


    [tex]a_x = \frac{d||v||}{dt} * uv_x + ||v|| * \frac{d(uv_x)}{dt} + \frac{vx ^ 2 }{ r}[/tex]
    [tex]a_y = \frac{d||v||}{dt} * uv_y + ||v|| * \frac{d(uv_y)}{dt} + \frac{vy ^ 2 }{ r}[/tex]
    [tex]a_z = \frac{d||v||}{dt} * uv_z + ||v|| * \frac{d(uv_z)}{dt} + \frac{vz ^ 2}{ r}[/tex]

    It does not sit well with me that the beautiful derivative of time of the magnitude yields velocity then the derivative of velocity does not yield the acceleration and needs an additional centripetal factor for the total acceleration.
     
  15. Sep 5, 2013 #14

    D H

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    You've ignored the motion of the primary, jhae. Working in a frame with a fixed central body is a non-inertial (accelerating) frame. You need to account for this in your equations of motion. Do that and you'll find that the gravitational parameter μ becomes G(M+m) rather than just GM.


    The problem with those formulae is that the fundamental theorem of space curves has little applicability to orbits. I've found but one paper (and it's a technical report rather than a journal paper) that attempts to use the Frenet-Serret frame to describe orbits, and that one paper has zero citations. That paper: Livieratos, Evangelos. "Intrinsic Parameters and Satellite Orbital Elements." ifp (2003): 295.

    Even the plethora of mechanisms used to describe orbits in the two body problem become a lot less useful as soon as one goes beyond a pair of point masses (e.g., the N body problem, non-spherical gravity, atmospheric perturbations, etc.). One has little choice but to revert to numerical integration.
     
  16. Sep 5, 2013 #15

    UltrafastPED

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    Frenet-Serret is not for finding the orbits; you use it after you have the path. The original question was how to find the unit normal vector in acceleration.
     
  17. Sep 5, 2013 #16

    D H

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    If this r is the distance to the primary, that is not the centripetal acceleration except in the case of a circular orbit. For any other type of orbit, the center of curvature is not located at the central body. Gravitation is a central force, not a centripetal force.

    Rather than trying to reinvent a 400 year old wheel, I strongly suggest you go to a library. Pick up or buy a copy of "Fundamentals of Astrodynamics and Applications" by David Vallado or "Fundamentals of Astrodynamics" by Bate, Mueller, and White. Vallado is very extensive but can be ridiculously expensive. It costs $150 for a new third edition in paperback. A used version of the older second edition is much more affordable. Bate, Mueller, and White is cheap but it's much less extensive in coverage.
     
  18. Sep 5, 2013 #17

    D H

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    There is a problem here: Why?

    This thread is a prototypical example of an X-Y problem. In an X-Y problem, someone tries to solve problem X, and thinks approach Y would work. The person then starts running into problems with this approach Y and asks for help on this non-viable approach Y rather than on the original problem X.
     
  19. Sep 5, 2013 #18

    jhae2.718

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    You're correct, of course. For the proper two-body equations of motion I should have constructed a relative position vector [itex]{\boldsymbol r} = {\boldsymbol r}_2 - {\boldsymbol r}_1[/itex] and worked with that. I ignored the motion of the primary object (approximating it as a fixed point in space) and focused on the motion of the satellite in order to simplify things to try to show the method more clearly. Perhaps that was a mistake to do so (and I certainly should have stated those assumptions in my earlier post!).
     
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