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Coordinate transformation

  1. Nov 18, 2015 #1
    1. The problem statement, all variables and given/known data
    I'm trying to find the direction and magnitude of Earth's gravity on some projectile. The question states that I can ignore z, and that the origins of the x and y axes should be on the surface of the planet. I should then use Newton's law of Gravity to find the direction and magnitude of the acceleration due to gravity, but Newton's law assumes the origin at the centre of the planet.

    2. Relevant equations


    3. The attempt at a solution
    ##g = \frac{-GM\overrightarrow{r}}{r^3}##
    I'm splitting g into x and y components, I think that makes it easier. So for the y component, ##\overrightarrow{r}## is in the -y direction, so that's fine. And to find r, just add earth's radius onto the y-value, so

    ##g_{y}= \frac{-GM}{(radius+y)^3}## and I hope setting this equal to the y component means I don't need to multiply by ##\overrightarrow{y}##.

    The x bit I'm finding a bit harder. The distance r in Newton's law is going to be ##\sqrt{x^2+radius^2}##. The direction ##\overrightarrow{r}## I'm finding it hard to get my head round, because I think it's going to involve bits that point in the x direction and bits in y!
    So I think that rcos(theta) is the bit in the y direction and rsin(theta) the bit in x. So the bit that points in the y direction, do I add that to my already calculated y component then?
     
  2. jcsd
  3. Nov 18, 2015 #2

    SteamKing

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    For altitudes where the projectile remains relatively close to the ground, the value of g calculated at the surface can be used to determine the motion of the projectile with good accuracy. If you are launching a ballistic missile which travels halfway around the world, then things get more complicated.

    The direction of acceleration due to gravity is always between the object being considered and the center of the earth, i.e., in other words, gravity acts straight down, in a single direction only.

    To calculate the magnitude of the acceleration due to gravity, the distance between an object and the center of the earth is used in the formula

    F = GMm/r2

    because all of the mass of the earth can be considered to act at the center of the planet.

    You should not necessarily assume that the coordinate system convenient for your projectile motion must have its origin at the center of the earth, unless you are trying to analyze orbital or other long range ballistic trajectories.

    It's not clear what you are doing here. g always has only one component: from the object to the center of the earth.

    It's not clear what the purpose of this exercise is. If you want to calculate the trajectories of projectiles subject to certain limitations, then using plain old g is just as good as anything else you will derive (not to mention much simpler).

    If you want to account for the earth revolving about its axis while the projectile is in flight and other things, the physics becomes much more difficult.
     
  4. Nov 18, 2015 #3
    Neither clear nor concise really. Essentially, I want to find ##\overrightarrow{r}##, as a single vector, but I'm having trouble because of the different co-ordinate origins. And I definitely don't want to account for earth's rotation. As you probably guessed, this is still me trying to get a handle on the coding problem I've been set.

    I have to use axes on the earth's surface, I've been told that. And Newton's law assumes the origin at the centre of the planet. So I have to find what values my x and y, which will be measured on the axes at the surface, have in a frame at earth's centre.

    Gravity does always point to the centre of the earth, but it's expressing that direction in my co-ordinate system that I'm finding tricky.
     
  5. Nov 18, 2015 #4

    SteamKing

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    What I'm saying is that you are taking a simple problem and turning it into one which is needlessly more complex.

    The coordinate system which is used to calculate a value for g need not be the same coordinate system which is used in the projectile problem. Once a value for g is determined, the coordinate system for that calculation can be discarded in favor of one which is more convenient for analyzing projectile motion.
     
  6. Nov 18, 2015 #5
    Just to clarify, when I said 'neither clear nor concise' I was referring to MY original post! :) Sorry, thought I'd better make that clearer, just in case.

    If I calculate g in one co-ordinate system, then its direction will have been fixed in the co-ordinate system the calculation was made in, won't it? I can't just switch, and use the
    ##\overrightarrow{r}## from one frame in another frame. I thought that the vectors describing the direction would be different for both choices of origin.
     
  7. Nov 18, 2015 #6

    SteamKing

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    No, you are conflating how the distance r is calculated for use in Newton's formula for g with the choice of coordinate system used for the projectile calculations. The vectors which you use for projectile motion calculations don't necessarily have to share the same coordinate system or origin as that used to determine the value of g.

    g does not always need to be used as a vector quantity; it can be used as a scalar which represents the acceleration due to gravitation attraction between the earth and some other object. The direction of attraction is understood to be always between the center of the earth and the center of mass of this other object; for us who live on the surface of the earth, this direction is called 'down' colloquially, if imprecisely.

    To see examples of how g is used for many projectile motion problems, take a cruise through the Intro Physics HW forum here at PF.
     
  8. Nov 18, 2015 #7
    OK, thanks!
     
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