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Calculating direction of pull on Earth

  1. Apr 22, 2007 #1
    Hi there, I have been reading a book in which the author gives me a result of his calculation and I am trying to work it backwards to prove whether he is correct. On July 20, 2005 he states that the combined forces pulling on Earth was 285.8 degrees. He mentions the gravitational formula G*m1*m2/R^2, but instead of squaring he cubes the distance. Its refered to as Resultant Direction or Net Tidal Force. He does this for each planet Sun, Moon out to Saturn only. Can somebody please tell me the steps he took to achieve this result? I think it has something to do with summing the vectors. Any help would be appreciated.

    Kindest regards,
    Martin.
     
  2. jcsd
  3. Apr 22, 2007 #2
    I don't understand what means this angle. What is the reference?
    There are two main forces on earth: the sun, of course, and the moon. Other objects contribute in a lesser degree.
     
  4. Apr 22, 2007 #3

    D H

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    Regarding the use of the cube: Tidal forces result from the spacial gradient of the gravitational force. To first order, the tidal force varies as the inverse cube of distance rather than the inverse square.

    Regarding the answer being expressed in degrees: I haven't the foggiest.
     
  5. Apr 22, 2007 #4
    Thanks, but since I am trying to calculate the direction in which the earth is being pulled, I think I need to take the individual planets into account, calculate them individually and Net the result of all planets out to Saturn. I realize the effects of the Sun and Moon are major, but it will be a computer doing the work. Surely I should be able to break the direction down into individual components and sum them, right? Could somebody help with this? I think it has to do with vectors.

    Warmest regards,
    Martin.
     
  6. Apr 22, 2007 #5

    D H

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    I see. You are trying to compute the net gravitational acceleration of the Earth due to the Sun, Moon, and major planets. As you noted in post #1,
    the magnitude of the gravitational force of one object on another is

    [tex]||\vec F_{1,2}|| = \frac {G m_1 m_2}{||\vec r_{1\to2}||^2}[/tex]

    where [itex]\vec r_{1,\to2}[/itex] is the vector from body 1 to body 2 and [itex]\vec F_{1,2}[/itex] is the force exerted on body 1 by body 2.

    The direction of the force is toward body 2. Given any non-zero vector [itex]\vec u[/tex], the unit vector pointing along [itex]\vec u[/tex] is [itex]\vec u/||\vec u||[/tex]. Thus the gravitational force vector is

    [tex]\vec F_{1,2} = -\,\frac {G m_1 m_2}{||\vec r_{1\to2}||^2}\,\frac {\vec r_{1\to2}}{||\vec r_{1\to2}||}[/tex]

    or more simply,

    [tex]\vec F_{1,2} = -\,\frac {G m_1 m_2}{||\vec r_{1\to2}||^3}\,\vec r_{1\to2}[/tex]
     
  7. Apr 22, 2007 #6

    russ_watters

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    What is the reference axis for the angle? The calculations are a lot easier if you use the sun as zero...
     
  8. Apr 22, 2007 #7
    This is what I have: The way to determine it by using vectors of force. Plot out all of the forces on Earth, then break those forces down to their component parts, add the components up, then recombine to get a new vector that shows you the net result of all major forces. If we have a square being pulled in different directions on 3 corners A, B, C we can figure out by breaking each force down to their component parts, which tells us how much each force is pulling horizontally and vertically. So if we have 3 forces pulling up and 2 pushing down, this would result in 1 downward force and if we had 2 pulling left and 1 pulling right we would 1 force pulling left. Our result would be 2 down and 1 left. Each planet does exert a force on Earth, with the Sun being the major component, however I do need to include the Sun, Moon out to Saturn. Add all these up for a net result in a single direction.
    We can measure the force of an individual planet using Newtons law. So if m1 and m2 represent the masses of 2 objects, and r represents the distance between then, then the force can be expressed as (G*m1*m2)/R^2, where G is the universal constant.
    My tide generating force is similiar to the above, except I want to cube the distance. This equation provides me a way to measure the tidal force of every body in the solar system. As we know where these bodies are located in relation to Earth, and we know their masses, we can go through a similair exercise as my example and calculate the direction that all planets combined are pulling us.
    The Sun and the Moon are the major components of the formula and govern the location of this direction.

    Kindest regards to all who can shed some light on the above,
    Martin.
    PS On July 20, 2005 the result was that the direction was 285.8 degrees on the compass of 360 degrees. I need to verify this result by working backwards.
     
  9. Apr 22, 2007 #8

    D H

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    martindelica,
    Post #5 tells you how to calculate the force as a vector.

    Some questions:
    The gravity gradient terms (tidal forces) acts as a torque on the Earth. If you are treating the Earth as a point mass you don't need to worry about tidal effects.

    Firstly, what is your reference frame? Secondly, this is inherently a three dimensional problem. Stating the answer as one direction (rather than two) implies you are treating it as a two dimensional problem.
     
  10. Apr 23, 2007 #9
    Perhaps I need some time to think it through, given the above example , result and formula. This is not my formula or one from a testbook. It is merely a theory that I wish to prove. Yes this author does use some unconventional equations based on the ones found in a text book or have been proven from the 17th century. However this is what I have to work with. If it is not enough, Thanks for trying, however I still have to crack this theory. Thanks again.
     
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