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Math Help - n-body vector addition

  1. #1
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    n-body vector addition

    Hi,

    The Physics Help Forum not working today, at least not from my ISP, so this goes here. It's basically a math deal anyway:

    The formula to calculate the force of a point of mass, let's call them planets, that results from its being gravitationally attracted by another point of mass is Newton's:

    F=\frac{Gm_1m_2}{d^2}

    Where F is the force of the planet that results from the gravitational attraction exerted upon it by the other planet, G is Newton's gravity constant, m_1 and m_2 are the respecive masses of the planets, and d is the distance between them.

    For simplicity's sake let's say all the planets considered are of the same mass, so we can write m^2 instead of m_1m_2.

    Now, if I'm not mistaken, the formula for calculating the force of a planet resulting from the gravitational attraction of more than two planets is:

    F_j=\sum_{j\neq k}^{n}\frac{Gm^2}{d_{jk}^2}

    Where F_j is the force on the jth planet resulting from the gravitational attraction of the other n-j planets, and d_{jk} is the distance between the jth planet and the kth planet.

    My question is "where is the vector addition?" That is, when considering the force on one planet that results from the gravitational attraction of many other planets, we have to take into account not only the distance of the other planets from planet j but also their (x,y,z) position with respect to it (right?).

    Take for example the simple case of three planets in the same plane. Planet j is at the origin. Planet k is one unit to the right of j on the x axis, while planet l is one unit up the y axis. If the masses all equal 1, then, by the formula above, the force on planet j would be:

    F_j=\frac{G(1^2)}{1^2}+\frac{G(1^2)}{1^2}=2G

    But F_j is a function of both the distance and the position, right? So we must consider not only the Gravitational Forces individually exerted upon j by k and l, but also the angle at which these forces are exerted. That is, we must add the vectors. To add vectors you just plug in the x value and y value sums of the added vectors into pythagoras' formula. The force on planet j should therefore be:

    F_j= \sqrt{\frac{G(1^2)(\cos\theta_k)}{1^2}+\frac{G(1^2  )(\sin\theta_k)}{1^2}+\frac{G(1^2)(\cos\theta_l)}{  1^2}+\frac{G(1^2)(\sin\theta_l)}{1^2}}=\sqrt{2G}

    (Where \theta_x is the angle subtended by a line drawn from planet j to planet x. I.e. \theta_k=0 and \theta_l=\frac{\pi}{2})

    So, what am I missing here? I am fully aware that I, and not Newton, am missing something here. Someone please help point this out for me.

    Thanks
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  2. #2
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    Quote Originally Posted by rainer View Post
    Hi,

    The Physics Help Forum not working today, at least not from my ISP, so this goes here. It's basically a math deal anyway:

    The formula to calculate the force of a point of mass, let's call them planets, that results from its being gravitationally attracted by another point of mass is Newton's:

    F=\frac{Gm_1m_2}{d^2}

    Where F is the force of the planet that results from the gravitational attraction exerted upon it by the other planet, G is Newton's gravity constant, m_1 and m_2 are the respecive masses of the planets, and d is the distance between them.

    For simplicity's sake let's say all the planets considered are of the same mass, so we can write m^2 instead of m_1m_2.

    Now, if I'm not mistaken, the formula for calculating the force of a planet resulting from the gravitational attraction of more than two planets is:

    F_j=\sum_{j\neq k}^{n}\frac{Gm^2}{d_{jk}^2}

    Where F_j is the force on the jth planet resulting from the gravitational attraction of the other n-j planets, and d_{jk} is the distance between the jth planet and the kth planet.

    My question is "where is the vector addition?" That is, when considering the force on one planet that results from the gravitational attraction of many other planets, we have to take into account not only the distance of the other planets from planet j but also their (x,y,z) position with respect to it (right?).
    Yes, that's right. In order to make that a vector formula, you have to multiply the jth term by a unit vector pointing from the kth planet to the jth planet. You get such a unit vector by dividing the vector itself by its own length. Of course, the length of the vector from the kth planet to the jth planet is just d_{jk}. That gives
    F_j=\sum_{j\neq k}^{n}\frac{Gm^2}{d_{jk}^3}\vec{v_{jk}}<br />
    Notice that we now have a cube in the denominator. That third factior of d_{jk} cancels the length of \vec{v_{jk}}.

    Take for example the simple case of three planets in the same plane. Planet j is at the origin. Planet k is one unit to the right of j on the x axis, while planet l is one unit up the y axis. If the masses all equal 1, then, by the formula above, the force on planet j would be:

    F_j=\frac{G(1^2)}{1^2}+\frac{G(1^2)}{1^2}=2G

    But F_j is a function of both the distance and the position, right? So we must consider not only the Gravitational Forces individually exerted upon j by k and l, but also the angle at which these forces are exerted. That is, we must add the vectors. To add vectors you just plug in the x value and y value sums of the added vectors into pythagoras' formula. The force on planet j should therefore be:

    F_j= \sqrt{\frac{G(1^2)(\cos\theta_k)}{1^2}+\frac{G(1^2  )(\sin\theta_k)}{1^2}+\frac{G(1^2)(\cos\theta_l)}{  1^2}+\frac{G(1^2)(\sin\theta_l)}{1^2}}=\sqrt{2G}

    (Where \theta_x is the angle subtended by a line drawn from planet j to planet x. I.e. \theta_k=0 and \theta_l=\frac{\pi}{2})
    But you are making the same error as before: F_j must be a vector, not a number. Assuming angle \theta_j is the angle the line from the jth body to the kth makes with the x-axis we have
    F_{jx}=\sum_{j\neq k}^{n}\frac{Gm^2}{d_{jk}^2}cos(\theta)
    F_{jy}=\sum_{j\neq k}^{n}\frac{Gm^2}{d_{jk}^2}sin(\theta)


    So, what am I missing here? I am fully aware that I, and not Newton, am missing something here. Someone please help point this out for me.

    Thanks
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  3. #3
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    This is great. Could you please just define \vec{v_{jk}} for me? Then I will be able to properly digest your reply. Thanks!
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