n-body vector addition

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February 24th 2010, 06:51 AM
rainer

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
February 24th 2010, 08:43 AM
HallsofIvy

Quote:

Originally Posted by rainer

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 $\vec{v_{jk}}$ .

Quote:

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
$F_{jx}=\sum_{j\neq k}^{n}\frac{Gm^2}{d_{jk}^2}cos(\theta)$
$F_{jy}=\sum_{j\neq k}^{n}\frac{Gm^2}{d_{jk}^2}sin(\theta)$

Quote:

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
February 25th 2010, 11:23 AM
rainer

Halls-

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