# split gravitational force into x, y, and z componenets

I am writing a program for a computer science class in which I am doing an n-body simulation in 3-dimensional space. Currently, I have figured out the gravitational force along the hypotenuse between two bodies. Now I have to split these up into the x, y, and z components, and this is where I am having trouble.

Force between two objects (force_t)=G*m1*m2/r^2

Typically, force_x=force_t*cos(angle between x-axis and hypotenuse) force_y=force_t*sin(angle between x-axis and hypotenuse)

So, I can figure out my triangle. I know that my two bodies are located at (x1, y1, z1) and (x2, y2, z2), so I have my triangle. And I am fairly certain that the above equations will hold for x and y, even though I'm in three dimensions, but I'm just not sure what the corresponding equation for the z component will be. My thought was that the combination of the three forces has to equal the total force, so perhaps I solve for the z component algebraically using the pythagorean theorem, but I have no way to check whether or not this is true.

Anyway, any help would be appreciated. Thanks!

• What you want to read up on is the dot product: (en.wikipedia.org/wiki/Dot_product). $\vec{a}\cdot\vec{b}$ gives you the answer with $\vec{a}$ being your three base vectors and $\vec{b}$ the vector between the two bodies. Commented Nov 21, 2011 at 1:37

For this application, I'd suggest using the actual gravitational force equation, for vector quantities:

$$\vec{F} = \frac{Gm_1m_2}{r^3}\vec{r}$$

Here $$\vec{r}$$ is the vector pointing from the source object to the point where you are computing the force. So if you're computing the force on object 2 (caused by object 1), you'll have

$$\vec{r} = (x_2 - x_1, y_2 - y_1, z_2 - z_1)$$

or if you're computing the force on object 1 (caused by object 2), it'll be the other way around,

$$\vec{r} = (x_1 - x_2, y_1 - y_2, z_1 - z_2)$$

Since the force equation is a vector equation, you effectively get one copy of it for each of the three directions,

\begin{align}F_x &= \frac{Gm_1m_2}{r^3}r_x \\ F_y &= \frac{Gm_1m_2}{r^3}r_y \\ F_z &= \frac{Gm_1m_2}{r^3}r_z\end{align}

As a side note:

Typically, force_x=force_t*cos(angle between x-axis and hypotenuse)

force_y=force_t*sin(angle between x-axis and hypotenuse)

Those equations do not generally hold in 3D. Or rather, the first one does, but not the second one. The general rule is

$$F_s = |\vec{F}|\cos\bigl(\theta_{s,F}\bigr)$$

which means that the $$s$$-component of force ($$F_s$$), where $$s$$ can represent any axis, is equal to the magnitude of the force ($$|\vec{F}|$$) times the cosine of the angle between the force and the axis [$$\cos(\theta_{s,F})$$].

In the particular case of 2D space, the angle between the force and the x-axis is the complement of the angle between the force and the y-axis: $$\theta_{y,F} = 90^\circ - \theta_{x,F}$$. And since $$\cos(90^\circ - \theta) = \sin\theta$$ (in 2D only), you can replace $$\cos(\theta_{y,F})$$ with $$\sin(\theta_{x,F})$$.

Use $F_x = -x GmM/|r|^3$, similar for $y$ and $z$. This is obtained by taking $F=GmM/|r|^2$ where the direction is $-\hat{r} = -\vec{r}/|r| = -(x,y,z)/|r|$, so the $x$-component is $-x/|r|$.

In the above, $|r| = \sqrt{x^2+y^2+z^2}$.