# I Calculating the trajectory of an object into a force field

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1. Nov 29, 2017

### Vicara

How can I calculate the trajectory of an object knowing the initial conditions and the function F(x,y)?
For example, a object orbiting a planet or two electric charges orbiting each other.

2. Nov 29, 2017

### Staff: Mentor

You need to write the equations of motion and solve them. In some cases, an analytical solution is possible, but for an arbitrary potential they must be solved numerically.

3. Nov 30, 2017

### Vicara

For example, in the case of gravity, with the Newton's law of gravity, it can be integrated to obtain a formula for te velocity but it is not going to be vectorial (as far as I know) and I was going for a formula that gives you r(t) (being r the vector radious)...

4. Nov 30, 2017

### PeroK

This will be covered in any mechanics text book or course. There are online courses like the Khan Academy or try an undergraduate mechanics textbook.

5. Nov 30, 2017

### Ibix

The special thing about the gravitational field of a single spherically symmetric mass is that the force is always radial, so the angular momentum of a negligible-mass body in the field is constant and $\theta (t)$ is a function of $r (t)$ and the initial angular momentum and nothing else. The full answer is still a 2d curve (well, 3d, but you can always pick coordinates so that the orbit lies in their equatorial plane and the "latitude" coordinate is constant), but you only need the radial component (plus conservation laws) to completely characterise the orbit.

This is an example of a situation that can be reframed in a manageable way. As @DrClaude notes, in general this is not possible and you just need to go for a numeric solution.

6. Nov 30, 2017

### Delta²

Lets do a generic treatment of the problem in 3D space using Newtonian mechanics.

Suppose the vector field in cartesian coordinates is $\vec{F}(x,y,z,t)=(F_x(x,y,z,t),F_y(x,y,z,t),F_z(x,y,z,t))$. ( I took the field to have 3 components one in each axis and also that the field is time-dependent.).

then from newton's 2nd law $\vec{F}=m\frac{d^2\vec{r}}{dt^2}$ where $\vec{r}=(x(t),y(t),z(t))$ the position vector of the object at time t, and treating the law component-wise we get the following 3 equations (assuming for simplicity that m=1):

$\frac{d^2x(t)}{dt^2}=F_x(x(t),y(t),z(t),t) (1)$
$\frac{d^2y(t)}{dt^2}=F_y(x(t),y(t),z(t),t) (2)$
$\frac{d^2z(t)}{dt^2}=F_z(x(t),y(t),z(t),t) (3)$

As you can see the 3 differential equations (that are linear regarding their left hand side)

are coupled, meaning for example that x(t) depends also from y(t) and z(t),

and in the right hand side of the equations, that each coordinate appears as argument inside the function of each component of the field , for example in 1st equation x(t) appears inside $F_x$ as its first argument. This fact might make the differential equation non-linear depending what is the form of $F_x,F_y,F_z$ and non-linear differential equations are much harder to solve than linear ones.

These two things make the equations hard to solve even for relatively simple expressions for the components of the field, so quite often we have to resort to numerical solution techniques.

7. Nov 30, 2017

### Vicara

It looks harder to solve that I initially thought it would be... Thanks to all for answering