# Rewriting the equation of motion as a system of first order differential equation

I am required to rewrite the equation

$$\ddot{\mathbf{r}} = - \frac{\mu}{r^3} \mathbf{r}$$

as a first order differential equation. How do I go about doing this?

I understand how this equation was derived from the original

$$\mathbf{F}_{21}=-\frac{G m_1 m_2}{r^2}\hat{\mathbf{u}}_r$$

but I am struggling to find information on the process of attaining $$\dot{\mathbf{r}}$$ from the above equations.

• Who is requiring you? If this is a homework question make sure to consider these points. Which are also relevant for answers if this indeed a homework question. meta.stackoverflow.com/questions/334822/… Nov 21, 2022 at 13:52

Given a scalar function $$f(x)$$ that is described as an $$n^{th}$$ order differential equation can be converted to a first order $$n$$ dimensional vector differential equation. For example, suppose \begin{aligned} f'(x) &= g(f(x), x) \\ f''(x) &= h(f(x), f'(x), x) \end{aligned} The solution is to make an augmented function $$\vec u(x)$$ such that $$u[1](x) = f(x)$$ and $$u[2](x) = f'(x)$$. The same can be applied to a vector function $$\vec r(t)$$. In the case of gravitation, we the result is a six vector, where $$u[1](t) = r_x(t)$$, $$u[2](t) = r_y(t)$$, $$u[3](t) = r_z(t)$$, $$u[4](t) = v_x(t)$$, $$u[5](t) = v_y(t)$$, and $$u[6](t) = v_z(t)$$. The time derivatives of the first three elements are simply the velocity vector, while the time derivatives of the last three elements are given by Newton's law of gravitation.
$$\begin{bmatrix}\dot{r}_x\\ \dot{r}_y\\ \dot{r}_z\\ \dot{v}_x\\ \dot{v}_y\\ \dot{v}_z\end{bmatrix}= \begin{bmatrix}0 & 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 0 & 1\\ & \cdots & & & & \\ & & \cdots & & & \\ & & & \cdots & & \end{bmatrix} \begin{bmatrix}r_x\\ r_y\\ r_z\\ v_x\\ v_y\\ v_z\end{bmatrix}$$