# Multiple time derivatives of gravitational potential

• I
• Belginator

#### Belginator

Hello!

Let's say our gravitational potential is (as usual for 2 body),

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

Then the gradient of this is G,
$$\frac{\partial G}{\partial \mathbf{r}} = G = \frac{\mu}{r^3} [3 \hat{\mathbf{r}} \hat{\mathbf{r}}^\top - I]$$

Now if we take two time derivatives of G, we get

$$\ddot{G} = \frac{3\mu}{r^5} [\hat{\mathbf{r}}^\top \mathbf{v})^2 (7 \hat{\mathbf{r}} \hat{\mathbf{r}}^\top - I) - 10(\hat{\mathbf{r}}^\top \mathbf{v})(\mathbf{v} \hat{\mathbf{r}}^\top + \hat{\mathbf{r}} \mathbf{v}^\top) + 2 \mathbf{v} \mathbf{v}^\top - (\mathbf{v}^\top \mathbf{v})(5 \hat{\mathbf{r}} \hat{\mathbf{r}}^\top - I) + (\hat{\mathbf{r}}^\top \hat{\mathbf{r}}) G]$$

Also for completeness, $$\mathbf{r}$$ is the position vector. $$\hat{\mathbf{r}}$$ is the unit position vector. $$r$$ is the norm of the position vector. $$\mathbf{v}$$ is the velocity vector.

Now this is where it gets tricky, I need to take 3 more time derivatives of $$\ddot{G}$$ so that I have up to the 5th derivative. The problem is, it's getting so long and tedious that I keep making mistakes. Is there a quick way of taking these derivatives? I tried mathematica but it just gets really messy because it does it in components rather than a vector. Any help is appreciated!

Well, you may have started off with an error. I think the potential you want is a scalar field proportional to $\frac{1}{r}$ whose gradient is a vector field. From there, all the time derivatives would be vectors rather than matrices, and it might look simpler and be easier to manipulate algebraically.