Multiple time derivatives of gravitational potential

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  • Thread starter Belginator
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  • #1
12
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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!
 

Answers and Replies

  • #2
321
125
Well, you may have started off with an error. I think the potential you want is a scalar field proportional to [itex]\frac{1}{r}[/itex] 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.

Also, the itex tag lets you do the same as the tex tag, but in line with a sentence, in case you didn't know.
 
  • #3
12
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Thanks for your advice, I didn't know about itex, and I actually made a mistake by calling the first equation a potential. It's really the first partial of the potential. But the problem otherwise remains the same, the equations are correct.
 

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