# Thread: Is this equation correct? d/dt(-μr/|r|) = (μ/|r|³)(r·ṙ)r - μṙ/|r|

1. ## Is this equation correct? d/dt(-μr/|r|) = (μ/|r|³)(r·ṙ)r - μṙ/|r|

Is this equation correct?

d/dt(-μr/|r|) = (μ/|r|³)(r·ṙ)r - μ/|r|,

where μ is an arbitrary constant, and r = r(t) is a vector.

Apparently is it, but I don't see how, please explain...
Thanks.

2. Originally Posted by feyomi
Is this equation correct?

d/dt(-μr/|r|) = (μ/|r|³)(r·ṙ)r - μ/|r|,

where μ is an arbitrary constant, and r = r(t) is a vector.

Apparently is it, but I don't see how, please explain...
Thanks.
Pull out the $- \mu$ and consider $\frac{d}{dt} \frac{}{\sqrt{x^2+y^2+z^2}}$ $= \frac{<\dot{x},\dot{y},\dot{z}>}{\sqrt{x^2+y^2+z^2 }}$ $- \frac{ (x \dot{x} + y \dot{y} + z \dot{z})}{(x^2+y^2+z^2)^{3/2}}$.

You see it now?

3. Originally Posted by feyomi
Is this equation correct?

d/dt(-μr/|r|) = (μ/|r|³)(r·ṙ)r - μ/|r|,

where μ is an arbitrary constant, and r = r(t) is a vector.

Apparently is it, but I don't see how, please explain...
Thanks.
That equation is correct.

WLOG, suppose $\mathbf{r}(t)=\left$

Note that $\frac{d}{dt}|\mathbf{r}|=\frac{1}{2|\mathbf{r}|}\c dot (2x\dot{x}+2y\dot{y}+2z\dot{z})=\frac{\left\cdot\left<\dot{x},\dot{y},\dot{z}\right>}{| \mathbf{r}|}=\frac{\mathbf{r}\cdot\dot{\mathbf{r}} }{|\mathbf{r}|}$

So apply quotient rule to $\frac{d}{dt}\left[-\frac{\mu\mathbf{r}}{|\mathbf{r}|}\right]$ keeping in mind what I showed you above. You should then be able to justify this equation.

Can you take it from here?