Is this equation correct?
d/dt(-μr/|r|) = (μ/|r|³)(r·ṙ)r - μṙ/|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.
Is this equation correct?
d/dt(-μr/|r|) = (μ/|r|³)(r·ṙ)r - μṙ/|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 $\displaystyle - \mu$ and consider $\displaystyle \frac{d}{dt} \frac{<x,y,z>}{\sqrt{x^2+y^2+z^2}}$ $\displaystyle = \frac{<\dot{x},\dot{y},\dot{z}>}{\sqrt{x^2+y^2+z^2 }}$ $\displaystyle - \frac{<x,y,z> (x \dot{x} + y \dot{y} + z \dot{z})}{(x^2+y^2+z^2)^{3/2}}$.
You see it now?
That equation is correct.
WLOG, suppose $\displaystyle \mathbf{r}(t)=\left<x(t),y(t),z(t)\right>$
Note that $\displaystyle \frac{d}{dt}|\mathbf{r}|=\frac{1}{2|\mathbf{r}|}\c dot (2x\dot{x}+2y\dot{y}+2z\dot{z})=\frac{\left<x,y,z\ right>\cdot\left<\dot{x},\dot{y},\dot{z}\right>}{| \mathbf{r}|}=\frac{\mathbf{r}\cdot\dot{\mathbf{r}} }{|\mathbf{r}|}$
So apply quotient rule to $\displaystyle \frac{d}{dt}\left[-\frac{\mu\mathbf{r}}{|\mathbf{r}|}\right]$ keeping in mind what I showed you above. You should then be able to justify this equation.
Can you take it from here?