#### Answer

$|r|$ is constant.

#### Work Step by Step

Since, $r \dfrac{dr}{dt}=0$
or, $2r \dfrac{dr}{dt}=0$
or, $r \dfrac{dr}{dt}+r \dfrac{dr}{dt}=0$
This implies $\dfrac{d(r\cdot r)}{dt}=0$
or, $\dfrac{d(|r^2|)}{dt}=0$
This means that $|r^2|$ is constant , so $|r|$ is also constant.