= Circular-orbit energy gradient
{title2=$d\varepsilon=\boldsymbol\Omega\cdot d\mathbf h$}
In a spherically symmetric <gravitational potential> $\Phi(r)$, a <circular orbit> has $r\Omega^2=\Phi'(r)$, <specific orbital energy> $\varepsilon=\Phi+r^2\Omega^2/2$ and <specific angular momentum> $\mathbf h=r^2\Omega\mathbf n$, with $|\mathbf n|=1$. Differentiating gives $\varepsilon'=2r\Omega^2+r^2\Omega\Omega'=\Omega(r^2\Omega)'$. Changes in orbital orientation have $\mathbf n\cdot d\mathbf n=0$, so $\boldsymbol\Omega\cdot d\mathbf h=\Omega\,d(r^2\Omega)=d\varepsilon$. For a particle of variable <mass> $M$ and total <angular momentum> $\mathbf H=M\mathbf h$, this implies $d(M\varepsilon)=(\varepsilon-\Omega h)dM+\boldsymbol\Omega\cdot d\mathbf H$. Moreover $d(\varepsilon-\Omega h)/dr=-h\Omega'$, which is positive when the orbital <angular velocity> decreases outward.
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