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We consider a disk around a star of mass $M_s$. At a distance $r$ from that star is a planet of mass $M_p$. The planet interacts gravitationally with a fluid element at a radius $r_f$ such that $b = r - r_f \ll r$. Both the fluid element and the planet are moving along circular Keplerian orbits around the star, so the velocity difference between them is

$u = \sqrt{\frac{G M_s}{r}} - \sqrt{\frac{G M_s}{r - b}} \approx \sqrt{\frac{G M_s}{r}} \frac{b}{r}$

Most of the deflection occurs when the distance between the planet and the fluid element is about $b$, and the time interval when both are that close is about $b/u$. The velocity deflection perpendicular to the fluid element's original direction of motion is therefore

$\Delta v_{\perp} \approx \frac{G M_p}{b^2} \frac{b}{u} \approx \frac{G M_p}{b^2} / \sqrt{\frac{G M_s}{r^3}} \approx \sqrt{\frac{G M_s}{r}} \frac{M_p}{M_s} \left( \frac{b}{r} \right)^2$

Conservation of energy dictates that there must also be a change in the component of velocity parallel to the original trajectory

$\Delta v_{\parallel} \approx \sqrt{u^2-\Delta_{\perp}^2} - u \approx - \frac{\Delta v_{\perp}^2}{u} \approx - \sqrt{\frac{G M_s}{r}} \left( \frac{M_p}{M_s}\right)^2 \left( \frac{r}{b} \right)^5$

If the mass of the fluid element is $d m$, then the exchange of angular momentum is $\Delta L \approx dm \Delta v_{\parallel} r$. The mass of the fluid element is given by $dm = 2 \pi \Sigma r \cdot db$, where $db$ is the width of the annulus. The time it takes the annulus to interact with the planet is just

$t \approx \frac{2 \pi}{\Omega_p - \Omega_f} = \frac{2 \pi}{\sqrt{G M_s/r^3} - \sqrt{G M_s / \left( r - b\right)^3}} \approx 2 \pi \frac{r /b}{\sqrt{G M_s/r^3}}$

The torque between the annulus and the planet is given by

$dT = \frac{\Delta L}{t} \approx \sqrt{\frac{G M_s}{r}} \left( \frac{M_p}{M_s}\right)^2 \left( \frac{b}{r}\right)^5 r \cdot r \cdot \Sigma \cdot db \cdot \frac{\sqrt{G M_s / r^3}}{r/b}$

The total torque is

$T = \int_{b_{\min}}^{\infty} dT \approx G M_s \Sigma r \left(\frac{M_p}{M_s} \right)^2 \left( \frac{r}{b_{\min}} \right)^3$

The lower limit on the integration is the minimum of either the scale height of the disk or the radius of the Hill sphere. In the case of the latter,

$b_{\min} \approx r \left( \frac{M_p}{M_s} \right)^{1/3}$

and

$T = G M_p \Sigma r$