FANDOM


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